Add Haskell solution for Problem 17

.gitignore

@@ -1,2 +1,3 @@
 Python/__pycache__
 ProblemsOverviews/*
+Haskell/.stack-work

C/p001.c

@@ -2,36 +2,38 @@
  *
  * Find the sum of all the multiples of 3 or 5 below 1000.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
 
 int main(int argc, char **argv)
 {
-   int i, sum = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i, sum = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Simple brute-force approach: try every number between 3 and 999,
-    * check if it's a multiple of 3 or 5, if yes add it to the total.*/
-   for(i = 3; i < 1000; i++)
-   {
-      if (i % 3 == 0 || i % 5 == 0)
-      {
-         sum += i;
-      }
-   }
+    /* Simple brute-force approach: try every number between 3 and 999,
+     * check if it's a multiple of 3 or 5, if yes add it to the total.*/
+    for(i = 3; i < 1000; i++)
+    {
+        if (i % 3 == 0 || i % 5 == 0)
+        {
+            sum += i;
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 1\n");
-   printf("Answer: %d\n", sum);
+    printf("Project Euler, Problem 1\n");
+    printf("Answer: %d\n", sum);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p002.c

@@ -4,6 +4,8 @@
  *
  * By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -12,36 +14,36 @@
 
 int main(int argc, char **argv)
 {
-   int fib0 = 1, fib1 = 2, fib2, sum = 2;
-   double elapsed;
-   struct timespec start, end;
+    int fib0 = 1, fib1 = 2, fib2, sum = 2;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   fib2 = fib0 + fib1;
+    fib2 = fib0 + fib1;
 
-   /* Simple brute-force approach: generate every value in the Fibonacci
-    * sequence smaller than 4 million and if it's even add it to the total.*/
-   while(fib2 <= N)
-   {
-      if(fib2 % 2 == 0)
-      {
-         sum += fib2;
-      }
+    /* Simple brute-force approach: generate every value in the Fibonacci
+     * sequence smaller than 4 million and if it's even add it to the total.*/
+    while(fib2 <= N)
+    {
+        if(fib2 % 2 == 0)
+        {
+            sum += fib2;
+        }
 
-      fib0 = fib1;
-      fib1 = fib2;
-      fib2 = fib0 + fib1;
-   }
+        fib0 = fib1;
+        fib1 = fib2;
+        fib2 = fib0 + fib1;
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 2\n");
-   printf("Answer: %d\n", sum);
+    printf("Project Euler, Problem 2\n");
+    printf("Answer: %d\n", sum);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p003.c

@@ -2,6 +2,8 @@
  *
  * What is the largest prime factor of the number 600851475143?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -11,26 +13,26 @@ long int max_prime_factor(long int num);
 
 int main(int argc, char **argv)
 {
-   long int res;
-   long int num = 600851475143;
-   double elapsed;
-   struct timespec start, end;
+    long int res;
+    long int num = 600851475143;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Use a function to calculate the largest prime factor.*/
-   res = max_prime_factor(num);
+    /* Use a function to calculate the largest prime factor.*/
+    res = max_prime_factor(num);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 3\n");
-   printf("Answer: %ld\n", res);
+    printf("Project Euler, Problem 3\n");
+    printf("Answer: %ld\n", res);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 /* Recursive approach: if num is prime, return num, otherwise
@@ -38,39 +40,39 @@ int main(int argc, char **argv)
  * by its prime factors until only the largest remains.*/
 long int max_prime_factor(long int num)
 {
-   long int i;
+    long int i;
 
-   /* Use function defined in projecteuler.c to check if a number is prime.*/
-   if(is_prime(num))
-   {
-      return num;
-   }
+    /* Use function defined in projecteuler.c to check if a number is prime.*/
+    if(is_prime(num))
+    {
+        return num;
+    }
 
-   /* If num is even, find the largest prime factor of num/2.*/
-   if(num % 2 == 0)
-   {
-      return max_prime_factor(num/2);
-   }
-   else
-   {
-      i = 3;
-      
-      while(1)
-      {
-         /* If num is divisible by i and i is prime, find largest prime
-          * factor of num/i.*/
-         if(num % i == 0)
-         {
-            if(is_prime(i))
+    /* If num is even, find the largest prime factor of num/2.*/
+    if(num % 2 == 0)
+    {
+        return max_prime_factor(num/2);
+    }
+    else
+    {
+        i = 3;
+
+        while(1)
+        {
+            /* If num is divisible by i and i is prime, find largest prime
+             * factor of num/i.*/
+            if(num % i == 0)
             {
-               return max_prime_factor(num/i);
+                if(is_prime(i))
+                {
+                    return max_prime_factor(num/i);
+                }
             }
-         }
 
-         i += 2;
-      }
-   }
+            i += 2;
+        }
+    }
 
-   /* Should never get here.*/
-   return -1;
+    /* Should never get here.*/
+    return -1;
 }

C/p004.c

@@ -2,6 +2,8 @@
  *
  * Find the largest palindrome made from the product of two 3-digit numbers.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -9,36 +11,36 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, max = 0, num;
-   double elapsed;
-   struct timespec start, end;
+    int i, j, max = 0, num;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Using a brute-force approach: generate every product of 3-digit numbers
-    * and check if it's palindrome. If the product found is greater than the
-    * current maximum, save the current product.*/
-   for(i = 999; i >= 100; i--)
-   {
-      for(j = i; j >= 100; j--)
-      {
-         num = i * j;
-         /* Use the function defined in projecteuler.c to check if a number is palindrome.*/
-         if(num > max && is_palindrome(num, 10))
-         {
-            max = num;
-         }
-      }
-   }
+    /* Using a brute-force approach: generate every product of 3-digit numbers
+     * and check if it's palindrome. If the product found is greater than the
+     * current maximum, save the current product.*/
+    for(i = 999; i >= 100; i--)
+    {
+        for(j = i; j >= 100; j--)
+        {
+            num = i * j;
+            /* Use the function defined in projecteuler.c to check if a number is palindrome.*/
+            if(num > max && is_palindrome(num, 10))
+            {
+                max = num;
+            }
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 4\n");
-   printf("Answer: %d\n", max);
+    printf("Project Euler, Problem 4\n");
+    printf("Answer: %d\n", max);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p005.c

@@ -2,6 +2,8 @@
  *
  * What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -9,23 +11,23 @@
 
 int main(int argc, char **argv)
 {
-   long int res, n[20] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20};
-   double elapsed;
-   struct timespec start, end;
+    long int res, n[20] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20};
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Function define in projecteuler.c to find the least common multiple of multiple numbers.*/
-   res = lcmm(n, 20);
+    /* Function define in projecteuler.c to find the least common multiple of multiple numbers.*/
+    res = lcmm(n, 20);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 5\n");
-   printf("Answer: %ld\n", res);
+    printf("Project Euler, Problem 5\n");
+    printf("Answer: %ld\n", res);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p006.c

@@ -1,44 +1,46 @@
 /* The sum of the squares of the first ten natural numbers is,
  *
  * 1^2 + 2^2 + ... + 10^2 = 385
- * 
+ *
  * The square of the sum of the first ten natural numbers is,
  *
  * (1 + 2 + ... + 10)^2 = 55^2 = 3025
- * 
+ *
  * Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
  *
  * Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
 
 int main(int argc, char **argv)
 {
-   int i, sum_squares = 0, square_sum = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i, sum_squares = 0, square_sum = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Straightforward brute-force approach.*/
-   for(i = 1; i <= 100; i++)
-   {
-      sum_squares += i*i;
-      square_sum += i;
-   }
+    /* Straightforward brute-force approach.*/
+    for(i = 1; i <= 100; i++)
+    {
+        sum_squares += i*i;
+        square_sum += i;
+    }
 
-   square_sum *= square_sum;
+    square_sum *= square_sum;
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 6\n");
-   printf("Answer: %d\n", square_sum-sum_squares);
+    printf("Project Euler, Problem 6\n");
+    printf("Answer: %d\n", square_sum-sum_squares);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p007.c

@@ -2,6 +2,8 @@
  *
  * What is the 10 001st prime number?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -9,33 +11,33 @@
 
 int main(int argc, char **argv)
 {
-   int count = 1, n = 1, target = 10001;
-   double elapsed;
-   struct timespec start, end;
+    int count = 1, n = 1, target = 10001;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Brute force approach: start with count=1 and check every odd number
-    * (2 is the only even prime), if it's prime increment count, until the
-    * target prime is reached.*/
-   while(count != target)
-   {
-      n += 2;
-      /* Use the function in projecteuler.c to check if a number is prime.*/
-      if(is_prime(n))
-      {
-         count++;
-      }
-   }
+    /* Brute force approach: start with count=1 and check every odd number
+     * (2 is the only even prime), if it's prime increment count, until the
+     * target prime is reached.*/
+    while(count != target)
+    {
+        n += 2;
+        /* Use the function in projecteuler.c to check if a number is prime.*/
+        if(is_prime(n))
+        {
+            count++;
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 7\n");
-   printf("Answer: %d\n", n);
+    printf("Project Euler, Problem 7\n");
+    printf("Answer: %d\n", n);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p008.c

@@ -23,90 +23,92 @@
  *
  * Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
 
 int main(int argc, char **argv)
 {
-   char string[] = "73167176531330624919225119674426574742355349"
-                   "19493496983520312774506326239578318016984801"
-                   "86947885184385861560789112949495459501737958"
-                   "33195285320880551112540698747158523863050715"
-                   "69329096329522744304355766896648950445244523"
-                   "16173185640309871112172238311362229893423380"
-                   "30813533627661428280644448664523874930358907"
-                   "29629049156044077239071381051585930796086670"
-                   "17242712188399879790879227492190169972088809"
-                   "37766572733300105336788122023542180975125454"
-                   "05947522435258490771167055601360483958644670"
-                   "63244157221553975369781797784617406495514929"
-                   "08625693219784686224828397224137565705605749"
-                   "02614079729686524145351004748216637048440319"
-                   "98900088952434506585412275886668811642717147"
-                   "99244429282308634656748139191231628245861786"
-                   "64583591245665294765456828489128831426076900"
-                   "42242190226710556263211111093705442175069416"
-                   "58960408071984038509624554443629812309878799"
-                   "27244284909188845801561660979191338754992005"
-                   "24063689912560717606058861164671094050775410"
-                   "02256983155200055935729725716362695618826704"
-                   "28252483600823257530420752963450";
-   char cur, out;
-   long int max = 0, tmp = 1;
-   int i, j;
-   double elapsed;
-   struct timespec start, end;
+    char string[] = "73167176531330624919225119674426574742355349"
+                    "19493496983520312774506326239578318016984801"
+                    "86947885184385861560789112949495459501737958"
+                    "33195285320880551112540698747158523863050715"
+                    "69329096329522744304355766896648950445244523"
+                    "16173185640309871112172238311362229893423380"
+                    "30813533627661428280644448664523874930358907"
+                    "29629049156044077239071381051585930796086670"
+                    "17242712188399879790879227492190169972088809"
+                    "37766572733300105336788122023542180975125454"
+                    "05947522435258490771167055601360483958644670"
+                    "63244157221553975369781797784617406495514929"
+                    "08625693219784686224828397224137565705605749"
+                    "02614079729686524145351004748216637048440319"
+                    "98900088952434506585412275886668811642717147"
+                    "99244429282308634656748139191231628245861786"
+                    "64583591245665294765456828489128831426076900"
+                    "42242190226710556263211111093705442175069416"
+                    "58960408071984038509624554443629812309878799"
+                    "27244284909188845801561660979191338754992005"
+                    "24063689912560717606058861164671094050775410"
+                    "02256983155200055935729725716362695618826704"
+                    "28252483600823257530420752963450";
+    char cur, out;
+    long int max = 0, tmp = 1;
+    int i, j;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   for(i = 0; i < 1000; i++)
-   {
-      /* For the first 13 digits, just multiply them.*/
-      if(i < 13)
-      {
-         cur = string[i] - '0';
-         tmp *= (long int)cur;
-      }
-      else
-      {
-         /* If the current product is greater than the maximum, save the current as maximum.*/
-         if(tmp > max)
-         {
-            max = tmp;
-         }
-         /* Check the value of the first digit of the previous sequence, which will not be part
-          * of the next sequence.*/
-         out = string[i-13] - '0';
-         /* If the digit is zero, multiply all the 13 digits of the new sequence.*/
-         if(out == 0)
-         {
-            tmp = 1;
-            for(j = i - 12; j <= i; j++)
-            {
-               cur = string[j] - '0';
-               tmp *= (long int)cur;
-            }
-         }
-         /* If the digit not zero, instead of multiplying all the 13 digits of the new sequence,
-          * divide the current product by the remove digit and multiply it by the new digit.*/
-         else
-         {
+    for(i = 0; i < 1000; i++)
+    {
+        /* For the first 13 digits, just multiply them.*/
+        if(i < 13)
+        {
             cur = string[i] - '0';
-            tmp /= (long int)out;
             tmp *= (long int)cur;
-         }
-      }
-   }
+        }
+        else
+        {
+            /* If the current product is greater than the maximum, save the current as maximum.*/
+            if(tmp > max)
+            {
+                max = tmp;
+            }
+            /* Check the value of the first digit of the previous sequence, which will not be part
+             * of the next sequence.*/
+            out = string[i-13] - '0';
+            /* If the digit is zero, multiply all the 13 digits of the new sequence.*/
+            if(out == 0)
+            {
+                tmp = 1;
+                for(j = i - 12; j <= i; j++)
+                {
+                    cur = string[j] - '0';
+                    tmp *= (long int)cur;
+                }
+            }
+            /* If the digit not zero, instead of multiplying all the 13 digits of the new sequence,
+             * divide the current product by the remove digit and multiply it by the new digit.*/
+            else
+            {
+                cur = string[i] - '0';
+                tmp /= (long int)out;
+                tmp *= (long int)cur;
+            }
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 8\n");
-   printf("Answer: %ld\n", max);
+    printf("Project Euler, Problem 8\n");
+    printf("Answer: %ld\n", max);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p009.c

@@ -8,6 +8,8 @@
  *
  * Find the product abc.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -15,61 +17,61 @@
 
 int main(int argc, char **argv)
 {
-   int a, b, c, tmpa, tmpb, tmpc, i, m, n, found = 0;
-   double elapsed;
-   struct timespec start, end;
+    int a, b, c, tmpa, tmpb, tmpc, i, m, n, found = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Brute force approach: generate all the Pythagorean triplets using
-    * Euclid's formula, until the one where a+b+c=1000 is found.*/
-   for(m = 2; found == 0; m++)
-   {
-      for(n = 1; n < m && !found; n++)
-      {
-         if(gcd(m, n) == 1 && ((m % 2 == 0 && n % 2 != 0) || (m % 2 != 0 && n % 2 == 0)))
-         {
-            a = m * m - n * n;
-            b = 2 * m * n;
-            c = m * m + n * n;
-
-            if(a + b + c == 1000)
+    /* Brute force approach: generate all the Pythagorean triplets using
+     * Euclid's formula, until the one where a+b+c=1000 is found.*/
+    for(m = 2; found == 0; m++)
+    {
+        for(n = 1; n < m && !found; n++)
+        {
+            if(gcd(m, n) == 1 && ((m % 2 == 0 && n % 2 != 0) || (m % 2 != 0 && n % 2 == 0)))
             {
-               found = 1;
-               break;
+                a = m * m - n * n;
+                b = 2 * m * n;
+                c = m * m + n * n;
+
+                if(a + b + c == 1000)
+                {
+                    found = 1;
+                    break;
+                }
+
+                i = 2;
+
+                do
+                {
+                    tmpa = a * i;
+                    tmpb = b * i;
+                    tmpc = c * i;
+
+                    if(tmpa + tmpb + tmpc == 1000)
+                    {
+                        a = tmpa;
+                        b = tmpb;
+                        c = tmpc;
+                        found = 1;
+                        break;
+                    }
+
+                    i++;
+                } while(tmpa + tmpb + tmpc < 1000);
             }
+        }
+    }
 
-            i = 2;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-            do
-            {
-               tmpa = a * i;
-               tmpb = b * i;
-               tmpc = c * i;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-               if(tmpa + tmpb + tmpc == 1000)
-               {
-                  a = tmpa;
-                  b = tmpb;
-                  c = tmpc;
-                  found = 1;
-                  break;
-               }
-               
-               i++;
-            }while(tmpa + tmpb + tmpc < 1000);
-         }
-      }
-   }
+    printf("Project Euler, Problem 9\n");
+    printf("Answer: %d\n", a*b*c);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
-
-   printf("Project Euler, Problem 9\n");
-   printf("Answer: %d\n", a*b*c);
-   
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
-
-   return 0;
+    return 0;
 }

C/p010.c

@@ -2,6 +2,8 @@
  *
  * Find the sum of all the primes below two million.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -11,41 +13,41 @@
 
 int main(int argc, char **argv)
 {
-   int i; 
-   int *primes;
-   long int sum = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i;
+    int *primes;
+    long int sum = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Use the function in projecteuler.c implementing the 
-    * Sieve of Eratosthenes algorithm to generate primes.*/
-   if((primes = sieve(N)) == NULL)
-   {
-      fprintf(stderr, "Error! Sieve function returned NULL\n");
-      return 1;
-   }
+    /* Use the function in projecteuler.c implementing the
+     * Sieve of Eratosthenes algorithm to generate primes.*/
+    if((primes = sieve(N)) == NULL)
+    {
+        fprintf(stderr, "Error! Sieve function returned NULL\n");
+        return 1;
+    }
 
-   /* Sum all the primes.*/
-   for(i = 0; i < N; i++)
-   {
-      if(primes[i])
-      {
-         sum += i;
-      }
-   }
+    /* Sum all the primes.*/
+    for(i = 0; i < N; i++)
+    {
+        if(primes[i])
+        {
+            sum += i;
+        }
+    }
 
-   free(primes);
+    free(primes);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 10\n");
-   printf("Answer: %ld\n", sum);
+    printf("Project Euler, Problem 10\n");
+    printf("Answer: %ld\n", sum);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p011.c

@@ -25,110 +25,113 @@
  *
  * What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
 
 int main(int argc, char **argv)
 {
-   int grid[][20] = {{8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8},
-                    {49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 0},
-                    {81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65},
-                    {52, 70, 95, 23, 4, 60, 11, 42, 69, 24, 68, 56, 1, 32, 56, 71, 37, 2, 36, 91},
-                    {22, 31, 16, 71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33, 13, 80},
-                    {24, 47, 32, 60, 99, 3, 45, 2, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17, 12, 50},
-                    {32, 98, 81, 28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38, 64, 70},
-                    {67, 26, 20, 68, 2, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21},
-                    {24, 55, 58, 5, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72},
-                    {21, 36, 23, 9, 75, 0, 76, 44, 20, 45, 35, 14, 0, 61, 33, 97, 34, 31, 33, 95},
-                    {78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 3, 80, 4, 62, 16, 14, 9, 53, 56, 92},
-                    {16, 39, 5, 42, 96, 35, 31, 47, 55, 58, 88, 24, 0, 17, 54, 24, 36, 29, 85, 57},
-                    {86, 56, 0, 48, 35, 71, 89, 7, 5, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58},
-                    {19, 80, 81, 68, 5, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 4, 89, 55, 40},
-                    {4, 52, 8, 83, 97, 35, 99, 16, 7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66},
-                    {88, 36, 68, 87, 57, 62, 20, 72, 3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69},
-                    {4, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36},
-                    {20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 4, 36, 16},
-                    {20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54},
-                    {1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48}};
-   int i, j, k, w, max = 0, prod;
-   double elapsed;
-   struct timespec start, end;
+    int grid[][20] = {{8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8},
+        {49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 0},
+        {81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65},
+        {52, 70, 95, 23, 4, 60, 11, 42, 69, 24, 68, 56, 1, 32, 56, 71, 37, 2, 36, 91},
+        {22, 31, 16, 71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33, 13, 80},
+        {24, 47, 32, 60, 99, 3, 45, 2, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17, 12, 50},
+        {32, 98, 81, 28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38, 64, 70},
+        {67, 26, 20, 68, 2, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21},
+        {24, 55, 58, 5, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72},
+        {21, 36, 23, 9, 75, 0, 76, 44, 20, 45, 35, 14, 0, 61, 33, 97, 34, 31, 33, 95},
+        {78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 3, 80, 4, 62, 16, 14, 9, 53, 56, 92},
+        {16, 39, 5, 42, 96, 35, 31, 47, 55, 58, 88, 24, 0, 17, 54, 24, 36, 29, 85, 57},
+        {86, 56, 0, 48, 35, 71, 89, 7, 5, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58},
+        {19, 80, 81, 68, 5, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 4, 89, 55, 40},
+        {4, 52, 8, 83, 97, 35, 99, 16, 7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66},
+        {88, 36, 68, 87, 57, 62, 20, 72, 3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69},
+        {4, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36},
+        {20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 4, 36, 16},
+        {20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54},
+        {1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48}
+    };
+    int i, j, k, w, max = 0, prod;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Brute-force approach: for each number in the grid, try products with its three
-    * adjacent numbers in every direction (horizontal, vertical and the two diagonals).
-    * If the product is larger than the current  maximum, save it.*/
-   for(i = 0; i < 17; i++)
-   {
-      for(j = 0; j < 17; j++)
-      {
-         prod = 1;
-         /* Horizontal direction.*/
-         for(k = j; k < j + 4; k++)
-         {
-            prod *= grid[i][k];
-         }
+    /* Brute-force approach: for each number in the grid, try products with its three
+     * adjacent numbers in every direction (horizontal, vertical and the two diagonals).
+     * If the product is larger than the current  maximum, save it.*/
+    for(i = 0; i < 17; i++)
+    {
+        for(j = 0; j < 17; j++)
+        {
+            prod = 1;
+            /* Horizontal direction.*/
+            for(k = j; k < j + 4; k++)
+            {
+                prod *= grid[i][k];
+            }
 
-         if(prod > max)
-         {
-            max = prod;
-         }
-
-         prod = 1;
-         /* Vertical direction.*/
-         for(k = i; k < i + 4; k++)
-         {
-            prod *= grid[k][j];
-         }
-         if(prod > max)
-         {
-            max = prod;
-         }
-
-         prod = 1;
-         /* Diagonal direction, from top left to bottom right.*/
-         for(k = i, w = j; k < i + 4 && w < j + 4; k++, w++)
-         {
-            prod *= grid[k][w];
-         }
-         if(k == i + 4 && w == j + 4)
-         {
             if(prod > max)
             {
-               max = prod;
+                max = prod;
             }
-         }
-      }
-   }
 
-   /* The last diagonal is handled separately.*/ 
-   for(i = 0; i < 17; i++)
-   {
-      for(j = 3; j < 20; j++)
-      {
-         prod = 1;
-         /* Diagonal direction, from top right to bottom left.*/
-         for(k = i, w = j; k < i + 4 && w > j - 4; k++, w--)
-         {
-            prod *= grid[k][w];
-         }
-         if(prod > max)
-         {
-            max = prod;
-         }
-      }
-   }
+            prod = 1;
+            /* Vertical direction.*/
+            for(k = i; k < i + 4; k++)
+            {
+                prod *= grid[k][j];
+            }
+            if(prod > max)
+            {
+                max = prod;
+            }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+            prod = 1;
+            /* Diagonal direction, from top left to bottom right.*/
+            for(k = i, w = j; k < i + 4 && w < j + 4; k++, w++)
+            {
+                prod *= grid[k][w];
+            }
+            if(k == i + 4 && w == j + 4)
+            {
+                if(prod > max)
+                {
+                    max = prod;
+                }
+            }
+        }
+    }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    /* The last diagonal is handled separately.*/
+    for(i = 0; i < 17; i++)
+    {
+        for(j = 3; j < 20; j++)
+        {
+            prod = 1;
+            /* Diagonal direction, from top right to bottom left.*/
+            for(k = i, w = j; k < i + 4 && w > j - 4; k++, w--)
+            {
+                prod *= grid[k][w];
+            }
+            if(prod > max)
+            {
+                max = prod;
+            }
+        }
+    }
 
-   printf("Project Euler, Problem 11\n");
-   printf("Answer: %d\n", max);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   return 0;
+    printf("Project Euler, Problem 11\n");
+    printf("Answer: %d\n", max);
+
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }

C/p012.c

@@ -17,6 +17,8 @@
  *
  * What is the value of the first triangle number to have over five hundred divisors?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -24,34 +26,34 @@
 
 int main(int argc, char **argv)
 {
-   int i = 0, finished = 0, count, triang = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i = 0, finished = 0, count, triang = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Generate all triangle numbers until the first one with more than 500 divisors is found.*/
-   while(!finished)
-   {
-      i++;
-      triang += i;
-      /* Use the function implemented in projecteuler.c to count divisors of a number.*/
-      count = count_divisors(triang);
-      
-      if(count > 500)
-      {
-         finished = 1;
-      }
-   }
+    /* Generate all triangle numbers until the first one with more than 500 divisors is found.*/
+    while(!finished)
+    {
+        i++;
+        triang += i;
+        /* Use the function implemented in projecteuler.c to count divisors of a number.*/
+        count = count_divisors(triang);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+        if(count > 500)
+        {
+            finished = 1;
+        }
+    }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 12\n");
-   printf("Answer: %d\n", triang);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 12\n");
+    printf("Answer: %d\n", triang);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }

C/p013.c

@@ -101,6 +101,8 @@
  * 20849603980134001723930671666823555245252804609722
  * 53503534226472524250874054075591789781264330331690*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -108,94 +110,95 @@
 
 int main(int argc, char **argv)
 {
-   char n[100][51] = {"37107287533902102798797998220837590246510135740250", "46376937677490009712648124896970078050417018260538",
-      "74324986199524741059474233309513058123726617309629", "91942213363574161572522430563301811072406154908250",
-      "23067588207539346171171980310421047513778063246676", "89261670696623633820136378418383684178734361726757",
-      "28112879812849979408065481931592621691275889832738", "44274228917432520321923589422876796487670272189318",
-      "47451445736001306439091167216856844588711603153276", "70386486105843025439939619828917593665686757934951",
-      "62176457141856560629502157223196586755079324193331", "64906352462741904929101432445813822663347944758178",
-      "92575867718337217661963751590579239728245598838407", "58203565325359399008402633568948830189458628227828",
-      "80181199384826282014278194139940567587151170094390", "35398664372827112653829987240784473053190104293586",
-      "86515506006295864861532075273371959191420517255829", "71693888707715466499115593487603532921714970056938",
-      "54370070576826684624621495650076471787294438377604", "53282654108756828443191190634694037855217779295145",
-      "36123272525000296071075082563815656710885258350721", "45876576172410976447339110607218265236877223636045",
-      "17423706905851860660448207621209813287860733969412", "81142660418086830619328460811191061556940512689692",
-      "51934325451728388641918047049293215058642563049483", "62467221648435076201727918039944693004732956340691",
-      "15732444386908125794514089057706229429197107928209", "55037687525678773091862540744969844508330393682126",
-      "18336384825330154686196124348767681297534375946515", "80386287592878490201521685554828717201219257766954",
-      "78182833757993103614740356856449095527097864797581", "16726320100436897842553539920931837441497806860984",
-      "48403098129077791799088218795327364475675590848030", "87086987551392711854517078544161852424320693150332",
-      "59959406895756536782107074926966537676326235447210", "69793950679652694742597709739166693763042633987085",
-      "41052684708299085211399427365734116182760315001271", "65378607361501080857009149939512557028198746004375",
-      "35829035317434717326932123578154982629742552737307", "94953759765105305946966067683156574377167401875275",
-      "88902802571733229619176668713819931811048770190271", "25267680276078003013678680992525463401061632866526",
-      "36270218540497705585629946580636237993140746255962", "24074486908231174977792365466257246923322810917141",
-      "91430288197103288597806669760892938638285025333403", "34413065578016127815921815005561868836468420090470",
-      "23053081172816430487623791969842487255036638784583", "11487696932154902810424020138335124462181441773470",
-      "63783299490636259666498587618221225225512486764533", "67720186971698544312419572409913959008952310058822",
-      "95548255300263520781532296796249481641953868218774", "76085327132285723110424803456124867697064507995236",
-      "37774242535411291684276865538926205024910326572967", "23701913275725675285653248258265463092207058596522",
-      "29798860272258331913126375147341994889534765745501", "18495701454879288984856827726077713721403798879715",
-      "38298203783031473527721580348144513491373226651381", "34829543829199918180278916522431027392251122869539",
-      "40957953066405232632538044100059654939159879593635", "29746152185502371307642255121183693803580388584903",
-      "41698116222072977186158236678424689157993532961922", "62467957194401269043877107275048102390895523597457",
-      "23189706772547915061505504953922979530901129967519", "86188088225875314529584099251203829009407770775672",
-      "11306739708304724483816533873502340845647058077308", "82959174767140363198008187129011875491310547126581",
-      "97623331044818386269515456334926366572897563400500", "42846280183517070527831839425882145521227251250327",
-      "55121603546981200581762165212827652751691296897789", "32238195734329339946437501907836945765883352399886",
-      "75506164965184775180738168837861091527357929701337", "62177842752192623401942399639168044983993173312731",
-      "32924185707147349566916674687634660915035914677504", "99518671430235219628894890102423325116913619626622",
-      "73267460800591547471830798392868535206946944540724", "76841822524674417161514036427982273348055556214818",
-      "97142617910342598647204516893989422179826088076852", "87783646182799346313767754307809363333018982642090",
-      "10848802521674670883215120185883543223812876952786", "71329612474782464538636993009049310363619763878039",
-      "62184073572399794223406235393808339651327408011116", "66627891981488087797941876876144230030984490851411",
-      "60661826293682836764744779239180335110989069790714", "85786944089552990653640447425576083659976645795096",
-      "66024396409905389607120198219976047599490197230297", "64913982680032973156037120041377903785566085089252",
-      "16730939319872750275468906903707539413042652315011", "94809377245048795150954100921645863754710598436791",
-      "78639167021187492431995700641917969777599028300699", "15368713711936614952811305876380278410754449733078",
-      "40789923115535562561142322423255033685442488917353", "44889911501440648020369068063960672322193204149535",
-      "41503128880339536053299340368006977710650566631954", "81234880673210146739058568557934581403627822703280",
-      "82616570773948327592232845941706525094512325230608", "22918802058777319719839450180888072429661980811197",
-      "77158542502016545090413245809786882778948721859617", "72107838435069186155435662884062257473692284509516",
-      "20849603980134001723930671666823555245252804609722", "53503534226472524250874054075591789781264330331690"};
-   char result[100];
-   int i;
-   double elapsed;
-   struct timespec start, end;
-   mpz_t a, b;
+    char n[100][51] = {"37107287533902102798797998220837590246510135740250", "46376937677490009712648124896970078050417018260538",
+                       "74324986199524741059474233309513058123726617309629", "91942213363574161572522430563301811072406154908250",
+                       "23067588207539346171171980310421047513778063246676", "89261670696623633820136378418383684178734361726757",
+                       "28112879812849979408065481931592621691275889832738", "44274228917432520321923589422876796487670272189318",
+                       "47451445736001306439091167216856844588711603153276", "70386486105843025439939619828917593665686757934951",
+                       "62176457141856560629502157223196586755079324193331", "64906352462741904929101432445813822663347944758178",
+                       "92575867718337217661963751590579239728245598838407", "58203565325359399008402633568948830189458628227828",
+                       "80181199384826282014278194139940567587151170094390", "35398664372827112653829987240784473053190104293586",
+                       "86515506006295864861532075273371959191420517255829", "71693888707715466499115593487603532921714970056938",
+                       "54370070576826684624621495650076471787294438377604", "53282654108756828443191190634694037855217779295145",
+                       "36123272525000296071075082563815656710885258350721", "45876576172410976447339110607218265236877223636045",
+                       "17423706905851860660448207621209813287860733969412", "81142660418086830619328460811191061556940512689692",
+                       "51934325451728388641918047049293215058642563049483", "62467221648435076201727918039944693004732956340691",
+                       "15732444386908125794514089057706229429197107928209", "55037687525678773091862540744969844508330393682126",
+                       "18336384825330154686196124348767681297534375946515", "80386287592878490201521685554828717201219257766954",
+                       "78182833757993103614740356856449095527097864797581", "16726320100436897842553539920931837441497806860984",
+                       "48403098129077791799088218795327364475675590848030", "87086987551392711854517078544161852424320693150332",
+                       "59959406895756536782107074926966537676326235447210", "69793950679652694742597709739166693763042633987085",
+                       "41052684708299085211399427365734116182760315001271", "65378607361501080857009149939512557028198746004375",
+                       "35829035317434717326932123578154982629742552737307", "94953759765105305946966067683156574377167401875275",
+                       "88902802571733229619176668713819931811048770190271", "25267680276078003013678680992525463401061632866526",
+                       "36270218540497705585629946580636237993140746255962", "24074486908231174977792365466257246923322810917141",
+                       "91430288197103288597806669760892938638285025333403", "34413065578016127815921815005561868836468420090470",
+                       "23053081172816430487623791969842487255036638784583", "11487696932154902810424020138335124462181441773470",
+                       "63783299490636259666498587618221225225512486764533", "67720186971698544312419572409913959008952310058822",
+                       "95548255300263520781532296796249481641953868218774", "76085327132285723110424803456124867697064507995236",
+                       "37774242535411291684276865538926205024910326572967", "23701913275725675285653248258265463092207058596522",
+                       "29798860272258331913126375147341994889534765745501", "18495701454879288984856827726077713721403798879715",
+                       "38298203783031473527721580348144513491373226651381", "34829543829199918180278916522431027392251122869539",
+                       "40957953066405232632538044100059654939159879593635", "29746152185502371307642255121183693803580388584903",
+                       "41698116222072977186158236678424689157993532961922", "62467957194401269043877107275048102390895523597457",
+                       "23189706772547915061505504953922979530901129967519", "86188088225875314529584099251203829009407770775672",
+                       "11306739708304724483816533873502340845647058077308", "82959174767140363198008187129011875491310547126581",
+                       "97623331044818386269515456334926366572897563400500", "42846280183517070527831839425882145521227251250327",
+                       "55121603546981200581762165212827652751691296897789", "32238195734329339946437501907836945765883352399886",
+                       "75506164965184775180738168837861091527357929701337", "62177842752192623401942399639168044983993173312731",
+                       "32924185707147349566916674687634660915035914677504", "99518671430235219628894890102423325116913619626622",
+                       "73267460800591547471830798392868535206946944540724", "76841822524674417161514036427982273348055556214818",
+                       "97142617910342598647204516893989422179826088076852", "87783646182799346313767754307809363333018982642090",
+                       "10848802521674670883215120185883543223812876952786", "71329612474782464538636993009049310363619763878039",
+                       "62184073572399794223406235393808339651327408011116", "66627891981488087797941876876144230030984490851411",
+                       "60661826293682836764744779239180335110989069790714", "85786944089552990653640447425576083659976645795096",
+                       "66024396409905389607120198219976047599490197230297", "64913982680032973156037120041377903785566085089252",
+                       "16730939319872750275468906903707539413042652315011", "94809377245048795150954100921645863754710598436791",
+                       "78639167021187492431995700641917969777599028300699", "15368713711936614952811305876380278410754449733078",
+                       "40789923115535562561142322423255033685442488917353", "44889911501440648020369068063960672322193204149535",
+                       "41503128880339536053299340368006977710650566631954", "81234880673210146739058568557934581403627822703280",
+                       "82616570773948327592232845941706525094512325230608", "22918802058777319719839450180888072429661980811197",
+                       "77158542502016545090413245809786882778948721859617", "72107838435069186155435662884062257473692284509516",
+                       "20849603980134001723930671666823555245252804609722", "53503534226472524250874054075591789781264330331690"
+                      };
+    char result[100];
+    int i;
+    double elapsed;
+    struct timespec start, end;
+    mpz_t a, b;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Using the GNU Multiple Precision Arithmetic Library (GMP)
-    * to sum the numbers and get the first 10 digits of the sum.*/
-   mpz_inits(a, b, NULL);
-   mpz_set_str(a, n[0], 10);
+    /* Using the GNU Multiple Precision Arithmetic Library (GMP)
+     * to sum the numbers and get the first 10 digits of the sum.*/
+    mpz_inits(a, b, NULL);
+    mpz_set_str(a, n[0], 10);
 
-   for(i = 1; i < 100; i++)
-   {
-      mpz_set_str(b, n[i], 10);
-      mpz_add(a, a, b);
-   }
+    for(i = 1; i < 100; i++)
+    {
+        mpz_set_str(b, n[i], 10);
+        mpz_add(a, a, b);
+    }
 
-   gmp_sprintf(result, "%Zd\n", a);
+    gmp_sprintf(result, "%Zd\n", a);
 
-   mpz_clears(a, b, NULL);
+    mpz_clears(a, b, NULL);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec-start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec-start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 13\n");
-   printf("Answer: ");
+    printf("Project Euler, Problem 13\n");
+    printf("Answer: ");
 
-   for(i = 0; i < 10; i++)
-   {
-      printf("%c", result[i]);
-   }
+    for(i = 0; i < 10; i++)
+    {
+        printf("%c", result[i]);
+    }
 
-   printf("\n");
+    printf("\n");
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p014.c

@@ -14,6 +14,8 @@
  *
  * NOTE: Once the chain starts the terms are allowed to go above one million.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -26,36 +28,36 @@ int collatz_found[N] = {0};
 
 int main(int argc, char **argv)
 {
-   int i, count, max = 0, max_l = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i, count, max = 0, max_l = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   for(i = 1; i < N; i++)
-   {
-      /* For each number from 1 to 1000000, find the length of the sequence
-       * and save its value, so that it can be used for the next numbers.*/
-      count = collatz_length(i);
-      collatz_found[i] = count;
-      
-      if(count > max_l)
-      {
-         max_l = count;
-         max = i;
-      }
-   }
+    for(i = 1; i < N; i++)
+    {
+        /* For each number from 1 to 1000000, find the length of the sequence
+         * and save its value, so that it can be used for the next numbers.*/
+        count = collatz_length(i);
+        collatz_found[i] = count;
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+        if(count > max_l)
+        {
+            max_l = count;
+            max = i;
+        }
+    }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 14\n");
-   printf("Answer: %d\n", max);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 14\n");
+    printf("Answer: %d\n", max);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 /* Recursive function to calculate the Collatz sequence for n.
@@ -63,16 +65,16 @@ int main(int argc, char **argv)
  * Collatz(n)=1+Collatz(3*n+1).*/
 int collatz_length(long int n)
 {
-   if(n == 1)
-      return 1;
+    if(n == 1)
+        return 1;
 
-   /* If Collatz(n) has been previously calculated,
-    * just return the value.*/
-   if(n < N && collatz_found[n])
-      return collatz_found[n];
+    /* If Collatz(n) has been previously calculated,
+     * just return the value.*/
+    if(n < N && collatz_found[n])
+        return collatz_found[n];
 
-   if(n % 2 == 0)
-      return 1 + collatz_length(n/2);
-   else
-      return 1 + collatz_length(3*n+1);
+    if(n % 2 == 0)
+        return 1 + collatz_length(n/2);
+    else
+        return 1 + collatz_length(3*n+1);
 }

C/p015.c

@@ -1,6 +1,8 @@
 /* Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner
  * How many such routes are there through a 20×20 grid?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -8,33 +10,33 @@
 
 int main(int argc, char **argv)
 {
-   double elapsed;
-   struct timespec start, end;
-   mpz_t count, tmp;
+    double elapsed;
+    struct timespec start, end;
+    mpz_t count, tmp;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Using a combinatorial solution: in a 20x20 grid there will always be
-    * 20 movements to the right and 20 movements down, that can be represented
-    * as a string of Rs and Ds. The number of routes is the number of combinations.
-    * This is obtained calculating n!/(k!*(n-k)!), where n=40 and k=20. The GMP
-    * Library is used to calculate the factorials.*/
-   mpz_inits(count, tmp, NULL);
-   mpz_fac_ui(count, 40);
-   mpz_fac_ui(tmp, 20);
-   mpz_mul(tmp, tmp, tmp);
-   mpz_tdiv_q(count, count, tmp);
+    /* Using a combinatorial solution: in a 20x20 grid there will always be
+     * 20 movements to the right and 20 movements down, that can be represented
+     * as a string of Rs and Ds. The number of routes is the number of combinations.
+     * This is obtained calculating n!/(k!*(n-k)!), where n=40 and k=20. The GMP
+     * Library is used to calculate the factorials.*/
+    mpz_inits(count, tmp, NULL);
+    mpz_fac_ui(count, 40);
+    mpz_fac_ui(tmp, 20);
+    mpz_mul(tmp, tmp, tmp);
+    mpz_tdiv_q(count, count, tmp);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 15\n");
-   gmp_printf("Answer: %Zd\n", count);
+    printf("Project Euler, Problem 15\n");
+    gmp_printf("Answer: %Zd\n", count);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   mpz_clears(count, tmp, NULL);
+    mpz_clears(count, tmp, NULL);
 
-   return 0;
+    return 0;
 }

C/p016.c

@@ -2,6 +2,8 @@
  *
  * What is the sum of the digits of the number 2^1000?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -9,37 +11,37 @@
 
 int main(int argc, char **argv)
 {
-   double elapsed;
-   struct timespec start, end;
-   mpz_t p, sum, r;
+    double elapsed;
+    struct timespec start, end;
+    mpz_t p, sum, r;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Simply calculate 2^1000 with the GMP Library
-    * and sum all the digits.*/
-   mpz_init_set_ui(p, 2);
-   mpz_init_set_ui(sum, 0);
-   mpz_init(r);
+    /* Simply calculate 2^1000 with the GMP Library
+     * and sum all the digits.*/
+    mpz_init_set_ui(p, 2);
+    mpz_init_set_ui(sum, 0);
+    mpz_init(r);
 
-   mpz_pow_ui(p, p, 1000);
+    mpz_pow_ui(p, p, 1000);
 
-   while(mpz_cmp_ui(p, 0))
-   {
-      /* To get each digit, simply get the reminder of the division by 10.*/
-      mpz_tdiv_qr_ui(p, r, p, 10);
-      mpz_add(sum, sum, r);
-   }
+    while(mpz_cmp_ui(p, 0))
+    {
+        /* To get each digit, simply get the reminder of the division by 10.*/
+        mpz_tdiv_qr_ui(p, r, p, 10);
+        mpz_add(sum, sum, r);
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 16\n");
-   gmp_printf("Answer: %Zd\n", sum);
+    printf("Project Euler, Problem 16\n");
+    gmp_printf("Answer: %Zd\n", sum);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   mpz_clears(p, sum, r, NULL);
+    mpz_clears(p, sum, r, NULL);
 
-   return 0;
+    return 0;
 }

C/p017.c

@@ -5,78 +5,80 @@
  * NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen)
  * contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
 
 int main(int argc, char **argv)
 {
-   /* Number of letters for numbers from 1 to 19.*/
-   int n1_19[19] = {3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8};
-   /* Number of letters for "twenty", "thirty", ..., "ninety".*/
-   int n20_90[8] = {6, 6, 5, 5, 5, 7, 6, 6};
-   /* Number of letters for "one hundred and", "two hundred and", ...,
-    * "nine hundred and".*/
-   int n100_900[9] = {13, 13, 15, 14, 14, 13, 15, 15, 14};
-   /* Number of letters for 1000.*/
-   int n1000 = 11;
-   int sum = 0, i, j;
-   double elapsed;
-   struct timespec start, end;
+    /* Number of letters for numbers from 1 to 19.*/
+    int n1_19[19] = {3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8};
+    /* Number of letters for "twenty", "thirty", ..., "ninety".*/
+    int n20_90[8] = {6, 6, 5, 5, 5, 7, 6, 6};
+    /* Number of letters for "one hundred and", "two hundred and", ...,
+     * "nine hundred and".*/
+    int n100_900[9] = {13, 13, 15, 14, 14, 13, 15, 15, 14};
+    /* Number of letters for 1000.*/
+    int n1000 = 11;
+    int sum = 0, i, j;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Sum the letters of the first 19 numbers.*/
-   for(i = 0; i < 19; i++)
-   {
-      sum += n1_19[i];
-   }
+    /* Sum the letters of the first 19 numbers.*/
+    for(i = 0; i < 19; i++)
+    {
+        sum += n1_19[i];
+    }
 
-   /* Add the letters of the numbers from 20 to 99.*/ 
-   for(i = 0; i < 8; i++)
-   {
-      /* "Twenty", "thirty", ... "ninety" are used ten times each
-       * (e.g. "twenty", "twenty one", "twenty two", ..., "twenty nine").*/
-      n20_90[i] *= 10;
-      /* Add "one", "two", ..., "nine".*/
-      for(j = 0; j < 9; j++)
-      {
-         n20_90[i] += n1_19[j];
-      }
-      sum += n20_90[i];
-   }
+    /* Add the letters of the numbers from 20 to 99.*/
+    for(i = 0; i < 8; i++)
+    {
+        /* "Twenty", "thirty", ... "ninety" are used ten times each
+         * (e.g. "twenty", "twenty one", "twenty two", ..., "twenty nine").*/
+        n20_90[i] *= 10;
+        /* Add "one", "two", ..., "nine".*/
+        for(j = 0; j < 9; j++)
+        {
+            n20_90[i] += n1_19[j];
+        }
+        sum += n20_90[i];
+    }
 
-   /* Add the letters of the numbers from 100 to 999.*/
-   for(i = 0; i < 9; i++)
-   {
-      /* "One hundred and", "two hundred and",... are used 100 times each.*/
-      n100_900[i] *= 100;
-      /* Add "one" to "nineteen".*/
-      for(j = 0; j < 19; j++)
-      {
-         n100_900[i] += n1_19[j];
-      }
-      /* Add "twenty" to "ninety nine", previously calculated.*/
-      for(j = 0; j < 8; j++)
-      {
-         n100_900[i] += n20_90[j];
-      }
-      /* "One hundred", "two hundred", ... don't have the "and", so remove 
-       * three letters for each of them.*/
-      sum += n100_900[i] - 3;
-   }
+    /* Add the letters of the numbers from 100 to 999.*/
+    for(i = 0; i < 9; i++)
+    {
+        /* "One hundred and", "two hundred and",... are used 100 times each.*/
+        n100_900[i] *= 100;
+        /* Add "one" to "nineteen".*/
+        for(j = 0; j < 19; j++)
+        {
+            n100_900[i] += n1_19[j];
+        }
+        /* Add "twenty" to "ninety nine", previously calculated.*/
+        for(j = 0; j < 8; j++)
+        {
+            n100_900[i] += n20_90[j];
+        }
+        /* "One hundred", "two hundred", ... don't have the "and", so remove
+         * three letters for each of them.*/
+        sum += n100_900[i] - 3;
+    }
 
-   /* Add "one thousand".*/
-   sum += n1000;
+    /* Add "one thousand".*/
+    sum += n1000;
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 17\n");
-   printf("Answer: %d\n", sum);
+    printf("Project Euler, Problem 17\n");
+    printf("Answer: %d\n", sum);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p018.c

@@ -1,5 +1,5 @@
 /* By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
- *    
+ *
  *    3
  *   7 4
  *  2 4 6
@@ -25,9 +25,11 @@
  *  63 66 04 68 89 53 67 30 73 16 69 87 40 31
  * 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
  *
- * NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge 
+ * NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge
  * with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -35,63 +37,63 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, max;
-   int **triang;
-   double elapsed;
-   FILE *fp;
-   struct timespec start, end;
+    int i, j, max;
+    int **triang;
+    double elapsed;
+    FILE *fp;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((triang = (int **)malloc(15*sizeof(int *))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((triang = (int **)malloc(15*sizeof(int *))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   for(i = 1; i <= 15; i++)
-   {
-      if((triang[i-1] = (int *)malloc(i*sizeof(int))) == NULL)
-      {
-         fprintf(stderr, "Error while allocating memory\n");
-         return 1;
-      }
-   }
+    for(i = 1; i <= 15; i++)
+    {
+        if((triang[i-1] = (int *)malloc(i*sizeof(int))) == NULL)
+        {
+            fprintf(stderr, "Error while allocating memory\n");
+            return 1;
+        }
+    }
 
-   if((fp = fopen("triang.txt", "r")) == NULL)
-   {
-      fprintf(stderr, "Error while opening file %s\n", "triang.txt");
-      return 1;
-   }
+    if((fp = fopen("triang.txt", "r")) == NULL)
+    {
+        fprintf(stderr, "Error while opening file %s\n", "triang.txt");
+        return 1;
+    }
 
-   for(i = 1; i <= 15; i++)
-   {
-      for(j = 0; j < i; j++)
-      {
-         fscanf(fp, "%d", &triang[i-1][j]);
-      }
-   }
+    for(i = 1; i <= 15; i++)
+    {
+        for(j = 0; j < i; j++)
+        {
+            fscanf(fp, "%d", &triang[i-1][j]);
+        }
+    }
 
-   fclose(fp);
+    fclose(fp);
 
-   /* Use the function implemented in projecteuler.c to find the maximum path.*/
-   max = find_max_path(triang, 15);
+    /* Use the function implemented in projecteuler.c to find the maximum path.*/
+    max = find_max_path(triang, 15);
 
-   for(i = 0; i < 15; i++)
-   {
-      free(triang[i]);
-   }
+    for(i = 0; i < 15; i++)
+    {
+        free(triang[i]);
+    }
 
-   free(triang);
+    free(triang);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 18\n");
-   printf("Answer: %d\n", max);
+    printf("Project Euler, Problem 18\n");
+    printf("Answer: %d\n", max);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p019.c

@@ -11,6 +11,8 @@
  *
  * How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -20,75 +22,75 @@ typedef enum {mon, tue, wed, thu, fri, sat, sun} days;
 
 int main(int argc, char **argv)
 {
-   int year, i, limit, count = 0;
-   double elapsed;
-   struct timespec start, end;
-   months month;
-   days day;
+    int year, i, limit, count = 0;
+    double elapsed;
+    struct timespec start, end;
+    months month;
+    days day;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   day = mon;
-   month = jan;
-   year = 1900;
+    day = mon;
+    month = jan;
+    year = 1900;
 
-   while(year < 2001)
-   {
-      /* February has 29 days on leap years, otherwise 28. Leap years are those
-       * divisible by 4, but not if they're divisible by 100, except when they're
-       * divisible by 400.*/
-      if(month == feb)
-      {
-         if(year % 400 == 0 || (year % 4 == 0 && year % 100 != 0))
-         {
-            limit = 29;
-         }
-         else
-         {
-            limit = 28;
-         }
-      }
-      /* April, June, September and November have 30 days.*/
-      else if(month == apr || month == jun || month == sep || month == nov)
-      {
-         limit = 30;
-      }
-      /* All other months have 31 days.*/
-      else
-      {
-         limit = 31;
-      }
-      /* Loop on every day of the month.*/
-      for(i = 1; i <= limit; i++)
-      {
-         /* If it's the first day of the month and it's Sunday, increase
-          * counter, except if year=1900 (we need to count Sundays from
-          * 1901 to 2000.*/
-         if(year > 1900 && i == 1 && day == sun)
-         {
-            count++;
-         }
-         /* Change day of the week.*/
-         day = (day + 1) % 7;
-      }
-      /* At the end of the month, go to next month.*/
-      month = (month + 1) % 12;
+    while(year < 2001)
+    {
+        /* February has 29 days on leap years, otherwise 28. Leap years are those
+         * divisible by 4, but not if they're divisible by 100, except when they're
+         * divisible by 400.*/
+        if(month == feb)
+        {
+            if(year % 400 == 0 || (year % 4 == 0 && year % 100 != 0))
+            {
+                limit = 29;
+            }
+            else
+            {
+                limit = 28;
+            }
+        }
+        /* April, June, September and November have 30 days.*/
+        else if(month == apr || month == jun || month == sep || month == nov)
+        {
+            limit = 30;
+        }
+        /* All other months have 31 days.*/
+        else
+        {
+            limit = 31;
+        }
+        /* Loop on every day of the month.*/
+        for(i = 1; i <= limit; i++)
+        {
+            /* If it's the first day of the month and it's Sunday, increase
+             * counter, except if year=1900 (we need to count Sundays from
+             * 1901 to 2000.*/
+            if(year > 1900 && i == 1 && day == sun)
+            {
+                count++;
+            }
+            /* Change day of the week.*/
+            day = (day + 1) % 7;
+        }
+        /* At the end of the month, go to next month.*/
+        month = (month + 1) % 12;
 
-      /* If we're back to january, increase the year.*/
-      if(month == jan)
-      {
-         year++;
-      }
-   }
+        /* If we're back to january, increase the year.*/
+        if(month == jan)
+        {
+            year++;
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 19\n");
-   printf("Answer: %d\n", count);
+    printf("Project Euler, Problem 19\n");
+    printf("Answer: %d\n", count);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p020.c

@@ -5,6 +5,8 @@
  *
  * Find the sum of the digits in the number 100!*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -12,35 +14,35 @@
 
 int main(int argc, char **argv)
 {
-   double elapsed;
-   struct timespec start, end;
-   mpz_t fact, r, sum;
+    double elapsed;
+    struct timespec start, end;
+    mpz_t fact, r, sum;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Calculate the factorial using the GMP Library and sum the digits.*/
-   mpz_inits(fact, r, sum, NULL);
+    /* Calculate the factorial using the GMP Library and sum the digits.*/
+    mpz_inits(fact, r, sum, NULL);
 
-   mpz_fac_ui(fact, 100);
+    mpz_fac_ui(fact, 100);
 
-   mpz_set_ui(sum, 0);
+    mpz_set_ui(sum, 0);
 
-   while(mpz_cmp_ui(fact, 0))
-   {
-      mpz_tdiv_qr_ui(fact, r, fact, 10);
-      mpz_add(sum, sum, r);
-   }
+    while(mpz_cmp_ui(fact, 0))
+    {
+        mpz_tdiv_qr_ui(fact, r, fact, 10);
+        mpz_add(sum, sum, r);
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 20\n");
-   gmp_printf("Answer: %Zd\n", sum);
+    printf("Project Euler, Problem 20\n");
+    gmp_printf("Answer: %Zd\n", sum);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   mpz_clears(fact, r, sum, NULL);
+    mpz_clears(fact, r, sum, NULL);
 
-   return 0;
+    return 0;
 }

C/p021.c

@@ -6,6 +6,8 @@
  *
  * Evaluate the sum of all the amicable numbers under 10000.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -14,35 +16,35 @@
 
 int main(int argc, char **argv)
 {
-   int i, n, sum = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i, n, sum = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   for(i = 2; i < 10000; i++)
-   {
-      /* Calculate the sum of proper divisors with the function
-       * implemented in projecteuler.c.*/
-      n = sum_of_divisors(i, 1);
-      /* If i!=n and the sum of proper divisors of n=i,
-       * sum the pair of numbers and add it to the total.*/
-      if(i != n && sum_of_divisors(n, 1) == i)
-      {
-         sum += i + n;
-      }
-   }
+    for(i = 2; i < 10000; i++)
+    {
+        /* Calculate the sum of proper divisors with the function
+         * implemented in projecteuler.c.*/
+        n = sum_of_divisors(i, 1);
+        /* If i!=n and the sum of proper divisors of n=i,
+         * sum the pair of numbers and add it to the total.*/
+        if(i != n && sum_of_divisors(n, 1) == i)
+        {
+            sum += i + n;
+        }
+    }
 
-   sum /= 2;
+    sum /= 2;
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 21\n"); 
-   printf("Answer: %d\n", sum);
+    printf("Project Euler, Problem 21\n");
+    printf("Answer: %d\n", sum);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p022.c

@@ -6,6 +6,8 @@
  *
  * What is the total of all the name scores in the file?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
@@ -16,86 +18,86 @@ int compare(void *string1, void *string2);
 
 int main(int argc, char **argv)
 {
-   FILE *fp;
-   int i, j, n, len, score, sum = 0;
-   double elapsed;
-   char *line, **names;
-   struct timespec start, end;
+    FILE *fp;
+    int i, j, n, len, score, sum = 0;
+    double elapsed;
+    char *line, **names;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((fp = fopen("names.txt", "r")) == NULL)
-   {
-      fprintf(stderr, "Error while opening file %s\n", "names.txt");
-      return 1;
-   }
+    if((fp = fopen("names.txt", "r")) == NULL)
+    {
+        fprintf(stderr, "Error while opening file %s\n", "names.txt");
+        return 1;
+    }
 
-   fscanf(fp, "%ms", &line);
+    fscanf(fp, "%ms", &line);
 
-   fclose(fp);
+    fclose(fp);
 
-   len = strlen(line);
-   n = 1;
+    len = strlen(line);
+    n = 1;
 
-   /* Count the names in the file.*/
-   for(i = 0; i < len; i++)
-   {
-      if(line[i] == ',')
-      {
-         n++;
-      }
-   }
+    /* Count the names in the file.*/
+    for(i = 0; i < len; i++)
+    {
+        if(line[i] == ',')
+        {
+            n++;
+        }
+    }
 
-   if((names = (char **)malloc(n*sizeof(char *))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((names = (char **)malloc(n*sizeof(char *))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   /* Save each name in a string.*/
-   names[0] = strtok(line, ",\"");
+    /* Save each name in a string.*/
+    names[0] = strtok(line, ",\"");
 
-   for(i = 1; i < n; i++)
-   {
-      names[i] = strtok(NULL, ",\"");
-   }
+    for(i = 1; i < n; i++)
+    {
+        names[i] = strtok(NULL, ",\"");
+    }
 
-   /* Use quick_sort algorithm implemented in projecteuler.c to sort the names.*/
-   quick_sort((void **)names, 0, n-1, compare);
+    /* Use quick_sort algorithm implemented in projecteuler.c to sort the names.*/
+    quick_sort((void **)names, 0, n-1, compare);
 
-   /* Calculate the score of each name an multiply by its position.*/
-   for(i = 0; i < n; i++)
-   {
-      len = strlen(names[i]);
-      score = 0;
-      for(j = 0; j < len; j++)
-      {
-         score += names[i][j] - 'A' + 1;
-      }
-      score *= (i + 1);
-      sum += score;
-   }
+    /* Calculate the score of each name an multiply by its position.*/
+    for(i = 0; i < n; i++)
+    {
+        len = strlen(names[i]);
+        score = 0;
+        for(j = 0; j < len; j++)
+        {
+            score += names[i][j] - 'A' + 1;
+        }
+        score *= (i + 1);
+        sum += score;
+    }
 
-   free(line);
+    free(line);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 22\n");
-   printf("Answer: %d\n", sum);
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 22\n");
+    printf("Answer: %d\n", sum);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 /* Function to compare two strings, to pass to the quick_sort function.*/
 int compare(void *string1, void *string2)
 {
-   char *s1, *s2;
+    char *s1, *s2;
 
-   s1 = (char *)string1;
-   s2 = (char *)string2;
+    s1 = (char *)string1;
+    s2 = (char *)string2;
 
-   return strcmp(s1, s2);
+    return strcmp(s1, s2);
 }

C/p023.c

@@ -1,4 +1,4 @@
-/* A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. 
+/* A perfect number is a number for which the sum of its proper divisors is exactly equal to the number.
  * For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
  *
  * A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.
@@ -10,6 +10,8 @@
  *
  * Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -20,69 +22,69 @@ int is_abundant(int n);
 
 int main(int argc, char **argv)
 {
-   int ab_nums[28123], sums[28123] = {0};
-   int i, j, sum;
-   double elapsed;
-   struct timespec start, end;
+    int ab_nums[28123], sums[28123] = {0};
+    int i, j, sum;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   for(i = 0; i < 28123; i++)
-   {
-      /* Find all abundant numbers smaller than 28123.*/
-      ab_nums[i] = is_abundant(i+1);
-   }
+    for(i = 0; i < 28123; i++)
+    {
+        /* Find all abundant numbers smaller than 28123.*/
+        ab_nums[i] = is_abundant(i+1);
+    }
 
-   /* For every abundant number, sum every other abundant number greater 
-    * than itself, until the sum exceeds 28123. Record that the resulting
-    * number is the sum of two abundant numbers.*/
-   for(i = 0; i < 28123; i++)
-   {
-      if(ab_nums[i])
-      {
-         for(j = i; j < 28123; j++)
-         {
-            if(ab_nums[j])
+    /* For every abundant number, sum every other abundant number greater
+     * than itself, until the sum exceeds 28123. Record that the resulting
+     * number is the sum of two abundant numbers.*/
+    for(i = 0; i < 28123; i++)
+    {
+        if(ab_nums[i])
+        {
+            for(j = i; j < 28123; j++)
             {
-               sum = i + j + 2;
+                if(ab_nums[j])
+                {
+                    sum = i + j + 2;
 
-               if(sum <= 28123)
-               {
-                  sums[sum-1] = 1;
-               }
-               else
-               {
-                  break;
-               }
+                    if(sum <= 28123)
+                    {
+                        sums[sum-1] = 1;
+                    }
+                    else
+                    {
+                        break;
+                    }
+                }
             }
-         }
-      }
-   }
+        }
+    }
 
-   sum = 0;
+    sum = 0;
 
-   /* Sum every number that was not found as a sum of two abundant numbers.*/
-   for(i = 0; i < 28123; i++)
-   {
-      if(!sums[i])
-      {
-         sum += i + 1;
-      }
-   }
+    /* Sum every number that was not found as a sum of two abundant numbers.*/
+    for(i = 0; i < 28123; i++)
+    {
+        if(!sums[i])
+        {
+            sum += i + 1;
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 23\n");
-   printf("Answer: %d\n", sum);
+    printf("Project Euler, Problem 23\n");
+    printf("Answer: %d\n", sum);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int is_abundant(int n)
 {
-   return sum_of_divisors(n, 1) > n;
+    return sum_of_divisors(n, 1) > n;
 }

C/p024.c

@@ -5,6 +5,8 @@
  *
  * What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -14,69 +16,69 @@ int compare(void *a, void *b);
 
 int main(int argc, char **argv)
 {
-   int i, res[10];
-   int **perm;
-   double elapsed;
-   struct timespec start, end;
+    int i, res[10];
+    int **perm;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((perm = (int **)malloc(10*sizeof(int *))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((perm = (int **)malloc(10*sizeof(int *))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   for(i = 0; i < 10; i++)
-   {
-      if((perm[i] = (int *)malloc(sizeof(int))) == NULL)
-      {
-         fprintf(stderr, "Error while allocating memory\n");
-         return 1;
-      }
-      *perm[i] = i;
-   }
+    for(i = 0; i < 10; i++)
+    {
+        if((perm[i] = (int *)malloc(sizeof(int))) == NULL)
+        {
+            fprintf(stderr, "Error while allocating memory\n");
+            return 1;
+        }
+        *perm[i] = i;
+    }
 
-   for(i = 0; i < 999999; i++)
-   {
-      /* Function that generates permutations in lexicographic order.
-       * Finish when the 1000000th is found.*/
-      next_permutation((void **)perm, 10, compare);
-   }
+    for(i = 0; i < 999999; i++)
+    {
+        /* Function that generates permutations in lexicographic order.
+         * Finish when the 1000000th is found.*/
+        next_permutation((void **)perm, 10, compare);
+    }
 
-   for(i = 0; i < 10; i++)
-   {
-      res[i] = *perm[i];
-      free(perm[i]);
-   }
+    for(i = 0; i < 10; i++)
+    {
+        res[i] = *perm[i];
+        free(perm[i]);
+    }
 
-   free(perm);
+    free(perm);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 24\n");
-   printf("Answer: ");
+    printf("Project Euler, Problem 24\n");
+    printf("Answer: ");
 
-   for(i = 0; i < 10; i++)
-   {
-      printf("%d", res[i]);
-   }
+    for(i = 0; i < 10; i++)
+    {
+        printf("%d", res[i]);
+    }
 
-   printf("\n");
+    printf("\n");
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int compare(void *a, void *b)
 {
-   int *n1, *n2;
+    int *n1, *n2;
 
-   n1 = (int *)a;
-   n2 = (int *)b;
+    n1 = (int *)a;
+    n2 = (int *)b;
 
-   return *n1 - *n2;
+    return *n1 - *n2;
 }

C/p025.c

@@ -19,6 +19,8 @@
  *
  * What is the index of the first term in the Fibonacci sequence to contain 1000 digits?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
@@ -27,61 +29,61 @@
 
 int main(int argc, char **argv)
 {
-   int i;
-   double elapsed;
-   struct timespec start, end;
-   mpz_t f1, f2, fn;
-   char *num;
-   size_t size;
+    int i;
+    double elapsed;
+    struct timespec start, end;
+    mpz_t f1, f2, fn;
+    char *num;
+    size_t size;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   mpz_init_set_ui(f1, 1);
-   mpz_init_set_ui(f2, 1);
-   mpz_init(fn);
+    mpz_init_set_ui(f1, 1);
+    mpz_init_set_ui(f2, 1);
+    mpz_init(fn);
 
-   i = 2;
+    i = 2;
 
-   while(1)
-   {
-      /* Use the GMP Library to calculate the Fibonacci numbers.*/
-      mpz_add(fn, f1, f2);
-      i++;
+    while(1)
+    {
+        /* Use the GMP Library to calculate the Fibonacci numbers.*/
+        mpz_add(fn, f1, f2);
+        i++;
 
-      /* The function mpz_sizeinbase gives the number of digits of
-       * the number in the given base, but the result is either exact
-       * or one too big. To check the exact size, the number is
-       * converted to string and the strlen function is used.*/
-      if((size = mpz_sizeinbase(fn, 10)) >= 1000)
-      {
-         if((num = (char *)malloc((2+size)*sizeof(char))) == NULL)
-         {
-            fprintf(stderr, "Error while allocating memory\n");
-            return 1;
-         }
-         gmp_sprintf(num, "%Zd", fn);
-         size = strlen(num);
-         free(num);
-         if(size == 1000)
-         {
-            break;
-         }
-      }
+        /* The function mpz_sizeinbase gives the number of digits of
+         * the number in the given base, but the result is either exact
+         * or one too big. To check the exact size, the number is
+         * converted to string and the strlen function is used.*/
+        if((size = mpz_sizeinbase(fn, 10)) >= 1000)
+        {
+            if((num = (char *)malloc((2+size)*sizeof(char))) == NULL)
+            {
+                fprintf(stderr, "Error while allocating memory\n");
+                return 1;
+            }
+            gmp_sprintf(num, "%Zd", fn);
+            size = strlen(num);
+            free(num);
+            if(size == 1000)
+            {
+                break;
+            }
+        }
 
-      mpz_set(f1, f2);
-      mpz_set(f2, fn);
-   }
+        mpz_set(f1, f2);
+        mpz_set(f2, fn);
+    }
 
-   mpz_clears(f1, f2, fn, NULL);
+    mpz_clears(f1, f2, fn, NULL);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 25\n");
-   printf("Answer: %d\n", i);
+    printf("Project Euler, Problem 25\n");
+    printf("Answer: %d\n", i);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p026.c

@@ -14,6 +14,8 @@
  *
  * Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -22,67 +24,67 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, n, max = 0, max_n = 0;
-   double elapsed;
-   struct timespec start, end;
-   mpz_t k, div;
+    int i, j, n, max = 0, max_n = 0;
+    double elapsed;
+    struct timespec start, end;
+    mpz_t k, div;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   mpz_init(k);
-   mpz_init(div);
+    mpz_init(k);
+    mpz_init(div);
 
-   for(i = 2; i < 1000; i++)
-   {
-      j = i;
+    for(i = 2; i < 1000; i++)
+    {
+        j = i;
 
-      /* The repeating cycle of 1/(2^a*5^b*p^c*...) is equal to
-       * that of 1/p^c*..., so factors 2 and 5 can be eliminated.*/
-      while(j % 2 == 0 && j > 1)
-         j /= 2;
+        /* The repeating cycle of 1/(2^a*5^b*p^c*...) is equal to
+         * that of 1/p^c*..., so factors 2 and 5 can be eliminated.*/
+        while(j % 2 == 0 && j > 1)
+            j /= 2;
 
-      while(j % 5 == 0 && j > 1)
-         j /= 5;
+        while(j % 5 == 0 && j > 1)
+            j /= 5;
 
-      /* If the denominator had only factors 2 and 5, there is no
-       * repeating cycle.*/
-      if(j == 1)
-         n = 0;
-      else
-      {
-         n = 1;
-         mpz_set_ui(k, 9);
-         mpz_set_ui(div, j);
+        /* If the denominator had only factors 2 and 5, there is no
+         * repeating cycle.*/
+        if(j == 1)
+            n = 0;
+        else
+        {
+            n = 1;
+            mpz_set_ui(k, 9);
+            mpz_set_ui(div, j);
 
-         /* After eliminating factors 2s and 5s, the length of the repeating cycle 
-          * of 1/d is the smallest n for which k=10^n-1/d is an integer. So we start 
-          * with k=9, then k=99, k=999 and  so on until k is divisible by d. 
-          * The number  of digits of k is the length of the repeating cycle.*/
-         while(!mpz_divisible_p(k, div))
-         {
-            n++;
-            mpz_mul_ui(k, k, 10);
-            mpz_add_ui(k, k, 9);
-         }
+            /* After eliminating factors 2s and 5s, the length of the repeating cycle
+             * of 1/d is the smallest n for which k=10^n-1/d is an integer. So we start
+             * with k=9, then k=99, k=999 and  so on until k is divisible by d.
+             * The number  of digits of k is the length of the repeating cycle.*/
+            while(!mpz_divisible_p(k, div))
+            {
+                n++;
+                mpz_mul_ui(k, k, 10);
+                mpz_add_ui(k, k, 9);
+            }
 
-         if(n > max)
-         {
-            max = n;
-            max_n = i;
-         }
-      }
-   }
+            if(n > max)
+            {
+                max = n;
+                max_n = i;
+            }
+        }
+    }
 
-   mpz_clears(k, div, NULL);
+    mpz_clears(k, div, NULL);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 26\n");
-   printf("Answer: %d\n", max_n);
+    printf("Project Euler, Problem 26\n");
+    printf("Answer: %d\n", max_n);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p027.c

@@ -17,6 +17,8 @@
  *
  * Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n=0.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -25,56 +27,56 @@
 
 int main(int argc, char **argv)
 {
-   int a, b, n, p, count, max = 0, save_a, save_b;
-   double elapsed;
-   struct timespec start, end;
+    int a, b, n, p, count, max = 0, save_a, save_b;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Brute force approach, optimized by checking only values of b where b is prime.*/
-   for(a = -999; a < 1000; a++)
-   {
-      for(b = 2; b <= 1000; b++)
-      {
-         /* For n=0, n^2+an+b=b, so b must be prime.*/
-         if(is_prime(b))
-         {
-            n = 0;
-            count = 0;
-
-            while(1)
+    /* Brute force approach, optimized by checking only values of b where b is prime.*/
+    for(a = -999; a < 1000; a++)
+    {
+        for(b = 2; b <= 1000; b++)
+        {
+            /* For n=0, n^2+an+b=b, so b must be prime.*/
+            if(is_prime(b))
             {
-               p = n * n + a * n + b;
+                n = 0;
+                count = 0;
 
-               if(p > 1 && is_prime(p))
-               {
-                  count++;
-                  n++;
-               }
-               else
-               {
-                  break;
-               }
+                while(1)
+                {
+                    p = n * n + a * n + b;
+
+                    if(p > 1 && is_prime(p))
+                    {
+                        count++;
+                        n++;
+                    }
+                    else
+                    {
+                        break;
+                    }
+                }
+
+                if(count > max)
+                {
+                    max = count;
+                    save_a = a;
+                    save_b = b;
+                }
             }
+        }
+    }
 
-            if(count > max)
-            {
-               max = count;
-               save_a = a;
-               save_b = b;
-            }
-         }
-      }
-   }
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    printf("Project Euler, Problem 27\n");
+    printf("Answer: %d\n", save_a * save_b);
 
-   printf("Project Euler, Problem 27\n");
-   printf("Answer: %d\n", save_a * save_b);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
-
-   return 0;
+    return 0;
 }

C/p028.c

@@ -10,6 +10,8 @@
  *
  * What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -18,35 +20,35 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, step = 0, limit = N * N, sum = 1;
-   double elapsed;
-   struct timespec start, end;
+    int i, j, step = 0, limit = N * N, sum = 1;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Starting with the central 1, it's easy to see that the next four numbers in the diagonal
-    * are 1+2, 1+2+2, 1+2+2+2 and 1+2+2+2+2, then for the next four number the step is increased
-    * by two, so from 9 to 9+4, 9+4+4 etc, for the next four number the step is again increased
-    * by two, and so on. We go on until the value is equal to N*N, with N=1001 for this problem.*/
-   for(i = 0, j = 1; j < limit; i = (i + 1) % 4)
-   {
-      if(i == 0)
-      {
-         step += 2;
-      }
+    /* Starting with the central 1, it's easy to see that the next four numbers in the diagonal
+     * are 1+2, 1+2+2, 1+2+2+2 and 1+2+2+2+2, then for the next four number the step is increased
+     * by two, so from 9 to 9+4, 9+4+4 etc, for the next four number the step is again increased
+     * by two, and so on. We go on until the value is equal to N*N, with N=1001 for this problem.*/
+    for(i = 0, j = 1; j < limit; i = (i + 1) % 4)
+    {
+        if(i == 0)
+        {
+            step += 2;
+        }
 
-      j += step;
-      sum += j;
-   }
+        j += step;
+        sum += j;
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 28\n");
-   printf("Answer: %d\n", sum);
+    printf("Project Euler, Problem 28\n");
+    printf("Answer: %d\n", sum);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p029.c

@@ -11,6 +11,8 @@
  *
  * How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -21,86 +23,86 @@ int compare(void *a, void *b);
 
 int main(int argc, char **argv)
 {
-   mpz_t a;
-   mpz_t **powers;
-   int i, j, count;
-   double elapsed;
-   struct timespec start, end;
+    mpz_t a;
+    mpz_t **powers;
+    int i, j, count;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((powers = (mpz_t **)malloc(9801*sizeof(mpz_t *))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((powers = (mpz_t **)malloc(9801*sizeof(mpz_t *))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   for(i = 0; i < 9801; i++)
-   {
-      if((powers[i] = (mpz_t *)malloc(sizeof(mpz_t))) == NULL)
-      {
-         fprintf(stderr, "Error while allocating memory\n");
-         return 1;
-      }
-   }
+    for(i = 0; i < 9801; i++)
+    {
+        if((powers[i] = (mpz_t *)malloc(sizeof(mpz_t))) == NULL)
+        {
+            fprintf(stderr, "Error while allocating memory\n");
+            return 1;
+        }
+    }
 
-   mpz_init(a);
+    mpz_init(a);
 
-   for(i = 0; i < 9801; i++)
-   {
-      mpz_init(*powers[i]);
-   }
+    for(i = 0; i < 9801; i++)
+    {
+        mpz_init(*powers[i]);
+    }
 
-   /* Using the GMP Library to calculate all the powers.*/
-   for(i = 2; i <= 100; i++)
-   {
-      mpz_set_ui(a, i);
-      for(j = 2; j <= 100; j++)
-      {
-         mpz_pow_ui(*powers[(i-2)*99+j-2], a, j);
-      }
-   }
+    /* Using the GMP Library to calculate all the powers.*/
+    for(i = 2; i <= 100; i++)
+    {
+        mpz_set_ui(a, i);
+        for(j = 2; j <= 100; j++)
+        {
+            mpz_pow_ui(*powers[(i-2)*99+j-2], a, j);
+        }
+    }
 
-   mpz_clear(a);
+    mpz_clear(a);
 
-   /* Sort the values and count the different values.*/
-   quick_sort((void **)powers, 0, 9800, compare);
-   count = 1;
+    /* Sort the values and count the different values.*/
+    quick_sort((void **)powers, 0, 9800, compare);
+    count = 1;
 
-   for(i = 1; i < 9801; i++)
-   {
-      if(mpz_cmp(*powers[i], *powers[i-1]))
-      {
-         count++;
-      }
-   }
+    for(i = 1; i < 9801; i++)
+    {
+        if(mpz_cmp(*powers[i], *powers[i-1]))
+        {
+            count++;
+        }
+    }
 
-   for(i = 0; i < 9801; i++)
-   {
-      mpz_clear(*powers[i]);
-      free(powers[i]);
-   }
+    for(i = 0; i < 9801; i++)
+    {
+        mpz_clear(*powers[i]);
+        free(powers[i]);
+    }
 
-   free(powers);
+    free(powers);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 29\n");
-   printf("Answer: %d\n", count);
+    printf("Project Euler, Problem 29\n");
+    printf("Answer: %d\n", count);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int compare(void *a, void *b)
 {
-   mpz_t *n1, *n2;
+    mpz_t *n1, *n2;
 
-   n1 = (mpz_t *)a;
-   n2 = (mpz_t *)b;
+    n1 = (mpz_t *)a;
+    n2 = (mpz_t *)b;
 
-   return mpz_cmp(*n1, *n2);
+    return mpz_cmp(*n1, *n2);
 }

C/p030.c

@@ -10,6 +10,8 @@
  *
  * Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -17,41 +19,41 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, digit, sum, sum_tot = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i, j, digit, sum, sum_tot = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Straightforward brute force approach. The limit is chosen considering that
-    * 6*9^5=354294, so no number larger than that can be expressed as sum
-    * of 5th power of its digits.*/
-   for(i = 10; i < 354295; i++)
-   {
-      j = i;
-      sum = 0;
+    /* Straightforward brute force approach. The limit is chosen considering that
+     * 6*9^5=354294, so no number larger than that can be expressed as sum
+     * of 5th power of its digits.*/
+    for(i = 10; i < 354295; i++)
+    {
+        j = i;
+        sum = 0;
 
-      while(j > 0)
-      {
-         digit = j % 10;
-         sum += (pow(digit, 5));
-         j /= 10;
-      }
+        while(j > 0)
+        {
+            digit = j % 10;
+            sum += (pow(digit, 5));
+            j /= 10;
+        }
 
-      if(sum == i)
-      {
-         sum_tot += i;
-      }
-   }
+        if(sum == i)
+        {
+            sum_tot += i;
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed=(end.tv_sec-start.tv_sec)+(double)(end.tv_nsec-start.tv_nsec)/1000000000;
+    elapsed=(end.tv_sec-start.tv_sec)+(double)(end.tv_nsec-start.tv_nsec)/1000000000;
 
-   printf("Project Euler, problem 30\n");
-   printf("Answer: %d\n", sum_tot);
+    printf("Project Euler, problem 30\n");
+    printf("Answer: %d\n", sum_tot);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p031.c

@@ -8,6 +8,8 @@
  *
  * How many different ways can £2 be made using any number of coins?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -18,49 +20,49 @@ int coins[8] = {1, 2, 5, 10, 20, 50, 100, 200};
 
 int main(int argc, char **argv)
 {
-   int n;
-   double elapsed;
-   struct timespec start, end;
+    int n;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   n = count(0, 0, 0);
+    n = count(0, 0, 0);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 31\n");
-   printf("Answer: %d\n", n);
+    printf("Project Euler, Problem 31\n");
+    printf("Answer: %d\n", n);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 /* Simple recursive function that tries every combination.*/
 int count(int value, int n, int i)
 {
-   int j;
+    int j;
 
-   for(j = i; j < 8; j++)
-   {
-      value += coins[j];
+    for(j = i; j < 8; j++)
+    {
+        value += coins[j];
 
-      if(value == 200)
-      {
-         return n + 1;
-      }
-      else if(value > 200)
-      {
-         return n;
-      }
-      else
-      {
-         n = count(value, n, j);
-         value -= coins[j];
-      }
-   }
+        if(value == 200)
+        {
+            return n + 1;
+        }
+        else if(value > 200)
+        {
+            return n;
+        }
+        else
+        {
+            n = count(value, n, j);
+            value -= coins[j];
+        }
+    }
 
-   return n;
+    return n;
 }

C/p032.c

@@ -7,6 +7,8 @@
  *
  * HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
@@ -17,129 +19,129 @@ int compare(void *a, void *b);
 
 int main(int argc, char **argv)
 {
-   int i, j, p, sum, n = 0, num;
-   int **products;
-   char num_s[10];
-   double elapsed;
-   struct timespec start, end;
+    int i, j, p, sum, n = 0, num;
+    int **products;
+    char num_s[10];
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Initially I used a bigger array, but printing the resulting products
-    * shows that 10 values are sufficient.*/
-   if((products = (int **)malloc(10*sizeof(int *))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    /* Initially I used a bigger array, but printing the resulting products
+     * shows that 10 values are sufficient.*/
+    if((products = (int **)malloc(10*sizeof(int *))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   for(i = 0; i < 10; i++)
-   {
-      if((products[i] = (int *)malloc(sizeof(int))) == NULL)
-      {
-         fprintf(stderr, "Error while allocating memory\n");
-         return 1;
-      }
-   }
+    for(i = 0; i < 10; i++)
+    {
+        if((products[i] = (int *)malloc(sizeof(int))) == NULL)
+        {
+            fprintf(stderr, "Error while allocating memory\n");
+            return 1;
+        }
+    }
 
-   /* To get a 1 to 9 pandigital concatenation of the two factors and product,
-    * we need to multiply a 1 digit number times a 4 digit numbers (the biggest
-    * one digit number 9 times the biggest 3 digit number 999 multiplied give
-    * 8991 and the total digit count is 8, which is not enough), or a 2 digit 
-    * number times a 3 digit number (the smallest two different 3 digits number,
-    * 100 and 101, multiplied give 10100, and the total digit count is 11, which
-    * is too many). The outer loop starts at 2 because 1 times any number gives
-    * the same number, so its digit will be repeated and the result can't be 
-    * pandigital. The nested loop starts from 1234 because it's the smallest 
-    * 4-digit number with no repeated digits, and it ends at 4987 because it's
-    * the biggest number without repeated digits that multiplied by 2 gives a 
-    * 4 digit number. */
-   for(i = 2; i < 9; i++)
-   {
-      for(j = 1234; j < 4987; j++)
-      {
-         p = i * j;
-         sprintf(num_s, "%d%d%d", i, j, p);
+    /* To get a 1 to 9 pandigital concatenation of the two factors and product,
+     * we need to multiply a 1 digit number times a 4 digit numbers (the biggest
+     * one digit number 9 times the biggest 3 digit number 999 multiplied give
+     * 8991 and the total digit count is 8, which is not enough), or a 2 digit
+     * number times a 3 digit number (the smallest two different 3 digits number,
+     * 100 and 101, multiplied give 10100, and the total digit count is 11, which
+     * is too many). The outer loop starts at 2 because 1 times any number gives
+     * the same number, so its digit will be repeated and the result can't be
+     * pandigital. The nested loop starts from 1234 because it's the smallest
+     * 4-digit number with no repeated digits, and it ends at 4987 because it's
+     * the biggest number without repeated digits that multiplied by 2 gives a
+     * 4 digit number. */
+    for(i = 2; i < 9; i++)
+    {
+        for(j = 1234; j < 4987; j++)
+        {
+            p = i * j;
+            sprintf(num_s, "%d%d%d", i, j, p);
 
-         if(strlen(num_s) > 9)
-         {
-            break;
-         }
+            if(strlen(num_s) > 9)
+            {
+                break;
+            }
 
-         num = atoi(num_s);
+            num = atoi(num_s);
 
-         if(is_pandigital(num, 9))
-         {
-            *products[n] = p;
-            n++;
-         }
-      }
-   }
+            if(is_pandigital(num, 9))
+            {
+                *products[n] = p;
+                n++;
+            }
+        }
+    }
 
-   /* The outer loop starts at 12 because 10 has a 0 and 11 has two 1s, so
-    * the result can't be pandigital. The nested loop starts at 123 because
-    * it's the smallest 3-digit number with no digit repetitions and ends at
-    * 833, because 834*12 has 5 digits.*/
-   for(i = 12; i < 99; i++)
-   {
-      for(j = 123; j < 834; j++)
-      {
-         p = i * j;
-         sprintf(num_s, "%d%d%d", i, j, p);
+    /* The outer loop starts at 12 because 10 has a 0 and 11 has two 1s, so
+     * the result can't be pandigital. The nested loop starts at 123 because
+     * it's the smallest 3-digit number with no digit repetitions and ends at
+     * 833, because 834*12 has 5 digits.*/
+    for(i = 12; i < 99; i++)
+    {
+        for(j = 123; j < 834; j++)
+        {
+            p = i * j;
+            sprintf(num_s, "%d%d%d", i, j, p);
 
-         if(strlen(num_s) > 9)
-         {
-            break;
-         }
+            if(strlen(num_s) > 9)
+            {
+                break;
+            }
 
-         num = atoi(num_s);
+            num = atoi(num_s);
 
-         if(is_pandigital(num, 9))
-         {
-            *products[n] = p;
-            n++;
-         }
-      }
-   }
-   
-   /* Sort the found products to easily see if there are duplicates.*/
-   insertion_sort((void **)products, 0, n-1, compare);
+            if(is_pandigital(num, 9))
+            {
+                *products[n] = p;
+                n++;
+            }
+        }
+    }
 
-   sum = *products[0];
+    /* Sort the found products to easily see if there are duplicates.*/
+    insertion_sort((void **)products, 0, n-1, compare);
 
-   for(i = 1; i < n; i++)
-   {
-      if(*products[i] != *products[i-1])
-      {
-         sum += *products[i];
-      }
-   }
+    sum = *products[0];
 
-   for(i = 0; i < 10; i++)
-   {
-      free(products[i]);
-   }
+    for(i = 1; i < n; i++)
+    {
+        if(*products[i] != *products[i-1])
+        {
+            sum += *products[i];
+        }
+    }
 
-   free(products);
+    for(i = 0; i < 10; i++)
+    {
+        free(products[i]);
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    free(products);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 32\n");
-   printf("Answer: %d\n", sum);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 32\n");
+    printf("Answer: %d\n", sum);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 int compare(void *a, void *b)
 {
-   int *n1, *n2;
+    int *n1, *n2;
 
-   n1 = (int *)a;
-   n2 = (int *)b;
+    n1 = (int *)a;
+    n2 = (int *)b;
 
-   return *n1 - *n2;
+    return *n1 - *n2;
 }

C/p033.c

@@ -8,6 +8,8 @@
  *
  * If the product of these four fractions is given in its lowest common terms, find the value of the denominator.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -15,47 +17,47 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, n, d, prod_n = 1, prod_d = 1, div;
-   float f1, f2;
-   double elapsed;
-   struct timespec start, end;
+    int i, j, n, d, prod_n = 1, prod_d = 1, div;
+    float f1, f2;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   for(i = 11; i < 100; i++)
-   {
-      for(j = 11; j < 100; j++)
-      {
-         /* If the example is non-trivial, check if cancelling the digit that's equal
-          * in numerator and denominator gives the same fraction.*/
-         if(i % 10 && j % 10 && i != j && i % 10 == j / 10)
-         {
-            n = i / 10;
-            d = j % 10;
-
-            f1 = (float)i / j;
-            f2 = (float)n / d;
-
-            if(f1 == f2)
+    for(i = 11; i < 100; i++)
+    {
+        for(j = 11; j < 100; j++)
+        {
+            /* If the example is non-trivial, check if cancelling the digit that's equal
+             * in numerator and denominator gives the same fraction.*/
+            if(i % 10 && j % 10 && i != j && i % 10 == j / 10)
             {
-               prod_n *= i;
-               prod_d *= j;
+                n = i / 10;
+                d = j % 10;
+
+                f1 = (float)i / j;
+                f2 = (float)n / d;
+
+                if(f1 == f2)
+                {
+                    prod_n *= i;
+                    prod_d *= j;
+                }
             }
-         }
-      }
-   }
+        }
+    }
 
-   /* Find the greater common divisor of the fraction found.*/
-   div = gcd(prod_n, prod_d);
+    /* Find the greater common divisor of the fraction found.*/
+    div = gcd(prod_n, prod_d);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 33\n"); 
-   printf("Answer: %d\n", prod_d/div);
+    printf("Project Euler, Problem 33\n");
+    printf("Answer: %d\n", prod_d/div);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p034.c

@@ -4,6 +4,8 @@
  *
  * Note: as 1! = 1 and 2! = 2 are not sums they are not included.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -11,66 +13,66 @@
 
 int main(int argc, char **argv)
 {
-   int i;
-   unsigned long int digit;
-   double elapsed;
-   struct timespec start, end;
-   mpz_t a, b, q, sum_f, sum, factorials[10];
+    int i;
+    unsigned long int digit;
+    double elapsed;
+    struct timespec start, end;
+    mpz_t a, b, q, sum_f, sum, factorials[10];
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   mpz_init_set_ui(a, 10);
-   mpz_init_set_ui(sum, 0);
-   mpz_inits(b, q, sum_f, sum, NULL);
+    mpz_init_set_ui(a, 10);
+    mpz_init_set_ui(sum, 0);
+    mpz_inits(b, q, sum_f, sum, NULL);
 
-   for(i = 0; i < 10; i++)
-   {
-      mpz_init_set_ui(factorials[i], 1);
-   }
+    for(i = 0; i < 10; i++)
+    {
+        mpz_init_set_ui(factorials[i], 1);
+    }
 
-   /* Pre-calculate factorials of each digit from 0 to 9.*/
-   for(i = 2; i < 10; i++)
-   {
-      mpz_fac_ui(factorials[i], i);
-   }
+    /* Pre-calculate factorials of each digit from 0 to 9.*/
+    for(i = 2; i < 10; i++)
+    {
+        mpz_fac_ui(factorials[i], i);
+    }
 
-   /* 9!*7<9999999, so 9999999 is certainly un upper bound.*/
-   while(mpz_cmp_ui(a, 9999999) < 0)
-   {
-      mpz_set(b, a);
-      mpz_set_ui(sum_f, 0);
+    /* 9!*7<9999999, so 9999999 is certainly un upper bound.*/
+    while(mpz_cmp_ui(a, 9999999) < 0)
+    {
+        mpz_set(b, a);
+        mpz_set_ui(sum_f, 0);
 
-      while(mpz_cmp_ui(b, 0))
-      {
-         digit = mpz_fdiv_qr_ui(b, q, b, 10);
-         mpz_add(sum_f, sum_f, factorials[digit]);
-      }
+        while(mpz_cmp_ui(b, 0))
+        {
+            digit = mpz_fdiv_qr_ui(b, q, b, 10);
+            mpz_add(sum_f, sum_f, factorials[digit]);
+        }
 
-      if(!mpz_cmp(a, sum_f))
-      {
-         mpz_add(sum, sum, a);
-      }
+        if(!mpz_cmp(a, sum_f))
+        {
+            mpz_add(sum, sum, a);
+        }
 
-      mpz_add_ui(a, a, 1);
-   }
+        mpz_add_ui(a, a, 1);
+    }
 
-   mpz_clears(a, b, q, sum_f, NULL);
+    mpz_clears(a, b, q, sum_f, NULL);
 
-   for(i = 0; i < 10; i++)
-   {
-      mpz_clear(factorials[i]);
-   }
+    for(i = 0; i < 10; i++)
+    {
+        mpz_clear(factorials[i]);
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 34\n");
-   gmp_printf("Answer: %Zd\n", sum);
+    printf("Project Euler, Problem 34\n");
+    gmp_printf("Answer: %Zd\n", sum);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   mpz_clear(sum);
+    mpz_clear(sum);
 
-   return 0;
+    return 0;
 }

C/p035.c

@@ -4,6 +4,8 @@
  *
  * How many circular primes are there below one million?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -18,83 +20,83 @@ int *primes;
 
 int main(int argc, char **argv)
 {
-   int i, count = 13;
-   double elapsed;
-   struct timespec start, end;
+    int i, count = 13;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Calculate all primes below one million, then check if they're circular.*/
-   if((primes = sieve(N)) == NULL)
-   {
-      fprintf(stderr, "Error! Sieve function returned NULL\n");
-      return 1;
-   }
+    /* Calculate all primes below one million, then check if they're circular.*/
+    if((primes = sieve(N)) == NULL)
+    {
+        fprintf(stderr, "Error! Sieve function returned NULL\n");
+        return 1;
+    }
 
-   /* Starting from 101 because we already know that there are 13 circular primes below 100.*/
-   for(i = 101; i < 1000000; i += 2)
-   {
-      if(is_circular_prime(i))
-      {
-         count++;
-      }
-   }
+    /* Starting from 101 because we already know that there are 13 circular primes below 100.*/
+    for(i = 101; i < 1000000; i += 2)
+    {
+        if(is_circular_prime(i))
+        {
+            count++;
+        }
+    }
 
-   free(primes);
+    free(primes);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 35\n");
-   printf("Answer: %d\n", count);
+    printf("Project Euler, Problem 35\n");
+    printf("Answer: %d\n", count);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int is_circular_prime(int n)
 {
-   int i, tmp, count;
+    int i, tmp, count;
 
-   /* If n is not prime, it's obviously not a circular prime.*/
-   if(primes[n] == 0)
-   {
-      return 0;
-   }
+    /* If n is not prime, it's obviously not a circular prime.*/
+    if(primes[n] == 0)
+    {
+        return 0;
+    }
 
-   /* The primes below 10 are circular primes.*/
-   if(primes[n] == 1 && n < 10)
-   {
-      return 1;
-   }
+    /* The primes below 10 are circular primes.*/
+    if(primes[n] == 1 && n < 10)
+    {
+        return 1;
+    }
 
-   tmp = n;
-   count = 0;
+    tmp = n;
+    count = 0;
 
-   while(tmp > 0)
-   {
-      /* If the number has one or more even digits, it can't be a circular prime.
-       * because at least one of the rotations will be even.*/
-      if(tmp % 2 == 0)
-      {
-         return 0;
-      }
-      /* Count the number of digits.*/
-      count++;
-      tmp /= 10;
-   }
+    while(tmp > 0)
+    {
+        /* If the number has one or more even digits, it can't be a circular prime.
+         * because at least one of the rotations will be even.*/
+        if(tmp % 2 == 0)
+        {
+            return 0;
+        }
+        /* Count the number of digits.*/
+        count++;
+        tmp /= 10;
+    }
 
-   for(i = 1; i < count; i++)
-   {
-      /* Generate rotations and check if they're prime.*/
-      n = n % (int)pow(10, count-1) * 10 + n / (int)pow(10, count-1);
-      if(primes[n] == 0)
-      {
-         return 0;
-      }
-   }
+    for(i = 1; i < count; i++)
+    {
+        /* Generate rotations and check if they're prime.*/
+        n = n % (int)pow(10, count-1) * 10 + n / (int)pow(10, count-1);
+        if(primes[n] == 0)
+        {
+            return 0;
+        }
+    }
 
-   return 1;
+    return 1;
 }

C/p036.c

@@ -4,6 +4,8 @@
  *
  * (Please note that the palindromic number, in either base, may not include leading zeros.)*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -13,31 +15,31 @@
 
 int main(int argc, char **argv)
 {
-   int i, sum = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i, sum = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Brute force approach. For every number below 1 million, 
-    * check if they're palindrome in base 2 and 10 using the
-    * function implemented in projecteuler.c.*/
-   for(i = 1; i < N; i += 2)
-   {
-      if(is_palindrome(i, 10) && is_palindrome(i, 2))
-      {
-         sum += i;
-      }
-   }
+    /* Brute force approach. For every number below 1 million,
+     * check if they're palindrome in base 2 and 10 using the
+     * function implemented in projecteuler.c.*/
+    for(i = 1; i < N; i += 2)
+    {
+        if(is_palindrome(i, 10) && is_palindrome(i, 2))
+        {
+            sum += i;
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 36\n");
-   printf("Answer: %d\n", sum);
+    printf("Project Euler, Problem 36\n");
+    printf("Answer: %d\n", sum);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p037.c

@@ -5,6 +5,8 @@
  *
  * NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -15,75 +17,75 @@ int is_tr_prime(int n);
 
 int main(int argc, char **argv)
 {
-   int i = 0, n = 1, sum = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i = 0, n = 1, sum = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Check every number until 11 truncatable primes are found.*/
-   while(i < 11)
-   {
-      if(is_tr_prime(n))
-      {
-         sum += n;
-         i++;
-      }
-      n++;
-   }
+    /* Check every number until 11 truncatable primes are found.*/
+    while(i < 11)
+    {
+        if(is_tr_prime(n))
+        {
+            sum += n;
+            i++;
+        }
+        n++;
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 37\n");
-   printf("Answer: %d\n", sum);
+    printf("Project Euler, Problem 37\n");
+    printf("Answer: %d\n", sum);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int is_tr_prime(int n)
 {
-   int i, tmp;
+    int i, tmp;
 
-   /* One-digit numbers and non-prime numbers are
-    * not truncatable primes.*/
-   if(n < 11 || !is_prime(n))
-   {
-      return 0;
-   }
+    /* One-digit numbers and non-prime numbers are
+     * not truncatable primes.*/
+    if(n < 11 || !is_prime(n))
+    {
+        return 0;
+    }
 
-   /* Remove one digit at a time from the right and check
-    * if the resulting number is prime. Return 0 if it isn't.*/
-   tmp = n / 10;
+    /* Remove one digit at a time from the right and check
+     * if the resulting number is prime. Return 0 if it isn't.*/
+    tmp = n / 10;
 
-   while(tmp > 0)
-   {
-      if(!is_prime(tmp))
-      {
-         return 0;
-      }
-      tmp /= 10;
-   }
+    while(tmp > 0)
+    {
+        if(!is_prime(tmp))
+        {
+            return 0;
+        }
+        tmp /= 10;
+    }
 
-   /* Starting from the last digit, check if it's prime, then
-    * add back one digit at a time on the left and check if it
-    * is prime. Return 0 when it isn't.*/
-   i = 10;
-   tmp = n % i;
+    /* Starting from the last digit, check if it's prime, then
+     * add back one digit at a time on the left and check if it
+     * is prime. Return 0 when it isn't.*/
+    i = 10;
+    tmp = n % i;
 
-   while(tmp != n)
-   {
-      if(!is_prime(tmp))
-      {
-         return 0;
-      }
-      i *= 10;
-      tmp = n % i;
-   }
+    while(tmp != n)
+    {
+        if(!is_prime(tmp))
+        {
+            return 0;
+        }
+        i *= 10;
+        tmp = n % i;
+    }
 
-   /* If it gets here, the number is truncatable prime.*/
-   return 1;
+    /* If it gets here, the number is truncatable prime.*/
+    return 1;
 }

C/p038.c

@@ -11,6 +11,8 @@
  *
  * What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -18,64 +20,64 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, tmp;
-   long int n, max = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i, j, tmp;
+    long int n, max = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* A brute force approach is used, starting with 1 and multiplying
-    * the number by 1, 2 etc., concatenating the results, checking if 
-    * it's 1 to 9 pandigital, and going to the next number when the
-    * concatenated result is greater than the greatest 9 digit pandigital
-    * value. The limit is set to 10000, since 1*10000=10000, 2*10000=20000 and
-    * concatenating this two numbers a 10-digit number is obtained.*/
-   for(i = 1; i < 10000; i++)
-   {
-      n = 0;
-      j = 1;
-      do
-      {
-         tmp = i * j;
-         n += tmp;
-         j++;
+    /* A brute force approach is used, starting with 1 and multiplying
+     * the number by 1, 2 etc., concatenating the results, checking if
+     * it's 1 to 9 pandigital, and going to the next number when the
+     * concatenated result is greater than the greatest 9 digit pandigital
+     * value. The limit is set to 10000, since 1*10000=10000, 2*10000=20000 and
+     * concatenating this two numbers a 10-digit number is obtained.*/
+    for(i = 1; i < 10000; i++)
+    {
+        n = 0;
+        j = 1;
+        do
+        {
+            tmp = i * j;
+            n += tmp;
+            j++;
 
-         if(n > max && is_pandigital(n, 9))
-         {
-            max = n;
-         }
-         if(i * j < 10)
-         {
-            n *= 10;
-         }
-         else if(i * j < 100)
-         {
-            n *= 100;
-         }
-         else if(i * j < 1000)
-         {
-            n *= 1000;
-         }
-         else if(i * j < 10000)
-         {
-            n *= 10000;
-         }
-         else if(i * j < 100000)
-         {
-            n *= 100000;
-         }
-      }while(n <= 987654321);
-   }
+            if(n > max && is_pandigital(n, 9))
+            {
+                max = n;
+            }
+            if(i * j < 10)
+            {
+                n *= 10;
+            }
+            else if(i * j < 100)
+            {
+                n *= 100;
+            }
+            else if(i * j < 1000)
+            {
+                n *= 1000;
+            }
+            else if(i * j < 10000)
+            {
+                n *= 10000;
+            }
+            else if(i * j < 100000)
+            {
+                n *= 100000;
+            }
+        } while(n <= 987654321);
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 38\n");
-   printf("Answer: %ld\n", max);
+    printf("Project Euler, Problem 38\n");
+    printf("Answer: %ld\n", max);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p039.c

@@ -4,87 +4,89 @@
  *
  * For which value of p ≤ 1000, is the number of solutions maximised?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
 
 int main(int argc, char **argv)
 {
-   int p, m, n, a, b, c, count, max = 0, res = 0, tmpa, tmpb, tmpc, i;
-   int savedc[1000] = {0};
-   double elapsed;
-   struct timespec start, end;
+    int p, m, n, a, b, c, count, max = 0, res = 0, tmpa, tmpb, tmpc, i;
+    int savedc[1000] = {0};
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Start with p=12 (the smallest pythagorean triplet is (3,4,5) and 3+4+5=12.*/
-   for(p = 12; p <= 1000; p++)
-   {
-      count = 0;
-      a = 0;
-      b = 0;
-      c = 0;
+    /* Start with p=12 (the smallest pythagorean triplet is (3,4,5) and 3+4+5=12.*/
+    for(p = 12; p <= 1000; p++)
+    {
+        count = 0;
+        a = 0;
+        b = 0;
+        c = 0;
 
-      /* Generate pythagorean triplets.*/
-      for(m = 2; m * m < p; m++)
-      {
-         for(n = 1; n < m; n++)
-         {
-            a = m * m - n * n;
-            b = 2 * m * n;
-            c = m * m + n * n;
-
-            /* Increase counter if a+b+c=p and the triplet is new,
-             * then save the value of c to avoid counting the same 
-             * triplet more than once.*/
-            if(a + b + c == p && !savedc[c])
+        /* Generate pythagorean triplets.*/
+        for(m = 2; m * m < p; m++)
+        {
+            for(n = 1; n < m; n++)
             {
-               savedc[c] = 1;
-               count++;
+                a = m * m - n * n;
+                b = 2 * m * n;
+                c = m * m + n * n;
+
+                /* Increase counter if a+b+c=p and the triplet is new,
+                 * then save the value of c to avoid counting the same
+                 * triplet more than once.*/
+                if(a + b + c == p && !savedc[c])
+                {
+                    savedc[c] = 1;
+                    count++;
+                }
+
+                i = 2;
+                tmpa = a;
+                tmpb = b;
+                tmpc = c;
+
+                /* Check all the triplets obtained multiplying a, b and c
+                 * for integer numbers, until the perimeters exceeds p.*/
+                while(tmpa + tmpb + tmpc < p)
+                {
+                    tmpa = a * i;
+                    tmpb = b * i;
+                    tmpc = c * i;
+
+                    /* Increase counter if the new a, b and c give a perimeter=p.*/
+                    if(tmpa + tmpb + tmpc == p && !savedc[tmpc])
+                    {
+                        savedc[tmpc] = 1;
+                        count++;
+                    }
+
+                    i++;
+                }
             }
+        }
 
-            i = 2;
-            tmpa = a;
-            tmpb = b;
-            tmpc = c;
+        /* If the current value is greater than the maximum,
+         * save the new maximum and the value of p.*/
+        if(count > max)
+        {
+            max = count;
+            res = p;
+        }
+    }
 
-            /* Check all the triplets obtained multiplying a, b and c
-             * for integer numbers, until the perimeters exceeds p.*/
-            while(tmpa + tmpb + tmpc < p)
-            {
-               tmpa = a * i;
-               tmpb = b * i;
-               tmpc = c * i;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-               /* Increase counter if the new a, b and c give a perimeter=p.*/
-               if(tmpa + tmpb + tmpc == p && !savedc[tmpc])
-               {
-                  savedc[tmpc] = 1;
-                  count++;
-               }
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-               i++;
-            }
-         }
-      }
+    printf("Project Euler, Problem 39\n");
+    printf("Answer: %d\n", res);
 
-      /* If the current value is greater than the maximum,
-       * save the new maximum and the value of p.*/
-      if(count > max)
-      {
-         max = count;
-         res = p;
-      }
-   }
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
-
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
-
-   printf("Project Euler, Problem 39\n");
-   printf("Answer: %d\n", res);
-
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
-
-   return 0;
+    return 0;
 }

C/p040.c

@@ -8,86 +8,88 @@
  *
  * d_1 × d_10 × d_100 × d_1000 × d_10000 × d_100000 × d_1000000*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
 
 int main(int argc, char **argv)
 {
-   int i, value, n;
-   int digits[1000005];
-   double elapsed;
-   struct timespec start, end;
+    int i, value, n;
+    int digits[1000005];
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   i = 1;
-   value = 1;
+    i = 1;
+    value = 1;
 
-   /* Loop on all numbers and put the digits in the right place
-    * in an array. Use modulo and division to get the digits
-    * for numbers with more than one digit.*/
-   while(i <= 1000000)
-   {
-      if(value < 10)
-      {
-         digits[i-1] = value;
-         i++;
-      }
-      else if(value < 100)
-      {
-         digits[i-1] = value / 10;
-         digits[i] = value % 10;
-         i += 2;
-      }
-      else if(value < 1000)
-      {
-         digits[i-1] = value / 100;
-         digits[i] = (value / 10) % 10;
-         digits[i+1] = value % 10;
-         i += 3;
-      }
-      else if(value < 10000)
-      {
-         digits[i-1] = value / 1000;
-         digits[i] = (value / 100) % 10;
-         digits[i+1] = (value / 10) % 10;
-         digits[i+2] = value % 10;
-         i += 4;
-      }
-      else if(value < 100000)
-      {
-         digits[i-1] = value / 10000;
-         digits[i] = (value / 1000) % 10;
-         digits[i+1] = (value / 100) % 10;
-         digits[i+2] = (value / 10) % 10;
-         digits[i+3] = value % 10;
-         i += 5;
-      }
-      else if(value < 1000000)
-      {
-         digits[i-1] = value / 100000;
-         digits[i] = (value / 10000) % 10;
-         digits[i+1] = (value / 1000) % 10;
-         digits[i+2] = (value / 100) % 10;
-         digits[i+3] = (value / 10) % 10;
-         digits[i+4] = value % 10;
-         i += 6;
-      }
-      value++;
-   }
+    /* Loop on all numbers and put the digits in the right place
+     * in an array. Use modulo and division to get the digits
+     * for numbers with more than one digit.*/
+    while(i <= 1000000)
+    {
+        if(value < 10)
+        {
+            digits[i-1] = value;
+            i++;
+        }
+        else if(value < 100)
+        {
+            digits[i-1] = value / 10;
+            digits[i] = value % 10;
+            i += 2;
+        }
+        else if(value < 1000)
+        {
+            digits[i-1] = value / 100;
+            digits[i] = (value / 10) % 10;
+            digits[i+1] = value % 10;
+            i += 3;
+        }
+        else if(value < 10000)
+        {
+            digits[i-1] = value / 1000;
+            digits[i] = (value / 100) % 10;
+            digits[i+1] = (value / 10) % 10;
+            digits[i+2] = value % 10;
+            i += 4;
+        }
+        else if(value < 100000)
+        {
+            digits[i-1] = value / 10000;
+            digits[i] = (value / 1000) % 10;
+            digits[i+1] = (value / 100) % 10;
+            digits[i+2] = (value / 10) % 10;
+            digits[i+3] = value % 10;
+            i += 5;
+        }
+        else if(value < 1000000)
+        {
+            digits[i-1] = value / 100000;
+            digits[i] = (value / 10000) % 10;
+            digits[i+1] = (value / 1000) % 10;
+            digits[i+2] = (value / 100) % 10;
+            digits[i+3] = (value / 10) % 10;
+            digits[i+4] = value % 10;
+            i += 6;
+        }
+        value++;
+    }
 
-   /* Calculate the product.*/
-   n = digits[0] * digits[9] * digits[99] * digits[999] * digits[9999] * digits[99999] * digits[999999];
+    /* Calculate the product.*/
+    n = digits[0] * digits[9] * digits[99] * digits[999] * digits[9999] * digits[99999] * digits[999999];
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 40\n");
-   printf("Answer: %d\n", n);
+    printf("Project Euler, Problem 40\n");
+    printf("Answer: %d\n", n);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p041.c

@@ -1,8 +1,10 @@
-/* We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital 
+/* We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital
  * and is also prime.
  *
  * What is the largest n-digit pandigital prime that exists?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -13,53 +15,53 @@ int count_digits(int n);
 
 int main(int argc, char **argv)
 {
-   /* 8- and 9-digit pandigital numbers can't be prime, because 
-    * 1+2+...+8=36, which is divisible by 3, and 36+9=45 which is
-    * also divisible by 3, and therefore the whole number is divisible
-    * by 3. So we can start from the largest 7-digit pandigital number,
-    * until we find a prime.*/
-   int i = 7654321;
-   double elapsed;
-   struct timespec start, end;
+    /* 8- and 9-digit pandigital numbers can't be prime, because
+     * 1+2+...+8=36, which is divisible by 3, and 36+9=45 which is
+     * also divisible by 3, and therefore the whole number is divisible
+     * by 3. So we can start from the largest 7-digit pandigital number,
+     * until we find a prime.*/
+    int i = 7654321;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   while(i > 0)
-   {
-      if(is_pandigital(i, count_digits(i)) && is_prime(i))
-      {
-         break;
-      }
-      /*Skipping the even numbers.*/
-      i -= 2;
-   }
+    while(i > 0)
+    {
+        if(is_pandigital(i, count_digits(i)) && is_prime(i))
+        {
+            break;
+        }
+        /*Skipping the even numbers.*/
+        i -= 2;
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 41\n");
-   printf("Answer: %d\n", i);
+    printf("Project Euler, Problem 41\n");
+    printf("Answer: %d\n", i);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int count_digits(int n)
 {
-   int i = 0;
-   
-   if(n == 0)
-   {
-      return 1;
-   }
+    int i = 0;
 
-   while(n > 0)
-   {
-      i++;
-      n /= 10;
-   }
+    if(n == 0)
+    {
+        return 1;
+    }
 
-   return i;
+    while(n > 0)
+    {
+        i++;
+        n /= 10;
+    }
+
+    return i;
 }

C/p042.c

@@ -8,6 +8,8 @@
  * Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words,
  * how many are triangle words?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
@@ -17,93 +19,93 @@ int is_triang(int n);
 
 int main(int argc, char **argv)
 {
-   int i, j, n, len, value, count = 0;
-   char *line, **words;
-   double elapsed;
-   FILE *fp;
-   struct timespec start, end;
+    int i, j, n, len, value, count = 0;
+    char *line, **words;
+    double elapsed;
+    FILE *fp;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((fp = fopen("words.txt", "r")) == NULL)
-   {
-      fprintf(stderr, "Error while opening file %s\n", "words.txt");
-      return 1;
-   }
+    if((fp = fopen("words.txt", "r")) == NULL)
+    {
+        fprintf(stderr, "Error while opening file %s\n", "words.txt");
+        return 1;
+    }
 
-   fscanf(fp, "%ms", &line);
+    fscanf(fp, "%ms", &line);
 
-   fclose(fp);
+    fclose(fp);
 
-   n = 1;
-   len = strlen(line);
+    n = 1;
+    len = strlen(line);
 
-   /* Count the words.*/
-   for(i = 0; i < len; i++)
-   {
-      if(line[i] == ',')
-      {
-         n++;
-      }
-   }
+    /* Count the words.*/
+    for(i = 0; i < len; i++)
+    {
+        if(line[i] == ',')
+        {
+            n++;
+        }
+    }
 
-   if((words = (char **)malloc(n*sizeof(char *))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((words = (char **)malloc(n*sizeof(char *))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   /* Save the words in an array of strings.*/
-   words[0] = strtok(line, ",\"");
+    /* Save the words in an array of strings.*/
+    words[0] = strtok(line, ",\"");
 
-   for(i = 1; i < n; i++)
-   {
-      words[i] = strtok(NULL, ",\"");
-   }
+    for(i = 1; i < n; i++)
+    {
+        words[i] = strtok(NULL, ",\"");
+    }
 
-   /* For each word, calculate its value and check if it's a triangle number.*/
-   for(i = 0; i < n; i++)
-   {
-      value = 0;
-      len = strlen(words[i]);
+    /* For each word, calculate its value and check if it's a triangle number.*/
+    for(i = 0; i < n; i++)
+    {
+        value = 0;
+        len = strlen(words[i]);
 
-      for(j = 0; j < len; j++)
-      {
-         value += (words[i][j] - 'A' + 1);
-      }
+        for(j = 0; j < len; j++)
+        {
+            value += (words[i][j] - 'A' + 1);
+        }
 
-      if(is_triang(value))
-      {
-         count++;
-      }
-   }
+        if(is_triang(value))
+        {
+            count++;
+        }
+    }
 
-   free(line);
-   free(words);
+    free(line);
+    free(words);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec-start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec-start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 42\n");
-   printf("Answer: %d\n", count);
+    printf("Project Euler, Problem 42\n");
+    printf("Answer: %d\n", count);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int is_triang(int n)
 {
-   int i, j;
+    int i, j;
 
-   for(i = 1, j = 1; j <= n; i++, j += i)
-   {
-      if(n == j)
-      {
-         return 1;
-      }
-   }
+    for(i = 1, j = 1; j <= n; i++, j += i)
+    {
+        if(n == j)
+        {
+            return 1;
+        }
+    }
 
-   return 0;
+    return 0;
 }

C/p043.c

@@ -13,6 +13,8 @@
  *
  * Find the sum of all 0 to 9 pandigital numbers with this property.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -23,116 +25,116 @@ int has_property(int **n);
 
 int main(int argc, char **argv)
 {
-   int i;
-   int **n;
-   long int sum = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i;
+    int **n;
+    long int sum = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((n = (int **)malloc(10*sizeof(int *))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
- 
-   for(i = 0; i < 10; i++)
-   {
-      if((n[i] = (int *)malloc(sizeof(int))) == NULL)
-      {
-         fprintf(stderr, "Error while allocating memory\n");
-         return 1;
-      }
-      *n[i] = i;
-   }
+    if((n = (int **)malloc(10*sizeof(int *))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   /* Find the next permutation and check if it has the required property.
-    * Repeat until all the permutations have been found.*/
-   while(next_permutation((void **)n, 10, compare) != -1)
-   {
-      if(has_property(n))
-      {
-         sum += *n[0] * 1e9 + *n[1] * 1e8 + *n[2] * 1e7 + *n[3] * 1e6 + *n[4] * 1e5 +
-                *n[5] * 1e4 + *n[6] * 1e3 + *n[7] * 1e2 + *n[8] * 1e1 + *n[9];
-      }
-   }
+    for(i = 0; i < 10; i++)
+    {
+        if((n[i] = (int *)malloc(sizeof(int))) == NULL)
+        {
+            fprintf(stderr, "Error while allocating memory\n");
+            return 1;
+        }
+        *n[i] = i;
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    /* Find the next permutation and check if it has the required property.
+     * Repeat until all the permutations have been found.*/
+    while(next_permutation((void **)n, 10, compare) != -1)
+    {
+        if(has_property(n))
+        {
+            sum += *n[0] * 1e9 + *n[1] * 1e8 + *n[2] * 1e7 + *n[3] * 1e6 + *n[4] * 1e5 +
+                   *n[5] * 1e4 + *n[6] * 1e3 + *n[7] * 1e2 + *n[8] * 1e1 + *n[9];
+        }
+    }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 43\n");
-   printf("Answer: %ld\n", sum);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 43\n");
+    printf("Answer: %ld\n", sum);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 int compare(void *a, void *b)
 {
-   int *n1, *n2;
+    int *n1, *n2;
 
-   n1 = (int *)a;
-   n2 = (int *)b;
+    n1 = (int *)a;
+    n2 = (int *)b;
 
-   return *n1 - *n2;
+    return *n1 - *n2;
 }
 
 /* Function to check if the value has the desired property.*/
 int has_property(int **n)
 {
-   long int value;
+    long int value;
 
-   value = *n[1] * 100 + *n[2] * 10 + *n[3];
+    value = *n[1] * 100 + *n[2] * 10 + *n[3];
 
-   if(value % 2 != 0)
-   {
-      return 0;
-   }
+    if(value % 2 != 0)
+    {
+        return 0;
+    }
 
-   value = *n[2] * 100 + *n[3] * 10 + *n[4];
+    value = *n[2] * 100 + *n[3] * 10 + *n[4];
 
-   if(value % 3 != 0)
-   {
-      return 0;
-   }
+    if(value % 3 != 0)
+    {
+        return 0;
+    }
 
-   value = *n[3] * 100 + *n[4] * 10 + *n[5];
+    value = *n[3] * 100 + *n[4] * 10 + *n[5];
 
-   if(value % 5 != 0)
-   {
-      return 0;
-   }
+    if(value % 5 != 0)
+    {
+        return 0;
+    }
 
-   value = *n[4] * 100 + *n[5] * 10 + *n[6];
+    value = *n[4] * 100 + *n[5] * 10 + *n[6];
 
-   if(value % 7 != 0)
-   {
-      return 0;
-   }
+    if(value % 7 != 0)
+    {
+        return 0;
+    }
 
-   value = *n[5] * 100 + *n[6] * 10 + *n[7];
+    value = *n[5] * 100 + *n[6] * 10 + *n[7];
 
-   if(value % 11 != 0)
-   {
-      return 0;
-   }
+    if(value % 11 != 0)
+    {
+        return 0;
+    }
 
-   value = *n[6] * 100 + *n[7] * 10 + *n[8];
+    value = *n[6] * 100 + *n[7] * 10 + *n[8];
 
-   if(value %13 != 0)
-   {
-      return 0;
-   }
+    if(value %13 != 0)
+    {
+        return 0;
+    }
 
-   value = *n[7] * 100 + *n[8] * 10 + *n[9];
+    value = *n[7] * 100 + *n[8] * 10 + *n[9];
 
-   if(value % 17 != 0)
-   {
-      return 0;
-   }
+    if(value % 17 != 0)
+    {
+        return 0;
+    }
 
-   return 1;
+    return 1;
 }

C/p044.c

@@ -7,6 +7,8 @@
  * Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference are pentagonal and D = |Pk − Pj| is minimised;
  * what is the value of D?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -15,42 +17,42 @@
 
 int main(int argc, char **argv)
 {
-   int n, m, pn, pm, found = 0;
-   double elapsed;
-   struct timespec start, end;
+    int n, m, pn, pm, found = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   n = 2;
+    n = 2;
 
-   /* Check all couples of pentagonal numbers until the right one
-    * is found. Use the function implemented in projecteuler.c to
-    * check if the sum and difference ot the two numbers is pentagonal.*/
-   while(!found)
-   {
-      pn = n * (3 * n - 1) / 2;
+    /* Check all couples of pentagonal numbers until the right one
+     * is found. Use the function implemented in projecteuler.c to
+     * check if the sum and difference ot the two numbers is pentagonal.*/
+    while(!found)
+    {
+        pn = n * (3 * n - 1) / 2;
 
-      for(m = 1; m < n; m++)
-      {
-         pm = m * (3 * m - 1) / 2;
+        for(m = 1; m < n; m++)
+        {
+            pm = m * (3 * m - 1) / 2;
 
-         if(is_pentagonal(pn+pm) && is_pentagonal(pn-pm))
-         {
-            found = 1;
-            break;
-         }
-      }
-      n++;
-   }
+            if(is_pentagonal(pn+pm) && is_pentagonal(pn-pm))
+            {
+                found = 1;
+                break;
+            }
+        }
+        n++;
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 44\n");
-   printf("Answer: %d\n", pn-pm);
+    printf("Project Euler, Problem 44\n");
+    printf("Answer: %d\n", pn-pm);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p045.c

@@ -8,6 +8,8 @@
  *
  * Find the next triangle number that is also pentagonal and hexagonal.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -16,37 +18,37 @@
 
 int main(int argc, char **argv)
 {
-   int found = 0;
-   long int i, n;
-   double elapsed;
-   struct timespec start, end;
+    int found = 0;
+    long int i, n;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   i = 143;
+    i = 143;
 
-   while(!found)
-   {
-      i++;
-      /* Hexagonal numbers are also triangle numbers, so it's sufficient
-       * to generate hexagonal numbers (starting from H_144) and check if
-       * they're also pentagonal.*/
-      n = i * (2 * i - 1);
-      
-      if(is_pentagonal(n))
-      {
-         found = 1;
-      }
-   }
+    while(!found)
+    {
+        i++;
+        /* Hexagonal numbers are also triangle numbers, so it's sufficient
+         * to generate hexagonal numbers (starting from H_144) and check if
+         * they're also pentagonal.*/
+        n = i * (2 * i - 1);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+        if(is_pentagonal(n))
+        {
+            found = 1;
+        }
+    }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 45\n");
-   printf("Answer: %ld\n", n);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 45\n");
+    printf("Answer: %ld\n", n);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }

C/p046.c

@@ -11,6 +11,8 @@
  *
  * What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -25,73 +27,73 @@ int *primes;
 
 int main(int argc, char **argv)
 {
-   int i, found = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i, found = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((primes = sieve(N)) == NULL)
-   {
-      fprintf(stderr, "Error! Sieve function returned NULL\n");
-      return 1;
-   }
+    if((primes = sieve(N)) == NULL)
+    {
+        fprintf(stderr, "Error! Sieve function returned NULL\n");
+        return 1;
+    }
 
-   /* For every odd number, check if it's prime, if it is check
-    * if it satisfies the Goldbach property. Continue until the 
-    * first number that doesn't is found.*/
-   for(i = 3; !found && i < N; i += 2)
-   {
-      if(!primes[i])
-      {
-         if(!goldbach(i))
-         {
-            found = 1;
-         }
-      }
-   }
+    /* For every odd number, check if it's prime, if it is check
+     * if it satisfies the Goldbach property. Continue until the
+     * first number that doesn't is found.*/
+    for(i = 3; !found && i < N; i += 2)
+    {
+        if(!primes[i])
+        {
+            if(!goldbach(i))
+            {
+                found = 1;
+            }
+        }
+    }
 
-   free(primes);
-   
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    free(primes);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 46\n");
-   printf("Answer: %d\n", i-2);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 46\n");
+    printf("Answer: %d\n", i-2);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 int goldbach(int n)
 {
-   int i, j, tmp;
+    int i, j, tmp;
 
-   /* Check every prime smaller than n.*/
-   for(i = 3; i < n; i += 2)
-   {
-      if(primes[i])
-      {
-         j = 1;
+    /* Check every prime smaller than n.*/
+    for(i = 3; i < n; i += 2)
+    {
+        if(primes[i])
+        {
+            j = 1;
 
-         /* Check if summing twice a square to the prime number
-          * gives n. Return 1 when succeeding.*/
-         do
-         {
-            tmp = i + 2 * j * j;
-
-            if(tmp == n)
+            /* Check if summing twice a square to the prime number
+             * gives n. Return 1 when succeeding.*/
+            do
             {
-               return 1;
-            }
+                tmp = i + 2 * j * j;
 
-            j++;
-         }while(tmp < n);
-      }
-   }
+                if(tmp == n)
+                {
+                    return 1;
+                }
 
-   /* Return 0 if no solution is found.*/
-   return 0;
+                j++;
+            } while(tmp < n);
+        }
+    }
+
+    /* Return 0 if no solution is found.*/
+    return 0;
 }

C/p047.c

@@ -11,6 +11,8 @@
  *
  * Find the first four consecutive integers to have four distinct prime factors each. What is the first of these numbers?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -22,74 +24,74 @@ int *count_factors(int n);
 
 int main(int argc, char **argv)
 {
-   int i, count, res;
-   int *factors;
-   double elapsed;
-   struct timespec start, end;
+    int i, count, res;
+    int *factors;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((factors = count_factors(N)) == NULL)
-   {
-      fprintf(stderr, "Error! Count_factors function returned NULL\n");
-      return 1;
-   }
+    if((factors = count_factors(N)) == NULL)
+    {
+        fprintf(stderr, "Error! Count_factors function returned NULL\n");
+        return 1;
+    }
 
-   count = 0;
+    count = 0;
 
-   for(i = 0; i < N; i++)
-   {
-      if(factors[i] == 4)
-      {
-         count++;
-      }
-      else
-      {
-         count = 0;
-      }
+    for(i = 0; i < N; i++)
+    {
+        if(factors[i] == 4)
+        {
+            count++;
+        }
+        else
+        {
+            count = 0;
+        }
 
-      if(count == 4)
-      {
-         res = i - 3;
-         break;
-      }
-   }
+        if(count == 4)
+        {
+            res = i - 3;
+            break;
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 47\n");
-   printf("Answer: %d\n", res);
+    printf("Project Euler, Problem 47\n");
+    printf("Answer: %d\n", res);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 /* Function using a modified sieve of Eratosthenes to count
  * the distinct prime factors of each number.*/
 int *count_factors(int n)
 {
-   int i = 2, j;
-   int *factors;
+    int i = 2, j;
+    int *factors;
 
-   if((factors = (int *)calloc(n, sizeof(int))) == NULL)
-   {
-      return NULL;
-   }
+    if((factors = (int *)calloc(n, sizeof(int))) == NULL)
+    {
+        return NULL;
+    }
 
-   while(i < n / 2)
-   {
-      if(factors[i] == 0)
-      {
-         for(j = i; j < n; j += i)
-         {
-            factors[j]++;
-         }
-      }
-      i++;
-   }
+    while(i < n / 2)
+    {
+        if(factors[i] == 0)
+        {
+            for(j = i; j < n; j += i)
+            {
+                factors[j]++;
+            }
+        }
+        i++;
+    }
 
-   return factors;
+    return factors;
 }

C/p048.c

@@ -8,38 +8,40 @@
 #include <time.h>
 #include <gmp.h>
 
+#define _POSIX_C_SOURCE 199309L
+
 int main(int argc, char **argv)
 {
-   int i;
-   double elapsed;
-   struct timespec start, end;
-   mpz_t power, sum;
-   char *res;
+    int i;
+    double elapsed;
+    struct timespec start, end;
+    mpz_t power, sum;
+    char *res;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   mpz_init_set_ui(sum, 0);
-   mpz_init(power);
+    mpz_init_set_ui(sum, 0);
+    mpz_init(power);
 
-   /* Simply calculate the sum of the powers using the GMP Library.*/
-   for(i = 1; i <= 1000; i++)
-   {
-      mpz_ui_pow_ui(power, i, i);
-      mpz_add(sum, sum, power);
-   }
+    /* Simply calculate the sum of the powers using the GMP Library.*/
+    for(i = 1; i <= 1000; i++)
+    {
+        mpz_ui_pow_ui(power, i, i);
+        mpz_add(sum, sum, power);
+    }
 
-   res = mpz_get_str(NULL, 10, sum);
-   
-   mpz_clears(power, sum, NULL);
+    res = mpz_get_str(NULL, 10, sum);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    mpz_clears(power, sum, NULL);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 48\n");
-   printf("Answer: %s\n", res+strlen(res)-10);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 48\n");
+    printf("Answer: %s\n", res+strlen(res)-10);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }

C/p049.c

@@ -6,6 +6,8 @@
  *
  * What 12-digit number do you form by concatenating the three terms in this sequence?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -20,86 +22,86 @@ int *primes;
 
 int main(int argc, char **argv)
 {
-   int i = 1489, j, found = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i = 1489, j, found = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((primes = sieve(N)) == NULL)
-   {
-      fprintf(stderr, "Error! Sieve function returned NULL\n");
-      return 1;
-   }
+    if((primes = sieve(N)) == NULL)
+    {
+        fprintf(stderr, "Error! Sieve function returned NULL\n");
+        return 1;
+    }
 
-   /* Starting from i=1489 (bigger than the first number in the sequence given in the problem),
-    * check odd numbers. If they're prime, loop on even numbers j (odd+even=odd, odd+odd=even and
-    * we need odd numbers because we're looking for primes) up to 4254 (1489+2*4256=10001 which has
-    * 5 digits.*/
-   while(i < N)
-   {
-      if(primes[i])
-      {
-         for(j = 2; j < 4255; j += 2)
-         {
-            /* If i, i+j and i+2*j are all primes and they have
-             * all the same digits, the result has been found.*/
-            if(i + 2 * j < N && primes[i+j] && primes[i+2*j] &&
-                  is_permutation(i, i+j) && is_permutation(i, i+2*j))
+    /* Starting from i=1489 (bigger than the first number in the sequence given in the problem),
+     * check odd numbers. If they're prime, loop on even numbers j (odd+even=odd, odd+odd=even and
+     * we need odd numbers because we're looking for primes) up to 4254 (1489+2*4256=10001 which has
+     * 5 digits.*/
+    while(i < N)
+    {
+        if(primes[i])
+        {
+            for(j = 2; j < 4255; j += 2)
             {
-               found = 1;
-               break;
+                /* If i, i+j and i+2*j are all primes and they have
+                 * all the same digits, the result has been found.*/
+                if(i + 2 * j < N && primes[i+j] && primes[i+2*j] &&
+                        is_permutation(i, i+j) && is_permutation(i, i+2*j))
+                {
+                    found = 1;
+                    break;
+                }
             }
-         }
-      }
-      if(found)
-      {
-         break;
-      }
-      i += 2;
-   }
+        }
+        if(found)
+        {
+            break;
+        }
+        i += 2;
+    }
 
-   free(primes);
+    free(primes);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 49\n");
-   printf("Answer: %d%d%d\n", i, i+j, i+2*j);
+    printf("Project Euler, Problem 49\n");
+    printf("Answer: %d%d%d\n", i, i+j, i+2*j);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int is_permutation(int a, int b)
 {
-   int i;
-   int digits1[10] = {0}, digits2[10] = {0};
+    int i;
+    int digits1[10] = {0}, digits2[10] = {0};
 
-   /* Get digits of a.*/
-   while(a > 0)
-   {
-      digits1[a%10]++;
-      a /= 10;
-   }
+    /* Get digits of a.*/
+    while(a > 0)
+    {
+        digits1[a%10]++;
+        a /= 10;
+    }
 
-   /* Get digits of b.*/
-   while(b > 0)
-   {
-      digits2[b%10]++;
-      b /= 10;
-   }
+    /* Get digits of b.*/
+    while(b > 0)
+    {
+        digits2[b%10]++;
+        b /= 10;
+    }
 
-   /* If they're not the same, return 0.*/
-   for(i = 0; i < 10; i++)
-   {
-      if(digits1[i] != digits2[i])
-      {
-         return 0;
-      }
-   }
+    /* If they're not the same, return 0.*/
+    for(i = 0; i < 10; i++)
+    {
+        if(digits1[i] != digits2[i])
+        {
+            return 0;
+        }
+    }
 
-   return 1;
+    return 1;
 }

C/p050.c

@@ -8,6 +8,8 @@
  *
  * Which prime, below one-million, can be written as the sum of the most consecutive primes?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -18,78 +20,78 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, max = 0, max_p = 0, sum, count;
-   int *primes;
-   double elapsed;
-   struct timespec start, end;
+    int i, j, max = 0, max_p = 0, sum, count;
+    int *primes;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((primes = sieve(N)) == NULL)
-   {
-      fprintf(stderr, "Error! Sieve function returned NULL\n");
-      return 1;
-   }
+    if((primes = sieve(N)) == NULL)
+    {
+        fprintf(stderr, "Error! Sieve function returned NULL\n");
+        return 1;
+    }
 
-   /* Starting from a prime i, add consecutive primes until the 
-    * sum exceeds the limit, every time the sum is also a prime
-    * save the value and the count if the count is larger than the
-    * current maximum. Repeat for all primes below N.
-    * A separate loop is used for i=2, so later only odd numbers are 
-    * checked for primality.*/
-   i = 2;
-   count = 1;
-   sum = i;
+    /* Starting from a prime i, add consecutive primes until the
+     * sum exceeds the limit, every time the sum is also a prime
+     * save the value and the count if the count is larger than the
+     * current maximum. Repeat for all primes below N.
+     * A separate loop is used for i=2, so later only odd numbers are
+     * checked for primality.*/
+    i = 2;
+    count = 1;
+    sum = i;
 
-   for(j = i + 1; j < N && sum < N; j++)
-   {
-      if(primes[j])
-      {
-         sum += j;
-         count++;
+    for(j = i + 1; j < N && sum < N; j++)
+    {
+        if(primes[j])
+        {
+            sum += j;
+            count++;
 
-         if(sum < N && primes[sum] && count > max)
-         {
-            max = count;
-            max_p = sum;
-         }
-      }
-   }
-
-   for(i = 3; i < N; i += 2)
-   {
-      if(primes[i])
-      {
-         count = 1;
-         sum = i;
-
-         for(j = i + 2; j < N && sum < N; j += 2)
-         {
-            if(primes[j])
+            if(sum < N && primes[sum] && count > max)
             {
-               sum += j;
-               count++;
-
-               if(sum < N && primes[sum] && count > max)
-               {
-                  max = count;
-                  max_p = sum;
-               }
+                max = count;
+                max_p = sum;
             }
-         }
-      }
-   }
+        }
+    }
 
-   free(primes);
+    for(i = 3; i < N; i += 2)
+    {
+        if(primes[i])
+        {
+            count = 1;
+            sum = i;
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+            for(j = i + 2; j < N && sum < N; j += 2)
+            {
+                if(primes[j])
+                {
+                    sum += j;
+                    count++;
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+                    if(sum < N && primes[sum] && count > max)
+                    {
+                        max = count;
+                        max_p = sum;
+                    }
+                }
+            }
+        }
+    }
 
-   printf("Project Euler, Problem 50\n");
-   printf("Answer: %d\n", max_p);
+    free(primes);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   return 0;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+
+    printf("Project Euler, Problem 50\n");
+    printf("Answer: %d\n", max_p);
+
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }

C/p051.c

@@ -7,6 +7,8 @@
  * Find the smallest prime which, by replacing part of the number (not necessarily adjacent digits) with the same digit, is part of an eight prime
  * value family.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
@@ -23,119 +25,119 @@ int *primes;
 
 int main(int argc, char **argv)
 {
-   int i;
-   double elapsed;
-   struct timespec start, end;
+    int i;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* N set to 1000000 as a reasonable limit, which turns out to be enough.*/
-   if((primes = sieve(N)) == NULL)
-   {
-      fprintf(stderr, "Error! Sieve function returned NULL\n");
-      return 1;
-   }
+    /* N set to 1000000 as a reasonable limit, which turns out to be enough.*/
+    if((primes = sieve(N)) == NULL)
+    {
+        fprintf(stderr, "Error! Sieve function returned NULL\n");
+        return 1;
+    }
 
-   /* Checking only odd numbers with at least 4 digits.*/
-   for(i = 1001; i < N; i += 2)
-   {
-      /* The family of numbers needs to have at least one of 0, 1 or 2 as
-       * repeated digits, otherwise we can't get a 8 number family (there 
-       * are only 7 other digits). Also, the number of repeated digits must
-       * be 3, otherwise at least 3 resulting numbers will be divisible by 3.
-       * So the smallest number of this family must have three 0s, three 1s or
-       * three 2s.*/
-      if(count_digit(i, 0) >= 3 || count_digit(i, 1) >= 3 ||
-            count_digit(i, 2) >= 3)
-      {
-      /* If i is prime, try replacing digits, if obtaining 8 primes, then 
-       * this is the result.*/
-         if(primes[i] && replace(i) == 8)
-         {
-            break;
-         }
-      }
-   }
+    /* Checking only odd numbers with at least 4 digits.*/
+    for(i = 1001; i < N; i += 2)
+    {
+        /* The family of numbers needs to have at least one of 0, 1 or 2 as
+         * repeated digits, otherwise we can't get a 8 number family (there
+         * are only 7 other digits). Also, the number of repeated digits must
+         * be 3, otherwise at least 3 resulting numbers will be divisible by 3.
+         * So the smallest number of this family must have three 0s, three 1s or
+         * three 2s.*/
+        if(count_digit(i, 0) >= 3 || count_digit(i, 1) >= 3 ||
+                count_digit(i, 2) >= 3)
+        {
+            /* If i is prime, try replacing digits, if obtaining 8 primes, then
+             * this is the result.*/
+            if(primes[i] && replace(i) == 8)
+            {
+                break;
+            }
+        }
+    }
 
-   free(primes);
+    free(primes);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 51\n");
-   printf("Answer: %d\n", i);
+    printf("Project Euler, Problem 51\n");
+    printf("Answer: %d\n", i);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int count_digit(int n, int d)
 {
-   int count = 0;
+    int count = 0;
 
-   while(n > 0)
-   {
-      if(n % 10 == d)
-      {
-         count++;
-      }
-      n /= 10;
-   }
+    while(n > 0)
+    {
+        if(n % 10 == d)
+        {
+            count++;
+        }
+        n /= 10;
+    }
 
-   return count;
+    return count;
 }
 
 int replace(int n)
 {
-   int i, j, k, w, l, count, max = 0, s_to_n;
-   char n_to_s[7];
+    int i, j, k, w, l, count, max = 0, s_to_n;
+    char n_to_s[7];
 
-   sprintf(n_to_s, "%d", n);
-   l = strlen(n_to_s);
+    sprintf(n_to_s, "%d", n);
+    l = strlen(n_to_s);
 
-   for(i = 0; i < l - 3; i++)
-   {
-      for(j = i + 1; j < l - 2; j++)
-      {
-         /* Replacing the last digit can't give 8 primes, because at least
-          * six of the numbers obtained will be divisible by 2 and/or 5.*/
-         for(k = j + 1; k < l - 1; k++)
-         {
-            count = 0;
-
-            for(w = 0; w < 10; w++)
+    for(i = 0; i < l - 3; i++)
+    {
+        for(j = i + 1; j < l - 2; j++)
+        {
+            /* Replacing the last digit can't give 8 primes, because at least
+             * six of the numbers obtained will be divisible by 2 and/or 5.*/
+            for(k = j + 1; k < l - 1; k++)
             {
-               if(i == 0 && w == 0)
-               {
-                  continue;
-               }
+                count = 0;
 
-               sprintf(n_to_s, "%d", n);
-               n_to_s[i] = w + '0';
-               n_to_s[j] = w + '0';
-               n_to_s[k] = w + '0';
-               s_to_n = atoi(n_to_s);
+                for(w = 0; w < 10; w++)
+                {
+                    if(i == 0 && w == 0)
+                    {
+                        continue;
+                    }
 
-               if(primes[s_to_n])
-               {
-                  if(count == 0 && s_to_n != n)
-                  {
-                     continue;
-                  }
+                    sprintf(n_to_s, "%d", n);
+                    n_to_s[i] = w + '0';
+                    n_to_s[j] = w + '0';
+                    n_to_s[k] = w + '0';
+                    s_to_n = atoi(n_to_s);
 
-                  count++;
-               }
+                    if(primes[s_to_n])
+                    {
+                        if(count == 0 && s_to_n != n)
+                        {
+                            continue;
+                        }
+
+                        count++;
+                    }
+                }
+
+                if(count > max)
+                {
+                    max = count;
+                }
             }
+        }
+    }
 
-            if(count > max)
-            {
-               max = count;
-            }
-         }
-      }
-   }
-
-   return max;
+    return max;
 }

C/p052.c

@@ -2,6 +2,8 @@
  *
  * Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -10,65 +12,65 @@ int is_permutation(int a, int b);
 
 int main(int argc, char **argv)
 {
-   int i;
-   double elapsed;
-   struct timespec start, end;
+    int i;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   i = 1;
+    i = 1;
 
-   /* Brute force approach, try every integer number until the desired one is found.*/
-   while(1)
-   {
-      if(is_permutation(i, 2*i) && is_permutation(i, 3*i) && is_permutation(i, 4*i) &&
-            is_permutation(i, 5*i) && is_permutation(i, 6*i))
-      {
-         break;
-      }
+    /* Brute force approach, try every integer number until the desired one is found.*/
+    while(1)
+    {
+        if(is_permutation(i, 2*i) && is_permutation(i, 3*i) && is_permutation(i, 4*i) &&
+                is_permutation(i, 5*i) && is_permutation(i, 6*i))
+        {
+            break;
+        }
 
-      i++;
-   }
+        i++;
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 52\n");
-   printf("Answer: %d\n", i);
+    printf("Project Euler, Problem 52\n");
+    printf("Answer: %d\n", i);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int is_permutation(int a, int b)
 {
-   int i;
-   int digits1[10] = {0}, digits2[10] = {0};
+    int i;
+    int digits1[10] = {0}, digits2[10] = {0};
 
-   /* Get digits of a.*/
-   while(a > 0)
-   {
-      digits1[a%10]++;
-      a /= 10;
-   }
+    /* Get digits of a.*/
+    while(a > 0)
+    {
+        digits1[a%10]++;
+        a /= 10;
+    }
 
-   /* Get digits of b.*/
-   while(b > 0)
-   {
-      digits2[b%10]++;
-      b /= 10;
-   }
+    /* Get digits of b.*/
+    while(b > 0)
+    {
+        digits2[b%10]++;
+        b /= 10;
+    }
 
-   /* If they're not the same, return 0.*/
-   for(i = 0; i < 10; i++)
-   {
-      if(digits1[i] != digits2[i])
-      {
-         return 0;
-      }
-   }
+    /* If they're not the same, return 0.*/
+    for(i = 0; i < 10; i++)
+    {
+        if(digits1[i] != digits2[i])
+        {
+            return 0;
+        }
+    }
 
-   return 1;
+    return 1;
 }

C/p053.c

@@ -10,6 +10,8 @@
  *
  * How many, not necessarily distinct, values of (n r) for 1≤n≤100, are greater than one-million?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -22,63 +24,63 @@ void binomial(mpz_t bin, unsigned long int n, unsigned long int k, mpz_t *factor
 
 int main(int argc, char **argv)
 {
-   int i, j, count = 0;
-   double elapsed;
-   struct timespec start, end;
-   mpz_t bin;
-   mpz_t factorials[N+1];
+    int i, j, count = 0;
+    double elapsed;
+    struct timespec start, end;
+    mpz_t bin;
+    mpz_t factorials[N+1];
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   mpz_init(bin);
+    mpz_init(bin);
 
-   mpz_init_set_ui(factorials[0], 1);
+    mpz_init_set_ui(factorials[0], 1);
 
-   /* Straightforward brute force approach, using the GMP Library. First
-    * calculate all factorials, then use them for the binomial coefficients.*/
-   for(i = 1; i <= N; i++)
-   {
-      mpz_init(factorials[i]);
-      mpz_fac_ui(factorials[i], i);
-   }
+    /* Straightforward brute force approach, using the GMP Library. First
+     * calculate all factorials, then use them for the binomial coefficients.*/
+    for(i = 1; i <= N; i++)
+    {
+        mpz_init(factorials[i]);
+        mpz_fac_ui(factorials[i], i);
+    }
 
-   for(i = 23; i <= N; i++)
-   {
-      for(j = 1; j <= i; j++)
-      {
-         binomial(bin, i, j, factorials);
+    for(i = 23; i <= N; i++)
+    {
+        for(j = 1; j <= i; j++)
+        {
+            binomial(bin, i, j, factorials);
 
-         if(mpz_cmp_ui(bin, LIMIT) > 0)
-            count++;
-      }
-   }
+            if(mpz_cmp_ui(bin, LIMIT) > 0)
+                count++;
+        }
+    }
 
-   mpz_clear(bin);
+    mpz_clear(bin);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 53\n");
-   printf("Answer: %d\n", count);
+    printf("Project Euler, Problem 53\n");
+    printf("Answer: %d\n", count);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 void binomial(mpz_t bin, unsigned long n, unsigned long int k, mpz_t *factorials)
 {
-   mpz_t num, d1, d2; 
+    mpz_t num, d1, d2;
 
-   mpz_inits(num, d1, d2, NULL);
+    mpz_inits(num, d1, d2, NULL);
 
-   mpz_set(num, factorials[n]);
-   mpz_set(d1, factorials[k]);
-   mpz_set(d2, factorials[n-k]);
+    mpz_set(num, factorials[n]);
+    mpz_set(d1, factorials[k]);
+    mpz_set(d2, factorials[n-k]);
 
-   mpz_mul(d1, d1, d2);
-   mpz_tdiv_q(bin, num, d1);
+    mpz_mul(d1, d1, d2);
+    mpz_tdiv_q(bin, num, d1);
 
-   mpz_clears(num, d1, d2, NULL);
+    mpz_clears(num, d1, d2, NULL);
 }

C/p055.c

@@ -9,8 +9,8 @@
  * That is, 349 took three iterations to arrive at a palindrome.
  *
  * Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome
- * through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, 
- * we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either 
+ * through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem,
+ * we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either
  * (i) become a palindrome in less than fifty iterations, or,
  * (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown
  * to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).
@@ -21,6 +21,8 @@
  *
  * NOTE: Wording was modified slightly on 24 April 2007 to emphasise the theoretical nature of Lychrel numbers.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -31,73 +33,73 @@ int is_lychrel(mpz_t n);
 
 int main(int argc, char **argv)
 {
-   int i, count = 0;
-   double elapsed;
-   struct timespec start, end;
-   mpz_t n;
+    int i, count = 0;
+    double elapsed;
+    struct timespec start, end;
+    mpz_t n;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   mpz_init(n);
+    mpz_init(n);
 
-   /* For each number, use the is_lychrel function to check if the number
-    * is a Lychrel number.*/
-   for(i = 1; i < 10000; i++)
-   {
-      mpz_set_ui(n, i);
+    /* For each number, use the is_lychrel function to check if the number
+     * is a Lychrel number.*/
+    for(i = 1; i < 10000; i++)
+    {
+        mpz_set_ui(n, i);
 
-      if(is_lychrel(n))
-         count++;
-   }
+        if(is_lychrel(n))
+            count++;
+    }
 
-   mpz_clear(n);
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    mpz_clear(n);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 55\n");
-   printf("Answer: %d\n", count);
+    printf("Project Euler, Problem 55\n");
+    printf("Answer: %d\n", count);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int is_lychrel(mpz_t n)
 {
-   int i;
-   mpz_t tmp, reverse, rem;
+    int i;
+    mpz_t tmp, reverse, rem;
 
-   mpz_inits(tmp, reverse, rem, NULL);
-   mpz_set(tmp, n);
+    mpz_inits(tmp, reverse, rem, NULL);
+    mpz_set(tmp, n);
 
-   /* Run for 50 iterations.*/
-   for(i = 0; i < 50; i++)
-   {
-      mpz_set_ui(reverse, 0);
+    /* Run for 50 iterations.*/
+    for(i = 0; i < 50; i++)
+    {
+        mpz_set_ui(reverse, 0);
 
-      /* Find the reverse of the given number.*/ 
-      while(mpz_cmp_ui(tmp, 0) > 0)
-      {
-         mpz_mul_ui(reverse, reverse, 10);
-         mpz_tdiv_qr_ui(tmp, rem, tmp, 10);
-         mpz_add(reverse, reverse, rem);
-      }
+        /* Find the reverse of the given number.*/
+        while(mpz_cmp_ui(tmp, 0) > 0)
+        {
+            mpz_mul_ui(reverse, reverse, 10);
+            mpz_tdiv_qr_ui(tmp, rem, tmp, 10);
+            mpz_add(reverse, reverse, rem);
+        }
 
-      /* Add the reverse to the original number.*/
-      mpz_add(tmp, n, reverse);
+        /* Add the reverse to the original number.*/
+        mpz_add(tmp, n, reverse);
 
-      /* If the sum is a palindrome, the number is not a Lychrel number.*/
-      if(is_palindrome_mpz(tmp, 10))
-      {
-         mpz_clears(tmp, reverse, rem, NULL);
-         return 0;
-      }
+        /* If the sum is a palindrome, the number is not a Lychrel number.*/
+        if(is_palindrome_mpz(tmp, 10))
+        {
+            mpz_clears(tmp, reverse, rem, NULL);
+            return 0;
+        }
 
-      mpz_set(n, tmp);
-   }
+        mpz_set(n, tmp);
+    }
 
-   mpz_clears(tmp, reverse, rem, NULL);
+    mpz_clears(tmp, reverse, rem, NULL);
 
-   return 1;
+    return 1;
 }

C/p056.c

@@ -3,6 +3,8 @@
  *
  * Considering natural numbers of the form, a^b, where a, b < 100, what is the maximum digital sum?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -10,41 +12,41 @@
 
 int main(int argc, char **argv)
 {
-   int a, b, sum, max = 0;
-   double elapsed;
-   struct timespec start, end;
-   mpz_t pow;
+    int a, b, sum, max = 0;
+    double elapsed;
+    struct timespec start, end;
+    mpz_t pow;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   mpz_init(pow);
+    mpz_init(pow);
 
-   /* Straightforward brute force approach using the GMP Library.*/
-   for(a = 1; a < 100; a++)
-   {
-      for(b = 1; b < 100; b++)
-      {
-         mpz_ui_pow_ui(pow, a, b);
-         sum = 0;
+    /* Straightforward brute force approach using the GMP Library.*/
+    for(a = 1; a < 100; a++)
+    {
+        for(b = 1; b < 100; b++)
+        {
+            mpz_ui_pow_ui(pow, a, b);
+            sum = 0;
 
-         while(mpz_cmp_ui(pow, 0))
-            sum += mpz_tdiv_q_ui(pow, pow, 10);
+            while(mpz_cmp_ui(pow, 0))
+                sum += mpz_tdiv_q_ui(pow, pow, 10);
 
-         if(sum > max)
-            max = sum;
-      }
-   }
+            if(sum > max)
+                max = sum;
+        }
+    }
 
-   mpz_clear(pow);
+    mpz_clear(pow);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 56\n");
-   printf("Answer: %d\n", max);
+    printf("Project Euler, Problem 56\n");
+    printf("Answer: %d\n", max);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p057.c

@@ -14,6 +14,8 @@
  *
  * In the first one-thousand expansions, how many fractions contain a numerator with more digits than the denominator?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -23,64 +25,64 @@ int count_digits(mpz_t n);
 
 int main(int argc, char **argv)
 {
-   int i, count = 0;
-   mpz_t n, d, d2;
-   double elapsed;
-   struct timespec start, end;
+    int i, count = 0;
+    mpz_t n, d, d2;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   mpz_init_set_ui(n, 1);
-   mpz_init_set_ui(d, 1);
-   mpz_init(d2);
+    mpz_init_set_ui(n, 1);
+    mpz_init_set_ui(d, 1);
+    mpz_init(d2);
 
-   /* If n/d is the current term of the expansion, the next term can be calculated as
-    * (n+2d)/(n+d). Using the GMP Library the problem becomes trivial.*/
-   for(i = 1; i < 1000; i++)
-   {
-      mpz_mul_ui(d2, d, 2);
-      mpz_add(d, n, d);
-      mpz_add(n, n, d2);
+    /* If n/d is the current term of the expansion, the next term can be calculated as
+     * (n+2d)/(n+d). Using the GMP Library the problem becomes trivial.*/
+    for(i = 1; i < 1000; i++)
+    {
+        mpz_mul_ui(d2, d, 2);
+        mpz_add(d, n, d);
+        mpz_add(n, n, d2);
 
-      if(count_digits(n) > count_digits(d))
-      {
-         count++;
-      }
-   }
+        if(count_digits(n) > count_digits(d))
+        {
+            count++;
+        }
+    }
 
-   mpz_clears(n, d, d2, NULL);
+    mpz_clears(n, d, d2, NULL);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 57\n");
-   printf("Answer: %d\n", count);
+    printf("Project Euler, Problem 57\n");
+    printf("Answer: %d\n", count);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int count_digits(mpz_t n)
 {
-   int count = 0;
-   mpz_t value;
+    int count = 0;
+    mpz_t value;
 
-   if(mpz_cmp_ui(n, 0) == 0)
-   {
-      return 1;
-   }
+    if(mpz_cmp_ui(n, 0) == 0)
+    {
+        return 1;
+    }
 
-   mpz_init_set(value, n);
+    mpz_init_set(value, n);
 
-   while(mpz_cmp_ui(value, 0))
-   {
-      mpz_tdiv_q_ui(value, value, 10);
-      count++;
-   }
+    while(mpz_cmp_ui(value, 0))
+    {
+        mpz_tdiv_q_ui(value, value, 10);
+        count++;
+    }
 
-   mpz_clear(value);
+    mpz_clear(value);
 
-   return count;
+    return count;
 }

C/p058.c

@@ -14,6 +14,8 @@
  * If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued,
  * what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -22,57 +24,57 @@
 
 int main(int argc, char **argv)
 {
-   int i = 1, l = 1, step = 2, count = 0, diag = 5;
-   double ratio, elapsed;
-   struct timespec start, end;
+    int i = 1, l = 1, step = 2, count = 0, diag = 5;
+    double ratio, elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Starting with 1, the next four numbers in the diagonal are 3 (1+2), 5 (1+2+2), 7 (1+2+2+2)
-    * and 9 (1+2+2+2+2). Check which are prime, increment the counter every time a new prime is 
-    * found, and divide by the number of elements of the diagonal, which are increase by 4 at 
-    * every cycle. The next four number added to the diagonal are 13 (9+4), 17 (9+4+4), 21 and 25.
-    * Then 25+6 etc., at every cycle the step is increased by 2. Continue until the ratio goes below 0.1.*/
-   do
-   {
-      i += step;
+    /* Starting with 1, the next four numbers in the diagonal are 3 (1+2), 5 (1+2+2), 7 (1+2+2+2)
+     * and 9 (1+2+2+2+2). Check which are prime, increment the counter every time a new prime is
+     * found, and divide by the number of elements of the diagonal, which are increase by 4 at
+     * every cycle. The next four number added to the diagonal are 13 (9+4), 17 (9+4+4), 21 and 25.
+     * Then 25+6 etc., at every cycle the step is increased by 2. Continue until the ratio goes below 0.1.*/
+    do
+    {
+        i += step;
 
-      if(is_prime(i))
-      {
-         count++;
-      }
+        if(is_prime(i))
+        {
+            count++;
+        }
 
-      i += step;
+        i += step;
 
-      if(is_prime(i))
-      {
-         count++;
-      }
+        if(is_prime(i))
+        {
+            count++;
+        }
 
-      i += step;
+        i += step;
 
-      if(is_prime(i))
-      {
-         count++;
-      }
+        if(is_prime(i))
+        {
+            count++;
+        }
 
-      i += step;
-      ratio = (double)count / diag;
+        i += step;
+        ratio = (double)count / diag;
 
-      step += 2;
-      diag += 4;
-      l += 2;
+        step += 2;
+        diag += 4;
+        l += 2;
 
-   }while(ratio >= 0.1);
+    } while(ratio >= 0.1);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 58\n");
-   printf("Answer: %d\n", l);
+    printf("Project Euler, Problem 58\n");
+    printf("Answer: %d\n", l);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p059.c

@@ -16,6 +16,8 @@
  * encrypted ASCII codes, and the knowledge that the plain text must contain common English words, decrypt the message and find the sum of the
  * ASCII values in the original text.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>

C/p060.c

@@ -4,6 +4,8 @@
  *
  * Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -18,97 +20,97 @@ int *primes;
 
 int main(int argc, char **argv)
 {
-   int found = 0, p1, p2, p3, p4, p5, n;
-   double elapsed;
-   struct timespec start, end;
+    int found = 0, p1, p2, p3, p4, p5, n;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((primes = sieve(N)) == NULL)
-   {
-      fprintf(stderr, "Error! Sieve function returned NULL\n");
-      return 1;
-   }
+    if((primes = sieve(N)) == NULL)
+    {
+        fprintf(stderr, "Error! Sieve function returned NULL\n");
+        return 1;
+    }
 
-   /* Straightforward brute force approach.*/
-   for(p1 = 3; p1 < N && !found; p1 += 2)
-   {
-      /* If p1 is not prime, go to the next number.*/
-      if(!primes[p1])
-      {
-         continue;
-      }
-
-      for(p2 = p1 + 2; p2 < N && !found; p2 += 2)
-      {
-         /* If p2 is not prime, or at least one of the possible concatenations of
-          * p1 and p2 is not prime, go to the next number.*/
-         if(!primes[p2] || !is_prime(cat(p1, p2)) || !is_prime(cat(p2, p1)))
-         {
+    /* Straightforward brute force approach.*/
+    for(p1 = 3; p1 < N && !found; p1 += 2)
+    {
+        /* If p1 is not prime, go to the next number.*/
+        if(!primes[p1])
+        {
             continue;
-         }
+        }
 
-         for(p3 = p2 + 2; p3 < N && !found; p3 += 2)
-         {
-            /* If p3 is not prime, or at least one of the possible concatenations of
-            * p1, p2 and p3 is not prime, go to the next number.*/
-            if(!primes[p3] || !is_prime(cat(p1, p3)) || !is_prime(cat(p3, p1)) ||
-                  !is_prime(cat(p2, p3)) || !is_prime(cat(p3, p2)))
+        for(p2 = p1 + 2; p2 < N && !found; p2 += 2)
+        {
+            /* If p2 is not prime, or at least one of the possible concatenations of
+             * p1 and p2 is not prime, go to the next number.*/
+            if(!primes[p2] || !is_prime(cat(p1, p2)) || !is_prime(cat(p2, p1)))
             {
-               continue;
+                continue;
             }
- 
-            for(p4 = p3 + 2; p4 < N && !found; p4 += 2)
+
+            for(p3 = p2 + 2; p3 < N && !found; p3 += 2)
             {
-               /* If p4 is not prime, or at least one of the possible concatenations of
-               * p1, p2, p3 and p4 is not prime, go to the next number.*/
-               if(!primes[p4] || !is_prime(cat(p1, p4)) || !is_prime(cat(p4, p1)) ||
-                     !is_prime(cat(p2, p4)) || !is_prime(cat(p4, p2)) || 
-                     !is_prime(cat(p3, p4)) || !is_prime(cat(p4, p3)))
-               {
-                  continue;
-               }
+                /* If p3 is not prime, or at least one of the possible concatenations of
+                * p1, p2 and p3 is not prime, go to the next number.*/
+                if(!primes[p3] || !is_prime(cat(p1, p3)) || !is_prime(cat(p3, p1)) ||
+                        !is_prime(cat(p2, p3)) || !is_prime(cat(p3, p2)))
+                {
+                    continue;
+                }
 
-               for(p5 = p4 + 2; p5 < N && !found; p5 += 2)
-               {
-                  /* If p5 is not prime, or at least one of the possible concatenations of
-                  * p1, p2, p3, p4 and p5 is not prime, go to the next number.*/
-                  if(!primes[p5] || !is_prime(cat(p1, p5)) || !is_prime(cat(p5, p1)) ||
-                        !is_prime(cat(p2, p5)) || !is_prime(cat(p5, p2)) ||
-                        !is_prime(cat(p3, p5)) || !is_prime(cat(p5, p3)) ||
-                        !is_prime(cat(p4, p5)) || !is_prime(cat(p5, p4)))
-                  {
-                     continue;
-                  }
+                for(p4 = p3 + 2; p4 < N && !found; p4 += 2)
+                {
+                    /* If p4 is not prime, or at least one of the possible concatenations of
+                    * p1, p2, p3 and p4 is not prime, go to the next number.*/
+                    if(!primes[p4] || !is_prime(cat(p1, p4)) || !is_prime(cat(p4, p1)) ||
+                            !is_prime(cat(p2, p4)) || !is_prime(cat(p4, p2)) ||
+                            !is_prime(cat(p3, p4)) || !is_prime(cat(p4, p3)))
+                    {
+                        continue;
+                    }
 
-                  /* If it gets here, the five values have been found.*/
-                  n = p1 + p2 + p3 + p4 + p5;
-                  found = 1;
-               }
+                    for(p5 = p4 + 2; p5 < N && !found; p5 += 2)
+                    {
+                        /* If p5 is not prime, or at least one of the possible concatenations of
+                        * p1, p2, p3, p4 and p5 is not prime, go to the next number.*/
+                        if(!primes[p5] || !is_prime(cat(p1, p5)) || !is_prime(cat(p5, p1)) ||
+                                !is_prime(cat(p2, p5)) || !is_prime(cat(p5, p2)) ||
+                                !is_prime(cat(p3, p5)) || !is_prime(cat(p5, p3)) ||
+                                !is_prime(cat(p4, p5)) || !is_prime(cat(p5, p4)))
+                        {
+                            continue;
+                        }
+
+                        /* If it gets here, the five values have been found.*/
+                        n = p1 + p2 + p3 + p4 + p5;
+                        found = 1;
+                    }
+                }
             }
-         }
-      }
-   }
+        }
+    }
 
-   free(primes);
+    free(primes);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
-   
-   elapsed=(end.tv_sec-start.tv_sec)+(double)(end.tv_nsec-start.tv_nsec)/1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 60\n");
-   printf("Answer: %d\n", n);
+    elapsed=(end.tv_sec-start.tv_sec)+(double)(end.tv_nsec-start.tv_nsec)/1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 60\n");
+    printf("Answer: %d\n", n);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 int cat(int i, int j)
 {
-   char n[10];
+    char n[10];
 
-   sprintf(n, "%d%d", i, j);
+    sprintf(n, "%d%d", i, j);
 
-   return atoi(n);
+    return atoi(n);
 }

C/p061.c

@@ -17,6 +17,8 @@
  * Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal,
  * hexagonal, heptagonal, and octagonal, is represented by a different number in the set.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -28,94 +30,94 @@ int chain[6], flags[6] = {0};
 
 int main(int argc, char **argv)
 {
-   int i, n, sum = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i, n, sum = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   i = 1;
-   n = 1;
+    i = 1;
+    n = 1;
 
-   /* Generate all triangle numbers smaller than 10000*/
-   do
-   {
-      polygonal[0][n] = 1;
-      i++;
-      n = i * (i + 1) / 2;
-   }while(n < 10000);
+    /* Generate all triangle numbers smaller than 10000*/
+    do
+    {
+        polygonal[0][n] = 1;
+        i++;
+        n = i * (i + 1) / 2;
+    } while(n < 10000);
 
-   i = 1;
-   n = 1;
+    i = 1;
+    n = 1;
 
-   /* Generate all square numbers smaller than 10000.*/
-   do
-   {
-      polygonal[1][n] = 1;
-      i++;
-      n = i * i;
-   }while(n < 10000);
+    /* Generate all square numbers smaller than 10000.*/
+    do
+    {
+        polygonal[1][n] = 1;
+        i++;
+        n = i * i;
+    } while(n < 10000);
 
-   i = 1;
-   n = 1;
+    i = 1;
+    n = 1;
 
-   /* Generate all pentagonal numbers smaller than 10000.*/
-   do
-   {
-      polygonal[2][n] = 1;
-      i++;
-      n = i * (3 * i - 1) / 2;
-   }while(n < 10000);
+    /* Generate all pentagonal numbers smaller than 10000.*/
+    do
+    {
+        polygonal[2][n] = 1;
+        i++;
+        n = i * (3 * i - 1) / 2;
+    } while(n < 10000);
 
-   i = 1;
-   n = 1;
+    i = 1;
+    n = 1;
 
-   /* Generate all hexagonal numbers smaller than 10000.*/
-   do
-   {
-      polygonal[3][n] = 1;
-      i++;
-      n = i * (2 * i - 1);
-   }while(n < 10000);
+    /* Generate all hexagonal numbers smaller than 10000.*/
+    do
+    {
+        polygonal[3][n] = 1;
+        i++;
+        n = i * (2 * i - 1);
+    } while(n < 10000);
 
-   i = 1;
-   n = 1;
+    i = 1;
+    n = 1;
 
-   /* Generate all heptagonal numbers smaller than 10000.*/
-   do
-   {
-      polygonal[4][n] = 1;
-      i++;
-      n = i * (5 * i - 3) / 2;
-   }while(n < 10000);
+    /* Generate all heptagonal numbers smaller than 10000.*/
+    do
+    {
+        polygonal[4][n] = 1;
+        i++;
+        n = i * (5 * i - 3) / 2;
+    } while(n < 10000);
 
-   i = 1;
-   n = 1;
+    i = 1;
+    n = 1;
 
-   /* Generate all octagonal numbers smaller than 10000.*/
-   do
-   {
-      polygonal[5][n] = 1;
-      i++;
-      n = i * (3 * i - 2);
-   }while(n < 10000);
+    /* Generate all octagonal numbers smaller than 10000.*/
+    do
+    {
+        polygonal[5][n] = 1;
+        i++;
+        n = i * (3 * i - 2);
+    } while(n < 10000);
 
-   /* Find the requested set of numbers.*/
-   if(find_set(0, &sum) == 0)
-   {
-      printf("Set not found\n");
-   }
+    /* Find the requested set of numbers.*/
+    if(find_set(0, &sum) == 0)
+    {
+        printf("Set not found\n");
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 61\n");
-   printf("Answer: %d\n", sum);
-   
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 61\n");
+    printf("Answer: %d\n", sum);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 /* Recursive function to find the required set. It finds a polygonal number,
@@ -124,72 +126,72 @@ int main(int argc, char **argv)
  * backtracking and tries the next polygonal number.*/
 int find_set(int step, int *sum)
 {
-   int i, j;
+    int i, j;
 
-   /* Use one polygonal number per type, starting from triangular.*/
-   for(i = 0; i < 6; i++)
-   {
-      /* If the current type has not been used yet, try it.*/
-      if(!flags[i])
-      {
-         /* Set a flag to record that the current polygonal type has been used.*/
-         flags[i] = 1;
+    /* Use one polygonal number per type, starting from triangular.*/
+    for(i = 0; i < 6; i++)
+    {
+        /* If the current type has not been used yet, try it.*/
+        if(!flags[i])
+        {
+            /* Set a flag to record that the current polygonal type has been used.*/
+            flags[i] = 1;
 
-         /* Start from 1010 because numbers finishing with 00, 01, ..., 09 can't
-          * be part of the chain.*/
-         for(j = 1010; j < 10000; j++)
-         {
-            /* If the number doesn't finish with 00, 01, ..., 09 and is poligonal,
-             * try adding it to the chain and add its value to the total sum.*/
-            if(j % 100 > 9 && polygonal[i][j])
+            /* Start from 1010 because numbers finishing with 00, 01, ..., 09 can't
+             * be part of the chain.*/
+            for(j = 1010; j < 10000; j++)
             {
-               /* If it's the first number, just add it as first step in the chain.*/
-               if(step == 0)
-               {
-                  chain[step] = j;
-                  *sum += j;
+                /* If the number doesn't finish with 00, 01, ..., 09 and is poligonal,
+                 * try adding it to the chain and add its value to the total sum.*/
+                if(j % 100 > 9 && polygonal[i][j])
+                {
+                    /* If it's the first number, just add it as first step in the chain.*/
+                    if(step == 0)
+                    {
+                        chain[step] = j;
+                        *sum += j;
 
-                  /* Recursively try to add other numbers to the chain. If a solution
-                   * is found, return 1.*/
-                  if(find_set(step+1, sum))
-                  {
-                     return 1;
-                  }
+                        /* Recursively try to add other numbers to the chain. If a solution
+                         * is found, return 1.*/
+                        if(find_set(step+1, sum))
+                        {
+                            return 1;
+                        }
 
-                  /* If a solution was not found, backtrack, subtracting the value of
-                   * the number from the total.*/
-                  *sum -= j;
-               }
-               /* If this is the last step and the current number can be added to the chain,
-                * add it, update the sum and return 1. A solution has been found.*/
-               else if(step == 5 && j % 100 == chain[0] / 100 && j / 100 == chain[step-1] % 100)
-               {
-                  chain[step] = j;
-                  *sum += j;
-                  return 1;
-               }
-               /* For every other step, add the number to the chain if possible, then recursively
-                * try to add other numbers.*/
-               else if(step < 5 && j / 100 == chain[step-1] % 100)
-               {
-                  chain[step] = j;
-                  *sum += j;
+                        /* If a solution was not found, backtrack, subtracting the value of
+                         * the number from the total.*/
+                        *sum -= j;
+                    }
+                    /* If this is the last step and the current number can be added to the chain,
+                     * add it, update the sum and return 1. A solution has been found.*/
+                    else if(step == 5 && j % 100 == chain[0] / 100 && j / 100 == chain[step-1] % 100)
+                    {
+                        chain[step] = j;
+                        *sum += j;
+                        return 1;
+                    }
+                    /* For every other step, add the number to the chain if possible, then recursively
+                     * try to add other numbers.*/
+                    else if(step < 5 && j / 100 == chain[step-1] % 100)
+                    {
+                        chain[step] = j;
+                        *sum += j;
 
-                  if(find_set(step+1, sum))
-                  {
-                     return 1;
-                  }
+                        if(find_set(step+1, sum))
+                        {
+                            return 1;
+                        }
 
-                  /* If a solution was not found, backtrack.*/
-                  *sum -= j;
-               }
+                        /* If a solution was not found, backtrack.*/
+                        *sum -= j;
+                    }
+                }
             }
-         }
 
-         /* Remove the flag for the current polygonal type.*/
-         flags[i] = 0;
-      }
-   }
+            /* Remove the flag for the current polygonal type.*/
+            flags[i] = 0;
+        }
+    }
 
-   return 0;
+    return 0;
 }

C/p062.c

@@ -1,8 +1,10 @@
-/* The cube, 41063625 (345^3), can be permuted to produce two other cubes: 56623104 (384^3) and 66430125 (405^3). 
+/* The cube, 41063625 (345^3), can be permuted to produce two other cubes: 56623104 (384^3) and 66430125 (405^3).
  * In fact, 41063625 is the smallest cube which has exactly three permutations of its digits which are also cube.
  *
  * Find the smallest cube for which exactly five permutations of its digits are cube.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -14,104 +16,104 @@ int count_digits(long int a);
 
 int main(int argc, char **argv)
 {
-   int count;
-   long int i, j, cubes[N];
-   double elapsed;
-   struct timespec start, end;
+    int count;
+    long int i, j, cubes[N];
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Calculate i^3 for all i smaller than 10000.*/
-   for(i = 0; i < N; i++)
-   {
-      cubes[i] = i * i * i;
-   }
+    /* Calculate i^3 for all i smaller than 10000.*/
+    for(i = 0; i < N; i++)
+    {
+        cubes[i] = i * i * i;
+    }
 
-   /* For each cube, check if there are four other cubes which are also 
-    * a permutation of the first cube.*/
-   for(i = 0; i < N - 5; i++)
-   {
-      count = 1;
+    /* For each cube, check if there are four other cubes which are also
+     * a permutation of the first cube.*/
+    for(i = 0; i < N - 5; i++)
+    {
+        count = 1;
 
-      /* Stop when the limit has been reached, when 5 values have been found or
-       * when j^3 has more digits than i^3 (if they don't have the same digits,
-       * they can't be permutations).*/
-      for(j = i + 1; j < N && count_digits(cubes[j]) == count_digits(cubes[i]); j++)
-      {
-         if(is_permutation(cubes[i], cubes[j]))
-         {
-            count++;
-         }
+        /* Stop when the limit has been reached, when 5 values have been found or
+         * when j^3 has more digits than i^3 (if they don't have the same digits,
+         * they can't be permutations).*/
+        for(j = i + 1; j < N && count_digits(cubes[j]) == count_digits(cubes[i]); j++)
+        {
+            if(is_permutation(cubes[i], cubes[j]))
+            {
+                count++;
+            }
 
-         if(count == 5)
-         {
+            if(count == 5)
+            {
+                break;
+            }
+        }
+
+        if(count == 5)
+        {
             break;
-         }
-      }
+        }
+    }
 
-      if(count == 5)
-      {
-         break;
-      }
-   }
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    printf("Project Euler, Problem 62\n");
+    printf("Answer: %ld\n", cubes[i]);
 
-   printf("Project Euler, Problem 62\n");
-   printf("Answer: %ld\n", cubes[i]);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
-
-   return 0;
+    return 0;
 }
 
 int count_digits(long int a)
 {
-   int count = 0;
+    int count = 0;
 
-   if(a == 0)
-   {
-      return 1;
-   }
+    if(a == 0)
+    {
+        return 1;
+    }
 
-   while(a > 0)
-   {
-      count++;
-      a /= 10;
-   }
+    while(a > 0)
+    {
+        count++;
+        a /= 10;
+    }
 
-   return count;
+    return count;
 }
 
 int is_permutation(long int a, long int b)
 {
-   int i;
-   int digits1[10] = {0}, digits2[10] = {0};
+    int i;
+    int digits1[10] = {0}, digits2[10] = {0};
 
-   /* Get digits of a.*/
-   while(a > 0)
-   {
-      digits1[a%10]++;
-      a /= 10;
-   }
+    /* Get digits of a.*/
+    while(a > 0)
+    {
+        digits1[a%10]++;
+        a /= 10;
+    }
 
-   /* Get digits of b.*/
-   while(b > 0)
-   {
-      digits2[b%10]++;
-      b /= 10;
-   }
+    /* Get digits of b.*/
+    while(b > 0)
+    {
+        digits2[b%10]++;
+        b /= 10;
+    }
 
-   /* If they're not the same, return 0.*/
-   for(i = 0; i < 10; i++)
-   {
-      if(digits1[i] != digits2[i])
-      {
-         return 0;
-      }
-   }
+    /* If they're not the same, return 0.*/
+    for(i = 0; i < 10; i++)
+    {
+        if(digits1[i] != digits2[i])
+        {
+            return 0;
+        }
+    }
 
-   return 1;
+    return 1;
 }

C/p063.c

@@ -2,6 +2,8 @@
  *
  * How many n-digit positive integers exist which are also an nth power?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -12,70 +14,70 @@ int count_digits(mpz_t n);
 
 int main(int argc, char **argv)
 {
-   int i, j, count = 0, finished = 0;
-   double elapsed;
-   struct timespec start, end;
-   mpz_t p;
+    int i, j, count = 0, finished = 0;
+    double elapsed;
+    struct timespec start, end;
+    mpz_t p;
 
-   mpz_init(p);
+    mpz_init(p);
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   i = 1;
+    i = 1;
 
-   while(!finished) 
-   {
-      /* When j=10, j^i will have i+1 digits (e.g if i=3, 10^3=1000).*/
-      for(j = 1; j < 10; j++)
-      {
-         mpz_ui_pow_ui(p, j, i);
+    while(!finished)
+    {
+        /* When j=10, j^i will have i+1 digits (e.g if i=3, 10^3=1000).*/
+        for(j = 1; j < 10; j++)
+        {
+            mpz_ui_pow_ui(p, j, i);
 
-         if(count_digits(p) == i)
-         {
-            count++;
-         }
-      }
+            if(count_digits(p) == i)
+            {
+                count++;
+            }
+        }
 
-      /* When 9^i has less than i digits, all the numbers have been found.*/
-      if(count_digits(p) < i)
-      {
-         finished = 1;
-      }
+        /* When 9^i has less than i digits, all the numbers have been found.*/
+        if(count_digits(p) < i)
+        {
+            finished = 1;
+        }
 
-      i++;
-   }
+        i++;
+    }
 
-   mpz_clear(p);
+    mpz_clear(p);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 63\n");
-   printf("Answer: %d\n", count);
-   
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 63\n");
+    printf("Answer: %d\n", count);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 int count_digits(mpz_t n)
 {
-   int count = 0;
-   mpz_t tmp;
+    int count = 0;
+    mpz_t tmp;
 
-   if(mpz_cmp_ui(n, 0) == 0)
-   {
-      return 1;
-   }
+    if(mpz_cmp_ui(n, 0) == 0)
+    {
+        return 1;
+    }
 
-   mpz_init_set(tmp, n);
+    mpz_init_set(tmp, n);
 
-   while(mpz_cmp_ui(tmp, 0))
-   {
-      mpz_tdiv_q_ui(tmp, tmp, 10);
-      count++;
-   }
+    while(mpz_cmp_ui(tmp, 0))
+    {
+        mpz_tdiv_q_ui(tmp, tmp, 10);
+        count++;
+    }
 
-   return count;
+    return count;
 }

C/p064.c

@@ -40,6 +40,8 @@
  *
  * How many continued fractions for N≤10000 have an odd period?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -50,60 +52,60 @@ int is_square(int n);
 
 int main(int argc, char **argv)
 {
-   int i, count = 0, period;
-   int *fraction;
-   double elapsed;
-   struct timespec start, end;
+    int i, count = 0, period;
+    int *fraction;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   for(i = 2; i <= 10000; i++)
-   {
-      /* Perfect squares are obviously not represented as continued fractions.
-       * For all other numbers, calculate their period and check if it's odd.*/
-      if(!is_square(i))
-      {
-         if((fraction = build_sqrt_cont_fraction(i, &period, 300)) == NULL)
-         {
-            fprintf(stderr, "Error! Build_cont_fraction function returned NULL\n");
-            return 1;
-         }
+    for(i = 2; i <= 10000; i++)
+    {
+        /* Perfect squares are obviously not represented as continued fractions.
+         * For all other numbers, calculate their period and check if it's odd.*/
+        if(!is_square(i))
+        {
+            if((fraction = build_sqrt_cont_fraction(i, &period, 300)) == NULL)
+            {
+                fprintf(stderr, "Error! Build_cont_fraction function returned NULL\n");
+                return 1;
+            }
 
-         if(period % 2 != 0)
-         {
-            count++;
-         }
-      }
-   }
+            if(period % 2 != 0)
+            {
+                count++;
+            }
+        }
+    }
 
-   free(fraction);
+    free(fraction);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 64\n");
-   printf("Answer: %d\n", count);
+    printf("Project Euler, Problem 64\n");
+    printf("Answer: %d\n", count);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int is_square(int n)
 {
-   int m;
-   double p;
+    int m;
+    double p;
 
-   p = sqrt(n);
-   m = p;
+    p = sqrt(n);
+    m = p;
 
-   if(p == m)
-   {
-      return 1;
-   }
-   else
-   {
-      return 0;
-   }
+    if(p == m)
+    {
+        return 1;
+    }
+    else
+    {
+        return 0;
+    }
 }

C/p065.c

@@ -27,6 +27,8 @@
  *
  * Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -34,55 +36,55 @@
 
 int main(int argc, char **argv)
 {
-   int i, ai[3] = {1, 2, 1}, count = 4, sum = 0;
-   mpz_t n0, n1, n2;
-   double elapsed;
-   struct timespec start, end;
+    int i, ai[3] = {1, 2, 1}, count = 4, sum = 0;
+    mpz_t n0, n1, n2;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   mpz_init_set_ui(n0, 3);
-   mpz_init_set_ui(n1, 8);
-   mpz_init_set_ui(n2, 11);
+    mpz_init_set_ui(n0, 3);
+    mpz_init_set_ui(n1, 8);
+    mpz_init_set_ui(n2, 11);
 
-   /* For a continued fractions [a_0; a_1, a_2, ...], the numerator of the 
-    * next convergent N_n=a_n*N_(n-1)+N_(n-2). The first three values for e are
-    * 3, 8 and 11, the next ones are easily calculated, considering that a_n
-    * follows a simple pattern: 
-    * a_1=1, a_2=2, a_3=1
-    * a_4=1, a_5=4, a_6=1
-    * a_7=1, a_8=6, a_9=1
-    * and so on.*/
-   while(count < 100)
-   {
-      ai[1] += 2;
+    /* For a continued fractions [a_0; a_1, a_2, ...], the numerator of the
+     * next convergent N_n=a_n*N_(n-1)+N_(n-2). The first three values for e are
+     * 3, 8 and 11, the next ones are easily calculated, considering that a_n
+     * follows a simple pattern:
+     * a_1=1, a_2=2, a_3=1
+     * a_4=1, a_5=4, a_6=1
+     * a_7=1, a_8=6, a_9=1
+     * and so on.*/
+    while(count < 100)
+    {
+        ai[1] += 2;
 
-      for(i = 0; i < 3 && count < 100; i++)
-      {
-         mpz_set(n0, n1);
-         mpz_set(n1, n2);
-         mpz_mul_ui(n2, n1, ai[i]);
-         mpz_add(n2, n2, n0);
-         count++;
-      }
-   }
+        for(i = 0; i < 3 && count < 100; i++)
+        {
+            mpz_set(n0, n1);
+            mpz_set(n1, n2);
+            mpz_mul_ui(n2, n1, ai[i]);
+            mpz_add(n2, n2, n0);
+            count++;
+        }
+    }
 
-   /* Sum the digits.*/
-   while(mpz_cmp_ui(n2, 0))
-   {
-      sum += mpz_tdiv_q_ui(n2, n2, 10);
-   }
+    /* Sum the digits.*/
+    while(mpz_cmp_ui(n2, 0))
+    {
+        sum += mpz_tdiv_q_ui(n2, n2, 10);
+    }
 
-   mpz_clears(n0, n1, n2, NULL);
+    mpz_clears(n0, n1, n2, NULL);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 65\n");
-   printf("Answer: %d\n", sum);
+    printf("Project Euler, Problem 65\n");
+    printf("Answer: %d\n", sum);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p066.c

@@ -18,6 +18,8 @@
  *
  * Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -29,64 +31,63 @@ int is_square(int n);
 
 int main(int argc, char **argv)
 {
-   int i, max_d = -1;
-   mpz_t x, max;
-   double elapsed;
-   struct timespec start, end;
-   
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    int i, max_d = -1;
+    mpz_t x, max;
+    double elapsed;
+    struct timespec start, end;
+
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
 //   mpz_init(x);
-   mpz_init_set_ui(max, 0);
+    mpz_init_set_ui(max, 0);
 
-   for(i = 2; i <= 1000; i++)
-   {
-      if(!is_square(i))
-      {
-         /* Solve the Diophantine equation x^2-D*y^2=1 (Pell equation)*/
-         if(pell_eq(i, x) == -1)
-         {
-            fprintf(stderr, "Error! Pell_eq function failed\n");
-            return 1;
-         }
-         
-         if(mpz_cmp(x, max) > 0)
-         {
-            mpz_set(max, x);
-            max_d = i;
-         }
-      }
-   }
+    for(i = 2; i <= 1000; i++)
+    {
+        if(!is_square(i))
+        {
+            /* Solve the Diophantine equation x^2-D*y^2=1 (Pell equation)*/
+            if(pell_eq(i, x) == -1)
+            {
+                fprintf(stderr, "Error! Pell_eq function failed\n");
+                return 1;
+            }
 
-   mpz_clears(x, max, NULL);
-   
-   clock_gettime(CLOCK_MONOTONIC, &end);
+            if(mpz_cmp(x, max) > 0)
+            {
+                mpz_set(max, x);
+                max_d = i;
+            }
+        }
+    }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    mpz_clears(x, max, NULL);
 
-   printf("Project Euler, Problem 66\n");
-   printf("Answer: %d\n", max_d);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   return 0;
+    printf("Project Euler, Problem 66\n");
+    printf("Answer: %d\n", max_d);
+
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 int is_square(int n)
 {
-   int m;
-   double p;
+    int m;
+    double p;
 
-   p = sqrt(n);
-   m = p;
+    p = sqrt(n);
+    m = p;
 
-   if(p == m)
-   {
-      return 1;
-   }
-   else
-   {
-      return 0;
-   }
+    if(p == m)
+    {
+        return 1;
+    }
+    else
+    {
+        return 0;
+    }
 }
-

C/p067.c

@@ -13,6 +13,8 @@
  * If you could check one trillion (1012) routes every second it would take over twenty billion years to check them all.
  * There is an efficient algorithm to solve it. ;o)*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -20,63 +22,63 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, max;
-   int **triang;
-   double elapsed;
-   FILE *fp;
-   struct timespec start, end;
+    int i, j, max;
+    int **triang;
+    double elapsed;
+    FILE *fp;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((triang = (int **)malloc(100*sizeof(int *))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((triang = (int **)malloc(100*sizeof(int *))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   for(i = 1; i <= 100; i++)
-   {
-      if((triang[i-1] = (int *)malloc(i*sizeof(int))) == NULL)
-      {
-         fprintf(stderr, "Error while allocating memory\n");
-         return 1;
-      }
-   }
+    for(i = 1; i <= 100; i++)
+    {
+        if((triang[i-1] = (int *)malloc(i*sizeof(int))) == NULL)
+        {
+            fprintf(stderr, "Error while allocating memory\n");
+            return 1;
+        }
+    }
 
-   if((fp = fopen("triangle.txt", "r")) == NULL)
-   {
-      fprintf(stderr, "Error while opening file %s\n", "triangle.txt");
-      return 1;
-   }
+    if((fp = fopen("triangle.txt", "r")) == NULL)
+    {
+        fprintf(stderr, "Error while opening file %s\n", "triangle.txt");
+        return 1;
+    }
 
-   for(i = 1; i <= 100; i++)
-   {
-      for(j = 0; j < i; j++)
-      {
-         fscanf(fp, "%d", &triang[i-1][j]);
-      }
-   }
+    for(i = 1; i <= 100; i++)
+    {
+        for(j = 0; j < i; j++)
+        {
+            fscanf(fp, "%d", &triang[i-1][j]);
+        }
+    }
 
-   fclose(fp);
+    fclose(fp);
 
-   /* Use the function implemented in projecteuler.c to find the maximum path.*/
-   max = find_max_path(triang, 100);
+    /* Use the function implemented in projecteuler.c to find the maximum path.*/
+    max = find_max_path(triang, 100);
 
-   for(i = 0; i < 100; i++)
-   {
-      free(triang[i]);
-   }
+    for(i = 0; i < 100; i++)
+    {
+        free(triang[i]);
+    }
 
-   free(triang);
+    free(triang);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 67\n");
-   printf("Answer: %d\n", max);
+    printf("Project Euler, Problem 67\n");
+    printf("Answer: %d\n", max);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p068.c

@@ -43,6 +43,8 @@
  *               O
  */
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
@@ -56,79 +58,79 @@ int compare(void *a, void *b);
 
 int main(int argc, char **argv)
 {
-   int i;
-   int *eval, **ring;
-   long int n, max = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i;
+    int *eval, **ring;
+    long int n, max = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((ring = (int **)malloc(10*sizeof(int *))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((ring = (int **)malloc(10*sizeof(int *))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   for(i = 0; i < 10; i++)
-   {
-      if((ring[i] = (int *)malloc(sizeof(int))) == NULL)
-      {
-         fprintf(stderr, "Error while allocating memory\n");
-         return 1;
-      }
-      *ring[i] = i + 1;
-   }
+    for(i = 0; i < 10; i++)
+    {
+        if((ring[i] = (int *)malloc(sizeof(int))) == NULL)
+        {
+            fprintf(stderr, "Error while allocating memory\n");
+            return 1;
+        }
+        *ring[i] = i + 1;
+    }
 
-   /* Check if the first permutation of values is a possible solution.*/
-   eval = eval_ring(ring, 5);
+    /* Check if the first permutation of values is a possible solution.*/
+    eval = eval_ring(ring, 5);
 
-   /* Convert the vector into an integer number.*/
-   n = array_to_long(eval, 15);
+    /* Convert the vector into an integer number.*/
+    n = array_to_long(eval, 15);
 
-   if(n > max)
-   {
-      max = n;
-   }
-   
-   free(eval);
+    if(n > max)
+    {
+        max = n;
+    }
 
-   /* Generate all possible permutations, for each one check if
-    * it's a possible solution for the ring and save the maximum.*/
-   while(next_permutation((void **)ring, 10, compare) != -1)
-   {
-      eval = eval_ring(ring, 5);
+    free(eval);
 
-      n = array_to_long(eval, 15);
+    /* Generate all possible permutations, for each one check if
+     * it's a possible solution for the ring and save the maximum.*/
+    while(next_permutation((void **)ring, 10, compare) != -1)
+    {
+        eval = eval_ring(ring, 5);
 
-      if(n > max)
-      {
-         max = n;
-      }
-      
-      free(eval);
-   }
+        n = array_to_long(eval, 15);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+        if(n > max)
+        {
+            max = n;
+        }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+        free(eval);
+    }
 
-   printf("Project Euler, Problem 68\n");
-   printf("Answer: %ld\n", max);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   return 0;
+    printf("Project Euler, Problem 68\n");
+    printf("Answer: %ld\n", max);
+
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 int compare(void *a, void *b)
 {
-   int *n1, *n2;
+    int *n1, *n2;
 
-   n1 = (int *)a;
-   n2 = (int *)b;
+    n1 = (int *)a;
+    n2 = (int *)b;
 
-   return *n1 - *n2;
+    return *n1 - *n2;
 }
 
 /* Function to evaluate the ring. The ring is represented as a vector of 2*n elements,
@@ -136,82 +138,82 @@ int compare(void *a, void *b)
  * represent the internal ring.*/
 int *eval_ring(int **ring, int n)
 {
-   int i, j, magic_val, val;
-   int *res;
+    int i, j, magic_val, val;
+    int *res;
 
-   for(i = 1; i < n; i++)
-   {
-      /* We need to start from the lowest external node, so if
-       * the first element in the vector is not the lowest of 
-       * the first n elements (the external elements), the configuration
-       * is not a valid one.*/
-      if(*ring[i] < *ring[0])
-      {
-         return NULL;
-      }
-   }
-   
-   if((res = (int *)malloc(3*n*sizeof(int))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return NULL;
-   }
+    for(i = 1; i < n; i++)
+    {
+        /* We need to start from the lowest external node, so if
+         * the first element in the vector is not the lowest of
+         * the first n elements (the external elements), the configuration
+         * is not a valid one.*/
+        if(*ring[i] < *ring[0])
+        {
+            return NULL;
+        }
+    }
 
-   /* Each group of three must have the same value.*/
-   magic_val = *ring[0] + *ring[n] + *ring[n+1];
+    if((res = (int *)malloc(3*n*sizeof(int))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return NULL;
+    }
 
-   for(i = 0, j = 0; i < n; i++, j += 3)
-   {
-      /* We need to find the maximum 16-digit string, this is
-       * possible only if the element "10" is used only once,
-       * i.e. if it's one of the external nodes.*/
-      if(*ring[n+i] == 10 || *ring[n+(i+1)%n] == 10)
-      {
-         return NULL;
-      }
+    /* Each group of three must have the same value.*/
+    magic_val = *ring[0] + *ring[n] + *ring[n+1];
 
-      /* Check that the value of the current three-element group
-       * is the "magic" value.*/
-      val = *ring[i] + *ring[n+i] + *ring[n+(i+1)%n];
+    for(i = 0, j = 0; i < n; i++, j += 3)
+    {
+        /* We need to find the maximum 16-digit string, this is
+         * possible only if the element "10" is used only once,
+         * i.e. if it's one of the external nodes.*/
+        if(*ring[n+i] == 10 || *ring[n+(i+1)%n] == 10)
+        {
+            return NULL;
+        }
 
-      if(val != magic_val)
-      {
-         return NULL;
-      }
+        /* Check that the value of the current three-element group
+         * is the "magic" value.*/
+        val = *ring[i] + *ring[n+i] + *ring[n+(i+1)%n];
 
-      /* Save the current element group in the result string.*/
-      res[j] = *ring[i];
-      res[j+1] = *ring[n+i];
-      res[j+2] = *ring[n+(i+1)%n];
-   }
+        if(val != magic_val)
+        {
+            return NULL;
+        }
 
-   return res;
+        /* Save the current element group in the result string.*/
+        res[j] = *ring[i];
+        res[j+1] = *ring[n+i];
+        res[j+2] = *ring[n+(i+1)%n];
+    }
+
+    return res;
 }
 
 /* Function to convert the vector into a long int.*/
 long int array_to_long(int *n, int len)
 {
-   int i, k = 0;
-   long int res = 0;
+    int i, k = 0;
+    long int res = 0;
 
-   if(n == NULL)
-   {
-      return 0;
-   }
+    if(n == NULL)
+    {
+        return 0;
+    }
 
-   for(i = len - 1; i >= 0; i--)
-   {
-      res += n[i] * pow(10, k);
+    for(i = len - 1; i >= 0; i--)
+    {
+        res += n[i] * pow(10, k);
 
-      if(n[i] >= 10)
-      {
-         k += 2;
-      }
-      else
-      {
-         k++;
-      }
-   }
+        if(n[i] >= 10)
+        {
+            k += 2;
+        }
+        else
+        {
+            k++;
+        }
+    }
 
-   return res;
+    return res;
 }

C/p069.c

@@ -16,6 +16,8 @@
  *
  * Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -27,7 +29,7 @@
 int main(int argc, char **argv)
 {
     int i, res = 1;
-	double elapsed;
+    double elapsed;
     struct timespec start, end;
 
     clock_gettime(CLOCK_MONOTONIC, &start);
@@ -43,7 +45,7 @@ int main(int argc, char **argv)
     while(res < N)
     {
         i++;
-        
+
         if(is_prime(i))
         {
             res *= i;
@@ -58,9 +60,9 @@ int main(int argc, char **argv)
     elapsed = (end.tv_sec-start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
     printf("Project Euler, Problem 69\n");
-	printf("Answer: %d\n", res);
+    printf("Answer: %d\n", res);
 
     printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-	return 0;
+    return 0;
 }

C/p070.c

@@ -7,6 +7,8 @@
  *
  * Find the value of n, 1 < n < 10^7, for which φ(n) is a permutation of n and the ratio n/φ(n) produces a minimum.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -20,78 +22,78 @@ int is_permutation(int a, int b);
 
 int main(int argc, char **argv)
 {
-   int i, a, b, p, n = -1;
-   int *primes;
-   double elapsed, min = DBL_MAX;
-   struct timespec start, end;
+    int i, a, b, p, n = -1;
+    int *primes;
+    double elapsed, min = DBL_MAX;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((primes = sieve(N)) == NULL)
-   {
-      fprintf(stderr, "Error! Sieve function returned NULL\n");
-      return 1;
-   }
+    if((primes = sieve(N)) == NULL)
+    {
+        fprintf(stderr, "Error! Sieve function returned NULL\n");
+        return 1;
+    }
 
-   for(i = 2; i < N; i++)
-   {
-      /* When n is prime, phi(n)=(n-1), so to minimize n/phi(n) we should
-       * use n prime. But n-1 can't be a permutation of n. The second best 
-       * bet is to use semiprimes. For a semiprime n=p*q, phi(n)=(p-1)(q-1).
-       * So we check if a number is semiprime, if yes calculate phi, finally
-       * check if phi(n) is a permutation of n and update the minimum if it's
-       * smaller.*/
-      if(is_semiprime(i, &a, &b, primes))
-      {
-         p = phi_semiprime(i, a, b);
+    for(i = 2; i < N; i++)
+    {
+        /* When n is prime, phi(n)=(n-1), so to minimize n/phi(n) we should
+         * use n prime. But n-1 can't be a permutation of n. The second best
+         * bet is to use semiprimes. For a semiprime n=p*q, phi(n)=(p-1)(q-1).
+         * So we check if a number is semiprime, if yes calculate phi, finally
+         * check if phi(n) is a permutation of n and update the minimum if it's
+         * smaller.*/
+        if(is_semiprime(i, &a, &b, primes))
+        {
+            p = phi_semiprime(i, a, b);
 
-         if(is_permutation(p, i) && (double)i / p < min)
-         {
-            n = i;
-            min = (double)i / p;
-         }
-      }
-   }
+            if(is_permutation(p, i) && (double)i / p < min)
+            {
+                n = i;
+                min = (double)i / p;
+            }
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 70\n");
-   printf("Answer: %d\n", n);
+    printf("Project Euler, Problem 70\n");
+    printf("Answer: %d\n", n);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int is_permutation(int a, int b)
 {
-   int i;
-   int digits1[10] = {0}, digits2[10] = {0};
+    int i;
+    int digits1[10] = {0}, digits2[10] = {0};
 
-   /* Get digits of a.*/
-   while(a > 0)
-   {
-      digits1[a%10]++;
-      a /= 10;
-   }
+    /* Get digits of a.*/
+    while(a > 0)
+    {
+        digits1[a%10]++;
+        a /= 10;
+    }
 
-   /* Get digits of b.*/
-   while(b > 0)
-   {
-      digits2[b%10]++;
-      b /= 10;
-   }
+    /* Get digits of b.*/
+    while(b > 0)
+    {
+        digits2[b%10]++;
+        b /= 10;
+    }
 
-   /* If they're not the same, return 0.*/
-   for(i = 0; i < 10; i++)
-   {
-      if(digits1[i] != digits2[i])
-      {
-         return 0;
-      }
-   }
+    /* If they're not the same, return 0.*/
+    for(i = 0; i < 10; i++)
+    {
+        if(digits1[i] != digits2[i])
+        {
+            return 0;
+        }
+    }
 
-   return 1;
+    return 1;
 }

C/p071.c

@@ -9,6 +9,8 @@
  * By listing the set of reduced proper fractions for d ≤ 1,000,000 in ascending order of size, find the numerator
  * of the fraction immediately to the left of 3/7.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -18,47 +20,47 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, n, d, max_n;
-   double elapsed, max = 0.0;
-   struct timespec start, end;
+    int i, j, n, d, max_n;
+    double elapsed, max = 0.0;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* For each denominator q, we need to find the biggest numerator p for which
-    * p/q<a/b, where a/b is 3/7 for this problem. So:
-    * pb<aq
-    * pb<=aq-1
-    * p<=(aq-1)/b
-    * So we can set p=(3*q-1)/7 (using integer division).*/
-   for(i = 2; i <= N; i++)
-   {
-      j = (3 * i - 1) / 7;
+    /* For each denominator q, we need to find the biggest numerator p for which
+     * p/q<a/b, where a/b is 3/7 for this problem. So:
+     * pb<aq
+     * pb<=aq-1
+     * p<=(aq-1)/b
+     * So we can set p=(3*q-1)/7 (using integer division).*/
+    for(i = 2; i <= N; i++)
+    {
+        j = (3 * i - 1) / 7;
 
-      if((double)j / i > max)
-      {
-         n = j;
-         d = i;
-         max = (double)n / d;
+        if((double)j / i > max)
+        {
+            n = j;
+            d = i;
+            max = (double)n / d;
 
-         /* Reduce the fraction if it's not already reduced.*/
-         if(gcd(i, j) > 1)
-         {
-            n /= gcd(i, j);
-            d /= gcd(i, j);
-         }
+            /* Reduce the fraction if it's not already reduced.*/
+            if(gcd(i, j) > 1)
+            {
+                n /= gcd(i, j);
+                d /= gcd(i, j);
+            }
 
-         max_n = n;
-      }
-   }
+            max_n = n;
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 71\n");
-   printf("Answer: %d\n", max_n);
+    printf("Project Euler, Problem 71\n");
+    printf("Answer: %d\n", max_n);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p072.c

@@ -8,6 +8,8 @@
  *
  * How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -18,37 +20,37 @@
 
 int main(int argc, char **argv)
 {
-   int i;
-   int *primes;
-   long int count = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i;
+    int *primes;
+    long int count = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((primes = sieve(N)) == NULL)
-   {
-      fprintf(stderr, "Error! Sieve function returned NULL\n");
-      return 1;
-   }
+    if((primes = sieve(N)) == NULL)
+    {
+        fprintf(stderr, "Error! Sieve function returned NULL\n");
+        return 1;
+    }
 
-   /* For any denominator d, the number of reduced proper fractions is 
-    * the number of fractions n/d where gcd(n, d)=1, which is the definition
-    * of Euler's Totient Function phi. It's sufficient to calculate phi for each
-    * denominator and sum the value.*/
-   for(i = 2; i < N; i++)
-   {
-      count += phi(i, primes);
-   }
-   
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    /* For any denominator d, the number of reduced proper fractions is
+     * the number of fractions n/d where gcd(n, d)=1, which is the definition
+     * of Euler's Totient Function phi. It's sufficient to calculate phi for each
+     * denominator and sum the value.*/
+    for(i = 2; i < N; i++)
+    {
+        count += phi(i, primes);
+    }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 72\n");
-   printf("Answer: %ld\n", count);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 72\n");
+    printf("Answer: %ld\n", count);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }

C/p073.c

@@ -8,6 +8,8 @@
  *
  * How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for d ≤ 12,000?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -15,37 +17,37 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, limit, count = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i, j, limit, count = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* For each denominator q, we need to find the fractions p/q for which
-    * 1/3<p/q<1/2. For the lower limit, if p=q/3, then p/q=1/3, so we take
-    * p=q/3+1. For the upper limit, if p=q/2 p/q=1/2, so we take p=(q-1)/2.*/
-   for(i = 2; i <= 12000; i++)
-   {
-      limit = (i - 1) / 2;
+    /* For each denominator q, we need to find the fractions p/q for which
+     * 1/3<p/q<1/2. For the lower limit, if p=q/3, then p/q=1/3, so we take
+     * p=q/3+1. For the upper limit, if p=q/2 p/q=1/2, so we take p=(q-1)/2.*/
+    for(i = 2; i <= 12000; i++)
+    {
+        limit = (i - 1) / 2;
 
-      for(j = i / 3 + 1; j <= limit; j++)
-      {
-         /* Increment the counter if the current fraction is reduced.*/
-         if(gcd(j, i) == 1)
-         {
-            count++;
-         }
-      }
-   }
-   
-   clock_gettime(CLOCK_MONOTONIC, &end);
+        for(j = i / 3 + 1; j <= limit; j++)
+        {
+            /* Increment the counter if the current fraction is reduced.*/
+            if(gcd(j, i) == 1)
+            {
+                count++;
+            }
+        }
+    }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 73\n");
-   printf("Answer: %d\n", count);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 73\n");
+    printf("Answer: %d\n", count);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }

C/p074.c

@@ -20,6 +20,8 @@
  *
  * How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -34,99 +36,99 @@ int chains[N] = {0};
 
 int main(int argc, char **argv)
 {
-   int i, count = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i, count = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Simple brute force approach, for every number calculate
-    * the length of the chain.*/
-   for(i = 3; i < N; i++)
-   {
-      if(len_chain(i) == 60)
-      {
-         count++;
-      }
-   }
-   
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    /* Simple brute force approach, for every number calculate
+     * the length of the chain.*/
+    for(i = 3; i < N; i++)
+    {
+        if(len_chain(i) == 60)
+        {
+            count++;
+        }
+    }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 74\n");
-   printf("Answer: %d\n", count);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 74\n");
+    printf("Answer: %d\n", count);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 /* Recursively calculate the factorial, of a number, saving the result
  * so that it can be reused later.*/
 int factorial(int n)
 {
-   if(n == 0 || n == 1)
-   {
-      return 1;
-   }
-   else if(factorials[n] != 0)
-   {
-      return factorials[n];
-   }
-   else
-   {
-      factorials[n]=n*factorial(n-1);
-      return factorials[n];
-   }
+    if(n == 0 || n == 1)
+    {
+        return 1;
+    }
+    else if(factorials[n] != 0)
+    {
+        return factorials[n];
+    }
+    else
+    {
+        factorials[n]=n*factorial(n-1);
+        return factorials[n];
+    }
 }
 
 int len_chain(int n)
 {
-   int i, count = 0, finished = 0, value, tmp;
-   int chain[60];
+    int i, count = 0, finished = 0, value, tmp;
+    int chain[60];
 
-   value = n;
-   chain[count] = value;
+    value = n;
+    chain[count] = value;
 
-   while(!finished)
-   {
-      tmp = 0;
-      count++;
+    while(!finished)
+    {
+        tmp = 0;
+        count++;
 
-      /* Generate the next number of the chain by taking
-       * the digits of the current value, calculating the
-       * factorials and adding them.*/
-      while(value != 0)
-      {
-         tmp += factorial(value % 10);
-         value /= 10;
-      }
+        /* Generate the next number of the chain by taking
+         * the digits of the current value, calculating the
+         * factorials and adding them.*/
+        while(value != 0)
+        {
+            tmp += factorial(value % 10);
+            value /= 10;
+        }
 
-      /* If the chain length for the new value has already been
-       * calculated before, use the saved value (only chains for
-       * values smaller than N are saved).*/
-      if(tmp < N && chains[tmp] != 0)
-      {
-         return count + chains[tmp];
-      }
+        /* If the chain length for the new value has already been
+         * calculated before, use the saved value (only chains for
+         * values smaller than N are saved).*/
+        if(tmp < N && chains[tmp] != 0)
+        {
+            return count + chains[tmp];
+        }
 
-      value = tmp;
+        value = tmp;
 
-      /* If the current value is already present in the chain,
-       * the chain is finished.*/
-      for(i = 0; i < count; i++)
-      {
-         if(chain[i] == value)
-         {
-            finished = 1;
-         }
-      }
+        /* If the current value is already present in the chain,
+         * the chain is finished.*/
+        for(i = 0; i < count; i++)
+        {
+            if(chain[i] == value)
+            {
+                finished = 1;
+            }
+        }
 
-      chain[count] = value;
-   }
+        chain[count] = value;
+    }
 
-   chains[n] = count;
+    chains[n] = count;
 
-   return count;
+    return count;
 }

C/p075.c

@@ -15,6 +15,8 @@
  *
  * Given that L is the length of the wire, for how many values of L ≤ 1,500,000 can exactly one integer sided right angle triangle be formed?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -24,79 +26,79 @@
 
 int main(int argc, char **argv)
 {
-   int i, a, b, c, tmpa, tmpb, tmpc, m, n, count = 0;
-   int *l; 
-   double elapsed;
-   struct timespec start, end;
+    int i, a, b, c, tmpa, tmpb, tmpc, m, n, count = 0;
+    int *l;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((l = (int *)calloc(N + 1, sizeof(int))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((l = (int *)calloc(N + 1, sizeof(int))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   /* Generate all Pythagorean triplets using Euclid's algorithm:
-    * For m>=2 and n<m:
-    * a=m*m-n*n
-    * b=2*m*n
-    * c=m*m+n*n
-    * This gives a primitive triple if gcd(m, n)=1 and exactly one
-    * of m and n is odd. To generate all the triples, generate all
-    * the primitive one and multiply them by i=2,3, ..., n until the
-    * perimeter is larger than the limit. The limit for m is 865, because
-    * when m=866 even with the smaller n (i.e. 1) the perimeter is greater
-    * than the given limit.*/
-   for(m = 2; m < 866; m++)
-   {
-      for(n = 1; n < m; n++)
-      {
-         if(gcd(m, n) == 1 && ((m % 2 == 0 && n % 2 != 0) || (m % 2 != 0 && n % 2 == 0)))
-         {
-            a = m * m - n * n;
-            b = 2 * m * n;
-            c = m * m + n * n;
-
-            if(a + b + c <= N)
+    /* Generate all Pythagorean triplets using Euclid's algorithm:
+     * For m>=2 and n<m:
+     * a=m*m-n*n
+     * b=2*m*n
+     * c=m*m+n*n
+     * This gives a primitive triple if gcd(m, n)=1 and exactly one
+     * of m and n is odd. To generate all the triples, generate all
+     * the primitive one and multiply them by i=2,3, ..., n until the
+     * perimeter is larger than the limit. The limit for m is 865, because
+     * when m=866 even with the smaller n (i.e. 1) the perimeter is greater
+     * than the given limit.*/
+    for(m = 2; m < 866; m++)
+    {
+        for(n = 1; n < m; n++)
+        {
+            if(gcd(m, n) == 1 && ((m % 2 == 0 && n % 2 != 0) || (m % 2 != 0 && n % 2 == 0)))
             {
-               l[a+b+c]++;
+                a = m * m - n * n;
+                b = 2 * m * n;
+                c = m * m + n * n;
+
+                if(a + b + c <= N)
+                {
+                    l[a+b+c]++;
+                }
+
+                i = 2;
+                tmpa = i * a;
+                tmpb = i * b;
+                tmpc = i * c;
+
+                while(tmpa + tmpb + tmpc <= N)
+                {
+                    l[tmpa+tmpb+tmpc]++;
+
+                    i++;
+                    tmpa = i * a;
+                    tmpb = i * b;
+                    tmpc = i * c;
+                }
             }
+        }
+    }
 
-            i = 2;
-            tmpa = i * a;
-            tmpb = i * b;
-            tmpc = i * c;
+    for(i = 0; i <= N; i++)
+    {
+        if(l[i] == 1)
+            count++;
+    }
 
-            while(tmpa + tmpb + tmpc <= N)
-            {
-               l[tmpa+tmpb+tmpc]++;
+    free(l);
 
-               i++;
-               tmpa = i * a;
-               tmpb = i * b;
-               tmpc = i * c;
-            }
-         }
-      }
-   }
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   for(i = 0; i <= N; i++)
-   {
-      if(l[i] == 1)
-         count++;
-   }
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   free(l);
+    printf("Project Euler, Problem 75\n");
+    printf("Answer: %d\n", count);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
-
-   printf("Project Euler, Problem 75\n");
-   printf("Answer: %d\n", count);
-
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
-
-   return 0;
+    return 0;
 }

C/p076.c

@@ -9,6 +9,8 @@
  *
  * How many different ways can one hundred be written as a sum of at least two positive integers?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -16,34 +18,34 @@
 
 int main(int argc, char **argv)
 {
-	int n;
-   long int *partitions;
-	double elapsed;
-	struct timespec start, end;
+    int n;
+    long int *partitions;
+    double elapsed;
+    struct timespec start, end;
 
-	clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((partitions = (long int *)calloc(100, sizeof(long int))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((partitions = (long int *)calloc(100, sizeof(long int))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   /* The number of ways a number can be written as a sum is given by the partition function
-    * (-1 because the partition function includes also the number itself).
-    * The function is implemented in projecteuler.c*/
-	n = partition_fn(100, partitions, -1) - 1;
+    /* The number of ways a number can be written as a sum is given by the partition function
+     * (-1 because the partition function includes also the number itself).
+     * The function is implemented in projecteuler.c*/
+    n = partition_fn(100, partitions, -1) - 1;
 
-   free(partitions);
+    free(partitions);
 
-	clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-	elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 76\n");
-	printf("Answer: %d\n", n);
+    printf("Project Euler, Problem 76\n");
+    printf("Answer: %d\n", n);
 
-	printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-	return 0;
+    return 0;
 }

C/p077.c

@@ -8,6 +8,8 @@
  *
  * What is the first value which can be written as the sum of primes in over five thousand different ways?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -15,69 +17,69 @@
 #include "projecteuler.h"
 
 int count(int value, int n, int i, int target);
-   
+
 int primes[100];
 
 int main(int argc, char **argv)
 {
-	int i, j, n;
-	double elapsed;
-	struct timespec start, end;
+    int i, j, n;
+    double elapsed;
+    struct timespec start, end;
 
-	clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Generate a list of the first 100 primes.*/
-	for(i = 0, j = 0; j < 100; i++)
-	{
-		if(is_prime(i))
-      {
-			primes[j++] = i;
-      }
-	}
+    /* Generate a list of the first 100 primes.*/
+    for(i = 0, j = 0; j < 100; i++)
+    {
+        if(is_prime(i))
+        {
+            primes[j++] = i;
+        }
+    }
 
-	i = 1;
+    i = 1;
 
-   /* Use a function to count the number of prime partitions for 
-    * each number >= 2 until the one that can be written in over
-    * 5000 ways is found.*/
-	while((n = count(0, 0, 0, ++i)) <= 5000);
+    /* Use a function to count the number of prime partitions for
+     * each number >= 2 until the one that can be written in over
+     * 5000 ways is found.*/
+    while((n = count(0, 0, 0, ++i)) <= 5000);
 
-	clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-	elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 77\n");
-	printf("Answer: %d\n", i);
+    printf("Project Euler, Problem 77\n");
+    printf("Answer: %d\n", i);
 
-	printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-	return 0;
+    return 0;
 }
 
 /* Function using a simple recursive brute force approach
  * to find all the partitions.*/
 int count(int value, int n, int i, int target)
 {
-	int j;
+    int j;
 
-	for(j = i; j < 100; j++)
-	{
-		value += primes[j];
+    for(j = i; j < 100; j++)
+    {
+        value += primes[j];
 
-		if(value == target)
-      {
-			return n + 1;
-      }
-		else if(value > target)
-      {
-			return n;
-      }
-		else
-		{
-			n = count(value, n, j, target);
-			value -= primes[j];
-		}
-	}
+        if(value == target)
+        {
+            return n + 1;
+        }
+        else if(value > target)
+        {
+            return n;
+        }
+        else
+        {
+            n = count(value, n, j, target);
+            value -= primes[j];
+        }
+    }
 
-	return n;
+    return n;
 }

C/p078.c

@@ -11,6 +11,8 @@
  *
  * Find the least value of n for which p(n) is divisible by one million.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -20,25 +22,25 @@
 
 int main(int argc, char **argv)
 {
-   int i = -1;
-   long int partitions[N] = {0};
-   double elapsed;
-   struct timespec start, end;
+    int i = -1;
+    long int partitions[N] = {0};
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Using the partition function to calculate the number of partitions,
-    * giving the result modulo N.*/   
-   while(partition_fn(++i, partitions, N) != 0);
-	
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    /* Using the partition function to calculate the number of partitions,
+     * giving the result modulo N.*/
+    while(partition_fn(++i, partitions, N) != 0);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 78\n");
-   printf("Answer: %d\n", i);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 78\n");
+    printf("Answer: %d\n", i);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }

C/p079.c

@@ -6,6 +6,8 @@
  * Given that the three characters are always asked for in order, analyse the file so as to determine the shortest possible
  * secret passcode of unknown length.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -16,174 +18,174 @@ int check_passcode(int **passcode, int len, int **logins, int n);
 
 int main(int argc, char **argv)
 {
-   int i, j, keylog, len = 4, found = 0, digits[10] = {0}, passcode_digits[10] = {0}, **passcode, **logins;
-   char line[5];
-   FILE *fp;
-   double elapsed;
-   struct timespec start, end;
+    int i, j, keylog, len = 4, found = 0, digits[10] = {0}, passcode_digits[10] = {0}, **passcode, **logins;
+    char line[5];
+    FILE *fp;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((fp = fopen("keylog.txt", "r")) == NULL)
-   {
-      fprintf(stderr, "Error while opening file %s\n", "keylog.txt");
-      return 1;
-   }
+    if((fp = fopen("keylog.txt", "r")) == NULL)
+    {
+        fprintf(stderr, "Error while opening file %s\n", "keylog.txt");
+        return 1;
+    }
 
-   if((logins = (int **)malloc(50*sizeof(int*))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((logins = (int **)malloc(50*sizeof(int*))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   for(i = 0; i < 50; i++)
-   {
-      if((logins[i] = (int *)malloc(3*sizeof(int))) == NULL)
-      {
-         fprintf(stderr, "Error while allocating memory\n");
-         return 1;
-      }
-   }
-
-   i = 0;
-
-   while(fscanf(fp, "%s", line) != EOF)
-   {
-      j = 2;
-      keylog = atoi(line);
-
-      while(keylog > 0)
-      {
-         logins[i][j--] = keylog % 10;
-         /* Check which digits are present in the login attempts.*/
-         digits[keylog%10]++;
-         keylog /= 10;
-      }
-
-      i++;
-   }
-
-   fclose(fp);
-
-   j = 0;
-   for(i = 0; i < 10; i++)
-   {
-      /* To generate the passcode, only use the digits present in the
-       * login attempts.*/
-      if(digits[i] > 0)
-      {
-         passcode_digits[j++] = i;
-      }
-   }
-
-   while(!found)
-   {
-      if((passcode = (int **)malloc(len*sizeof(int *))) == NULL)
-      {
-         fprintf(stderr, "Error while allocating memory\n");
-         return 1;
-      }
-
-      for(i = 0; i < len; i++)
-      {
-         if((passcode[i] = (int *)malloc(sizeof(int))) == NULL)
-         {
+    for(i = 0; i < 50; i++)
+    {
+        if((logins[i] = (int *)malloc(3*sizeof(int))) == NULL)
+        {
             fprintf(stderr, "Error while allocating memory\n");
             return 1;
-         }
-         /* For the current length, generate the first passcode with the
-          * digits in order.*/
-         *passcode[i] = passcode_digits[i];
-      }
+        }
+    }
 
-      /* Check if the passcode is compatible with the login attempts.*/
-      if(check_passcode(passcode, len, logins, 50))
-      {
-         found = 1;
-         break;
-      }
+    i = 0;
 
-      /* For the given length, check every permutation until the correct
-       * passcode has been found, or all the permutations have been tried.*/
-      while(next_permutation((void **)passcode, len, compare) != -1)
-      {
-         if(check_passcode(passcode, len, logins, 50))
-         {
-            printf("Project Euler, Problem 79\n");
-            printf("Answer: ");
-            for(i = 0; i < len; i++)
-               printf("%d", *passcode[i]);
-            printf("\n");
+    while(fscanf(fp, "%s", line) != EOF)
+    {
+        j = 2;
+        keylog = atoi(line);
 
+        while(keylog > 0)
+        {
+            logins[i][j--] = keylog % 10;
+            /* Check which digits are present in the login attempts.*/
+            digits[keylog%10]++;
+            keylog /= 10;
+        }
+
+        i++;
+    }
+
+    fclose(fp);
+
+    j = 0;
+    for(i = 0; i < 10; i++)
+    {
+        /* To generate the passcode, only use the digits present in the
+         * login attempts.*/
+        if(digits[i] > 0)
+        {
+            passcode_digits[j++] = i;
+        }
+    }
+
+    while(!found)
+    {
+        if((passcode = (int **)malloc(len*sizeof(int *))) == NULL)
+        {
+            fprintf(stderr, "Error while allocating memory\n");
+            return 1;
+        }
+
+        for(i = 0; i < len; i++)
+        {
+            if((passcode[i] = (int *)malloc(sizeof(int))) == NULL)
+            {
+                fprintf(stderr, "Error while allocating memory\n");
+                return 1;
+            }
+            /* For the current length, generate the first passcode with the
+             * digits in order.*/
+            *passcode[i] = passcode_digits[i];
+        }
+
+        /* Check if the passcode is compatible with the login attempts.*/
+        if(check_passcode(passcode, len, logins, 50))
+        {
             found = 1;
             break;
-         }
-      }
+        }
 
-      for(i = 0; i < len; i++)
-      {
-         free(passcode[i]);
-      }
-      free(passcode);
+        /* For the given length, check every permutation until the correct
+         * passcode has been found, or all the permutations have been tried.*/
+        while(next_permutation((void **)passcode, len, compare) != -1)
+        {
+            if(check_passcode(passcode, len, logins, 50))
+            {
+                printf("Project Euler, Problem 79\n");
+                printf("Answer: ");
+                for(i = 0; i < len; i++)
+                    printf("%d", *passcode[i]);
+                printf("\n");
 
-      /* If the passcode has not yet been found, try a longer passcode.*/ 
-      len++;
-   }
+                found = 1;
+                break;
+            }
+        }
 
-   for(i = 0; i < 50; i++)
-   {
-      free(logins[i]);
-   }
+        for(i = 0; i < len; i++)
+        {
+            free(passcode[i]);
+        }
+        free(passcode);
 
-   free(logins);
-   
-   clock_gettime(CLOCK_MONOTONIC, &end);
+        /* If the passcode has not yet been found, try a longer passcode.*/
+        len++;
+    }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    for(i = 0; i < 50; i++)
+    {
+        free(logins[i]);
+    }
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    free(logins);
 
-   return 0;
+    clock_gettime(CLOCK_MONOTONIC, &end);
+
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 int compare(void *a, void *b)
 {
-   int *n1, *n2;
+    int *n1, *n2;
 
-   n1 = (int *)a;
-   n2 = (int *)b;
+    n1 = (int *)a;
+    n2 = (int *)b;
 
-   return *n1 - *n2;
+    return *n1 - *n2;
 }
 
 int check_passcode(int **passcode, int len, int **logins, int n)
 {
-   int i, j, k;
+    int i, j, k;
 
-   /* For every login attempt, check if all the digits appear in the
-    * passcode in the correct order. Return 0 if a login attempt 
-    * incompatible with the current passcode is found.*/
-   for(i = 0; i < n; i++)
-   {
-      k = 0;
-      for(j = 0; j < len; j++)
-      {
-         if(*passcode[j] == logins[i][k])
-         {
-            k++;
-            
-            if(k == 3)
+    /* For every login attempt, check if all the digits appear in the
+     * passcode in the correct order. Return 0 if a login attempt
+     * incompatible with the current passcode is found.*/
+    for(i = 0; i < n; i++)
+    {
+        k = 0;
+        for(j = 0; j < len; j++)
+        {
+            if(*passcode[j] == logins[i][k])
             {
-               break;
+                k++;
+
+                if(k == 3)
+                {
+                    break;
+                }
             }
-         }
-      }
+        }
 
-      if(k < 3)
-      {
-         return 0;
-      }
-   }
+        if(k < 3)
+        {
+            return 0;
+        }
+    }
 
-   return 1;
+    return 1;
 }

C/p080.c

@@ -6,6 +6,8 @@
  * For the first one hundred natural numbers, find the total of the digital sums of the first one hundred decimal digits
  * for all the irrational square roots*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -16,65 +18,65 @@ int is_square(int n);
 
 int main(int argc, char **argv)
 {
-   int i, j, sum = 0;
-   char sqrt_digits[104];
-   double elapsed;
-   struct timespec start, end;
-   mpf_t sqrt;
+    int i, j, sum = 0;
+    char sqrt_digits[104];
+    double elapsed;
+    struct timespec start, end;
+    mpf_t sqrt;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Set the precision to 333 bits (should be enough for 100 decimal digits.*/
-   mpf_set_default_prec(333);
-   mpf_init(sqrt);
+    /* Set the precision to 333 bits (should be enough for 100 decimal digits.*/
+    mpf_set_default_prec(333);
+    mpf_init(sqrt);
 
-   for(i = 2; i < 100; i++)
-   {
-      if(is_square(i))
-      {
-         continue;
-      }
+    for(i = 2; i < 100; i++)
+    {
+        if(is_square(i))
+        {
+            continue;
+        }
 
-      /* Calculate the square root of the current number with the given precision
-       * and sum the digits to the total.*/
-      mpf_sqrt_ui(sqrt, i);
-      gmp_sprintf(sqrt_digits, "%.101Ff\n", sqrt);
-      sum += (sqrt_digits[0] - '0');
+        /* Calculate the square root of the current number with the given precision
+         * and sum the digits to the total.*/
+        mpf_sqrt_ui(sqrt, i);
+        gmp_sprintf(sqrt_digits, "%.101Ff\n", sqrt);
+        sum += (sqrt_digits[0] - '0');
 
-      for(j = 2; j < 101; j++)
-      {
-         sum += (sqrt_digits[j] - '0');
-      }
-   }
+        for(j = 2; j < 101; j++)
+        {
+            sum += (sqrt_digits[j] - '0');
+        }
+    }
 
-   mpf_clear(sqrt);
-   
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    mpf_clear(sqrt);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 80\n");
-   printf("Answer: %d\n", sum);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 80\n");
+    printf("Answer: %d\n", sum);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 int is_square(int n)
 {
-   int m;
-   double p;
+    int m;
+    double p;
 
-   p = sqrt(n);
-   m = p;
+    p = sqrt(n);
+    m = p;
 
-   if(p == m)
-   {
-      return 1;
-   }
-   else
-   {
-      return 0;
-   }
+    if(p == m)
+    {
+        return 1;
+    }
+    else
+    {
+        return 0;
+    }
 }

C/p081.c

@@ -1,6 +1,6 @@
 /* In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by only moving to the right and down,
  * is indicated in bold red and is equal to 2427.
- * 
+ *
  *       *131* 673   234   103    18
  *       *201* *96* *342*  965   150
  *        630  803  *746* *422*  111
@@ -10,6 +10,8 @@
  * Find the minimal path sum, in matrix.txt, a 31K text file containing a 80 by 80 matrix, from the top left to the bottom right
  * by only moving right and down.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -19,92 +21,92 @@ int sum_paths(int **matrix, int m, int n);
 
 int main(int argc, char **argv)
 {
-   int i, j, dist;
-   int **matrix;
-   double elapsed;
-   struct timespec start, end;
-   FILE *fp;
+    int i, j, dist;
+    int **matrix;
+    double elapsed;
+    struct timespec start, end;
+    FILE *fp;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((fp = fopen("matrix.txt", "r")) == NULL)
-   {
-      fprintf(stderr, "Error while opening file %s\n", "matrix.txt");
-      return 1;
-   }
+    if((fp = fopen("matrix.txt", "r")) == NULL)
+    {
+        fprintf(stderr, "Error while opening file %s\n", "matrix.txt");
+        return 1;
+    }
 
-   if((matrix = (int **)malloc(80*sizeof(int *))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((matrix = (int **)malloc(80*sizeof(int *))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   for(i = 0; i < 80; i++)
-   {
-      if((matrix[i] = (int *)malloc(80*sizeof(int))) == NULL)
-      {
-         fprintf(stderr, "Error while allocating memory\n");
-         return 1;
-      }
-   }
+    for(i = 0; i < 80; i++)
+    {
+        if((matrix[i] = (int *)malloc(80*sizeof(int))) == NULL)
+        {
+            fprintf(stderr, "Error while allocating memory\n");
+            return 1;
+        }
+    }
 
-   for(i = 0; i < 80; i++)
-   {
-      for(j = 0; j < 80; j++)
-      {
-         fscanf(fp, "%d,", &matrix[i][j]);
-      }
-   }
+    for(i = 0; i < 80; i++)
+    {
+        for(j = 0; j < 80; j++)
+        {
+            fscanf(fp, "%d,", &matrix[i][j]);
+        }
+    }
 
-   fclose(fp);
+    fclose(fp);
 
-   dist = sum_paths(matrix, 80, 80);
+    dist = sum_paths(matrix, 80, 80);
 
-   for(i = 0; i < 80; i++)
-   {
-      free(matrix[i]);
-   }
+    for(i = 0; i < 80; i++)
+    {
+        free(matrix[i]);
+    }
 
-   free(matrix);
-   
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    free(matrix);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 81\n");
-   printf("Answer: %d\n", dist);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 81\n");
+    printf("Answer: %d\n", dist);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 /* Get the shortest path starting from the bottom right corner
  * and going backwards.*/
 int sum_paths(int **matrix, int m, int n)
 {
-   int i, j;
+    int i, j;
 
-   for(i = m - 2; i >= 0; i--)
-   {
-      matrix[m-1][i] += matrix[m-1][i+1];
-      matrix[i][m-1] += matrix[i+1][m-1];
-   }
+    for(i = m - 2; i >= 0; i--)
+    {
+        matrix[m-1][i] += matrix[m-1][i+1];
+        matrix[i][m-1] += matrix[i+1][m-1];
+    }
 
-   for(i = m - 2; i >= 0; i--)
-   {
-      for(j = n - 2; j >= 0; j--)
-      {
-         if(matrix[i][j+1] > matrix[i+1][j])
-         {
-            matrix[i][j] += matrix[i+1][j];
-         }
-         else
-         {
-            matrix[i][j] += matrix[i][j+1];
-         }
-      }
-   }
+    for(i = m - 2; i >= 0; i--)
+    {
+        for(j = n - 2; j >= 0; j--)
+        {
+            if(matrix[i][j+1] > matrix[i+1][j])
+            {
+                matrix[i][j] += matrix[i+1][j];
+            }
+            else
+            {
+                matrix[i][j] += matrix[i][j+1];
+            }
+        }
+    }
 
-   return matrix[0][0];
+    return matrix[0][0];
 }

C/p082.c

@@ -7,10 +7,12 @@
  *       *201* *96* *342*  965   150
  *        630  803   746   422   111
  *        537  699   497   121   956
- *        805  732   524    37   331 
+ *        805  732   524    37   331
  *
  * Find the minimal path sum, in matrix.txt, a 31K text file containing a 80 by 80 matrix, from the left column to the right column.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -19,85 +21,85 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, min_path=INT_MAX;
-   int **matrix, **distances;
-   double elapsed;
-   struct timespec start, end;
-   FILE *fp;
+    int i, j, min_path=INT_MAX;
+    int **matrix, **distances;
+    double elapsed;
+    struct timespec start, end;
+    FILE *fp;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
-   
-   if((fp = fopen("matrix.txt", "r")) == NULL)
-   {
-      fprintf(stderr, "Error while opening file %s\n", "matrix.txt");
-      return 1;
-   }
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((matrix = (int **)malloc(80*sizeof(int *))) == NULL ||
-         (distances = (int **)malloc(80*sizeof(int *))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((fp = fopen("matrix.txt", "r")) == NULL)
+    {
+        fprintf(stderr, "Error while opening file %s\n", "matrix.txt");
+        return 1;
+    }
 
-   for(i = 0; i < 80; i++)
-   {
-      if((matrix[i] = (int *)malloc(80*sizeof(int))) == NULL ||
-            (distances[i] = (int *)malloc(80*sizeof(int))) == NULL)
-      {
-         fprintf(stderr, "Error while allocating memory\n");
-         return 1;
-      }
-   }
+    if((matrix = (int **)malloc(80*sizeof(int *))) == NULL ||
+            (distances = (int **)malloc(80*sizeof(int *))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   for(i = 0; i < 80; i++)
-   {
-      for(j = 0; j < 80; j++)
-      {
-         fscanf(fp, "%d,", &matrix[i][j]);
-      }
-   }
+    for(i = 0; i < 80; i++)
+    {
+        if((matrix[i] = (int *)malloc(80*sizeof(int))) == NULL ||
+                (distances[i] = (int *)malloc(80*sizeof(int))) == NULL)
+        {
+            fprintf(stderr, "Error while allocating memory\n");
+            return 1;
+        }
+    }
 
-   fclose(fp);
+    for(i = 0; i < 80; i++)
+    {
+        for(j = 0; j < 80; j++)
+        {
+            fscanf(fp, "%d,", &matrix[i][j]);
+        }
+    }
 
-   /* Use Dijkstra's algorithm starting from all possible nodes
-   * in the first column.*/
-   for(i = 0; i < 80; i++)
-   {
-      if(dijkstra(matrix, distances, 80, 80, 1, 0, i) == -1)
-      {
-         fprintf(stderr, "Error! Dijkstra function returned -1\n");
-         return 1;
-      }
+    fclose(fp);
 
-      /* For the current starting node, check if there is an ending node
-       * with a smaller path cost than the current minimum.*/
-      for(j = 0; j < 80; j++)
-      {
-         if(distances[j][79] < min_path)
-         {
-            min_path = distances[j][79];
-         }
-      }
-   }
+    /* Use Dijkstra's algorithm starting from all possible nodes
+    * in the first column.*/
+    for(i = 0; i < 80; i++)
+    {
+        if(dijkstra(matrix, distances, 80, 80, 1, 0, i) == -1)
+        {
+            fprintf(stderr, "Error! Dijkstra function returned -1\n");
+            return 1;
+        }
 
-   for(i = 0; i < 80; i++)
-   {
-      free(matrix[i]);
-      free(distances[i]);
-   }
+        /* For the current starting node, check if there is an ending node
+         * with a smaller path cost than the current minimum.*/
+        for(j = 0; j < 80; j++)
+        {
+            if(distances[j][79] < min_path)
+            {
+                min_path = distances[j][79];
+            }
+        }
+    }
 
-   free(matrix);
-   free(distances);
-   
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    for(i = 0; i < 80; i++)
+    {
+        free(matrix[i]);
+        free(distances[i]);
+    }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    free(matrix);
+    free(distances);
 
-   printf("Project Euler, Problem 82\n");
-   printf("Answer: %d\n", min_path);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   return 0;
+    printf("Project Euler, Problem 82\n");
+    printf("Answer: %d\n", min_path);
+
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }

C/p083.c

@@ -2,7 +2,7 @@
  *
  * In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by moving left, right, up, and down,
  * is indicated in bold red and is equal to 2297.
- * 
+ *
  *       *131* 673  *234* *103*  *18*
  *       *201* *96* *342*  965  *150*
  *        630  803   746  *422* *111*
@@ -12,6 +12,8 @@
  * Find the minimal path sum, in matrix.txt, a 31K text file containing a 80 by 80 matrix,
  * from the top left to the bottom right by moving left, right, up, and down.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -20,73 +22,73 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, dist;
-   int **matrix, **distances;
-   double elapsed;
-   struct timespec start, end;
-   FILE *fp;
+    int i, j, dist;
+    int **matrix, **distances;
+    double elapsed;
+    struct timespec start, end;
+    FILE *fp;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((fp = fopen("matrix.txt", "r")) == NULL)
-   {
-      fprintf(stderr, "Error while opening file %s\n", "matrix.txt");
-      return 1;
-   }
+    if((fp = fopen("matrix.txt", "r")) == NULL)
+    {
+        fprintf(stderr, "Error while opening file %s\n", "matrix.txt");
+        return 1;
+    }
 
-   if((matrix = (int **)malloc(80*sizeof(int *))) == NULL ||
-         (distances = (int **)malloc(80*sizeof(int *))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((matrix = (int **)malloc(80*sizeof(int *))) == NULL ||
+            (distances = (int **)malloc(80*sizeof(int *))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   for(i = 0; i < 80; i++)
-   {
-      if((matrix[i] = (int *)malloc(80*sizeof(int))) == NULL ||
-            (distances[i] = (int *)malloc(80*sizeof(int))) == NULL)
-      {
-         fprintf(stderr, "Error while allocating memory\n");
-         return 1;
-      }
-   }
+    for(i = 0; i < 80; i++)
+    {
+        if((matrix[i] = (int *)malloc(80*sizeof(int))) == NULL ||
+                (distances[i] = (int *)malloc(80*sizeof(int))) == NULL)
+        {
+            fprintf(stderr, "Error while allocating memory\n");
+            return 1;
+        }
+    }
 
-   for(i = 0; i < 80; i++)
-   {
-      for(j = 0; j < 80; j++)
-      {
-         fscanf(fp, "%d,", &matrix[i][j]);
-      }
-   }
+    for(i = 0; i < 80; i++)
+    {
+        for(j = 0; j < 80; j++)
+        {
+            fscanf(fp, "%d,", &matrix[i][j]);
+        }
+    }
 
-   fclose(fp);
+    fclose(fp);
 
-   /* Use Dijkstra's algorithm to find the minimum path.*/
-   if(dijkstra(matrix, distances, 80, 80, 1, 1, 0) == -1)
-   {
-      fprintf(stderr, "Error! Dijkstra function returned -1\n");
-      return 1;
-   }
+    /* Use Dijkstra's algorithm to find the minimum path.*/
+    if(dijkstra(matrix, distances, 80, 80, 1, 1, 0) == -1)
+    {
+        fprintf(stderr, "Error! Dijkstra function returned -1\n");
+        return 1;
+    }
 
-   dist = distances[79][79];
+    dist = distances[79][79];
 
-   for(i = 0; i < 80; i++)
-   {
-      free(matrix[i]);
-      free(distances[i]);
-   }
+    for(i = 0; i < 80; i++)
+    {
+        free(matrix[i]);
+        free(distances[i]);
+    }
 
-   free(matrix);
-   free(distances);
-   
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    free(matrix);
+    free(distances);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 83\n");
-   printf("Answer: %d\n", dist);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 83\n");
+    printf("Answer: %d\n", dist);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }

C/p084.c

@@ -52,7 +52,7 @@
  * Go back 3 squares.
  *
  * The heart of this problem concerns the likelihood of visiting a particular square. That is, the probability of finishing
- * at that square after a roll. For this reason it should be clear that, with the exception of G2J for which the probability 
+ * at that square after a roll. For this reason it should be clear that, with the exception of G2J for which the probability
  * of finishing on it is zero, the CH squares will have the lowest probabilities, as 5/8 request a movement to another square,
  * and it is the final square that the player finishes at on each roll that we are interested in. We shall make no distinction
  * between "Just Visiting" and being sent to JAIL, and we shall also ignore the rule about requiring a double to "get out of jail",
@@ -66,6 +66,8 @@
  *
  * If, instead of using two 6-sided dice, two 4-sided dice are used, find the six-digit modal string.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
@@ -75,8 +77,9 @@
 
 /* Define the squares, the type of Community Chest cards and the type of Chance cards.*/
 typedef enum {GO, A1, CC1, A2, T1, R1, B1, CH1, B2, B3, JAIL,
-   C1, U1, C2, C3, R2, D1, CC2, D2, D3, FP, E1, CH2, E2, E3, R3, F1,
-   F2, U2, F3, G2J, G1, G2, CC3, G3, R4, CH3, H1, T2, H2} squares;
+              C1, U1, C2, C3, R2, D1, CC2, D2, D3, FP, E1, CH2, E2, E3, R3, F1,
+              F2, U2, F3, G2J, G1, G2, CC3, G3, R4, CH3, H1, T2, H2
+             } squares;
 typedef enum {GO_cc, JAIL_cc, NOP_cc} cc_card;
 typedef enum {GO_ch, JAIL_ch, C1_ch, E3_ch, H2_ch, R1_ch, R_ch1, R_ch2, U_ch, M3, NOP_ch} ch_card;
 
@@ -86,261 +89,261 @@ void play(long int *count_squares);
 
 int main(int argc, char **argv)
 {
-   int i, j, max, max_i;
-   long int count_squares[40] = {0};
-   char sq[3], *res;
-   double elapsed;
-   struct timespec start, end;
+    int i, j, max, max_i;
+    long int count_squares[40] = {0};
+    char sq[3], *res;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   srand(time(NULL));
+    srand(time(NULL));
 
-   /* Play Monopoly.*/
-   play(count_squares);
+    /* Play Monopoly.*/
+    play(count_squares);
 
-   if((res = (char *)malloc(7*sizeof(char))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((res = (char *)malloc(7*sizeof(char))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   /* Find the three squares with maximum count.*/ 
-   for(i = 0; i < 3; i++)
-   {
-      max = -1;
+    /* Find the three squares with maximum count.*/
+    for(i = 0; i < 3; i++)
+    {
+        max = -1;
 
-      for(j = 0; j < 40; j++)
-      {
-         if(count_squares[j] > max)
-         {
-            max = count_squares[j];
-            max_i = j;
-         }
-      }
+        for(j = 0; j < 40; j++)
+        {
+            if(count_squares[j] > max)
+            {
+                max = count_squares[j];
+                max_i = j;
+            }
+        }
 
-      if(max_i < 10)
-      {
-         sprintf(sq, "0%d", max_i);
-      }
-      else
-      {
-         sprintf(sq, "%d", max_i);
-      }
+        if(max_i < 10)
+        {
+            sprintf(sq, "0%d", max_i);
+        }
+        else
+        {
+            sprintf(sq, "%d", max_i);
+        }
 
-      res = strcat(res, sq);
+        res = strcat(res, sq);
 
-      count_squares[max_i] = 0;
-   }
+        count_squares[max_i] = 0;
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 84\n");
-   printf("Answer: %s\n", res);
+    printf("Project Euler, Problem 84\n");
+    printf("Answer: %s\n", res);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   free(res);
+    free(res);
 
-   return 0;
+    return 0;
 }
 
 /* Shuffle the Community Chest cards.*/
 void shuffle_cc(cc_card cc[])
 {
-   int i;
-   cc_card c;
+    int i;
+    cc_card c;
 
-   for(i = 0; i < 16; i++)
-   {
-      cc[i] = NOP_cc;
-   }
+    for(i = 0; i < 16; i++)
+    {
+        cc[i] = NOP_cc;
+    }
 
-   c = GO_cc;
+    c = GO_cc;
 
-   do
-   {
-      do
-      {
-         i = rand() % 16;
-      }while(cc[i] != NOP_cc);
+    do
+    {
+        do
+        {
+            i = rand() % 16;
+        } while(cc[i] != NOP_cc);
 
-      cc[i] = c;
-      c++;
-   }while(c < NOP_cc);
+        cc[i] = c;
+        c++;
+    } while(c < NOP_cc);
 }
 
 /* Shuffle the Chance cards.*/
 void shuffle_ch(ch_card ch[])
 {
-   int i;
-   ch_card h;
+    int i;
+    ch_card h;
 
-   for(i = 0; i < 16; i++)
-   {
-      ch[i] = NOP_ch;
-   }
+    for(i = 0; i < 16; i++)
+    {
+        ch[i] = NOP_ch;
+    }
 
-   h = GO_ch;
+    h = GO_ch;
 
-   do
-   {
-      do
-      {
-         i = rand() % 16;
-      }while(ch[i] != NOP_ch);
+    do
+    {
+        do
+        {
+            i = rand() % 16;
+        } while(ch[i] != NOP_ch);
 
-      ch[i] = h;
-      h++;
-   }while(h < NOP_ch);
+        ch[i] = h;
+        h++;
+    } while(h < NOP_ch);
 }
 
 /* Function to play Monopoly, rolling the dice N times.*/
 void play(long int *count_squares)
 {
-   int i, d1, d2, cc_i, ch_i, count_doubles;
-   squares current;
-   cc_card cc[16];
-   ch_card ch[16];
-  
-   /* Start from GO.*/
-   current = GO;
-   count_squares[GO]++;
-   cc_i = 0;
-   ch_i = 0;
-   count_doubles = 0;
+    int i, d1, d2, cc_i, ch_i, count_doubles;
+    squares current;
+    cc_card cc[16];
+    ch_card ch[16];
 
-   shuffle_cc(cc);
-   shuffle_ch(ch);
+    /* Start from GO.*/
+    current = GO;
+    count_squares[GO]++;
+    cc_i = 0;
+    ch_i = 0;
+    count_doubles = 0;
 
-   for(i = 0; i < N; i++)
-   {
-      /* Roll the dice.*/
-      d1 = 1 + rand() % 4;
-      d2 = 1 + rand() % 4;
+    shuffle_cc(cc);
+    shuffle_ch(ch);
 
-      /* Check if it's a double. Increment the doubles counter if yes,
-       * reset it if no.*/
-      if(d1 == d2)
-      {
-         count_doubles++;
-      }
-      else
-      {
-         count_doubles=0;
-      }
+    for(i = 0; i < N; i++)
+    {
+        /* Roll the dice.*/
+        d1 = 1 + rand() % 4;
+        d2 = 1 + rand() % 4;
 
-      /* If this is the third double in a row, go to JAIL and reset
-       * the doubles counter.*/
-      if(count_doubles == 3)
-      {
-         current = JAIL;
-         count_doubles = 0;
-      }
-      else
-      {
-         /* Move based on the dice roll.*/
-         current = (current + d1 + d2) % 40;
+        /* Check if it's a double. Increment the doubles counter if yes,
+         * reset it if no.*/
+        if(d1 == d2)
+        {
+            count_doubles++;
+        }
+        else
+        {
+            count_doubles=0;
+        }
 
-         /* If you get to the Go to JAIL square, go to JAIL.*/
-         if(current == G2J)
-         {
+        /* If this is the third double in a row, go to JAIL and reset
+         * the doubles counter.*/
+        if(count_doubles == 3)
+        {
             current = JAIL;
-         }
-         /* If you get to one of the Community Chest squares, get a CC card.*/
-         else if(current == CC1 || current == CC2 || current == CC3)
-         {
-            /* Go to GO.*/
-            if(cc[cc_i] == GO_cc)
-            {
-               current = GO;
-            }
-            /* Go to JAIL.*/
-            else if(cc[cc_i] == JAIL_cc)
-            {
-               current = JAIL;
-            }
-            /* Otherwise do nothing.*/
+            count_doubles = 0;
+        }
+        else
+        {
+            /* Move based on the dice roll.*/
+            current = (current + d1 + d2) % 40;
 
-            cc_i = (cc_i + 1) % 16;
-         }
-         /* If you get to one of the Chance squares, get a CH card.*/
-         else if(current == CH1 || current == CH2 || current == CH3)
-         {
-            /* Go to GO.*/
-            if(ch[ch_i] == GO_ch)
+            /* If you get to the Go to JAIL square, go to JAIL.*/
+            if(current == G2J)
             {
-               current = GO;
+                current = JAIL;
             }
-            /* Go to JAIL.*/
-            else if(ch[ch_i] == JAIL_ch)
+            /* If you get to one of the Community Chest squares, get a CC card.*/
+            else if(current == CC1 || current == CC2 || current == CC3)
             {
-               current = JAIL;
-            }
-            /* Go to C1.*/
-            else if(ch[ch_i] == C1_ch)
-            {
-               current = C1;
-            }
-            /* Go to E3.*/
-            else if(ch[ch_i] == E3_ch)
-            {
-               current = E3;
-            }
-            /* Go to H2.*/
-            else if(ch[ch_i] == H2_ch)
-            {
-               current = H2;
-            }
-            /* Go to R1.*/
-            else if(ch[ch_i] == R1_ch)
-            {
-               current = R1;
-            }
-            /* Go to the next Railway station.*/
-            else if(ch[ch_i] == R_ch1 || ch[ch_i] == R_ch2)
-            {
-               do
-               {
-                  current = (current + 1) % 40;
-               }while(current != R1 && current != R2 && current != R3 && current != R4);
-            }
-            /* Go to the next Utility company.*/
-            else if(ch[ch_i] == U_ch)
-            {
-               do
-               {
-                  current = (current + 1) % 40;
-               }while(current != U1 && current != U2);
-            }
-            /* Go back three squares.*/
-            else if(ch[ch_i] == M3)
-            {
-               current = current - 3;
+                /* Go to GO.*/
+                if(cc[cc_i] == GO_cc)
+                {
+                    current = GO;
+                }
+                /* Go to JAIL.*/
+                else if(cc[cc_i] == JAIL_cc)
+                {
+                    current = JAIL;
+                }
+                /* Otherwise do nothing.*/
 
-               /* If you get to Community Chest 3, get a card.*/
-               if(current == CC3)
-               {
-                  if(cc[cc_i] == GO_cc)
-                  {
-                     current = GO;
-                  }
-                  else if(cc[cc_i] == JAIL_cc)
-                  {
-                     current=JAIL;
-                  }
-
-                  cc_i = (cc_i + 1) % 16;
-               }
+                cc_i = (cc_i + 1) % 16;
             }
+            /* If you get to one of the Chance squares, get a CH card.*/
+            else if(current == CH1 || current == CH2 || current == CH3)
+            {
+                /* Go to GO.*/
+                if(ch[ch_i] == GO_ch)
+                {
+                    current = GO;
+                }
+                /* Go to JAIL.*/
+                else if(ch[ch_i] == JAIL_ch)
+                {
+                    current = JAIL;
+                }
+                /* Go to C1.*/
+                else if(ch[ch_i] == C1_ch)
+                {
+                    current = C1;
+                }
+                /* Go to E3.*/
+                else if(ch[ch_i] == E3_ch)
+                {
+                    current = E3;
+                }
+                /* Go to H2.*/
+                else if(ch[ch_i] == H2_ch)
+                {
+                    current = H2;
+                }
+                /* Go to R1.*/
+                else if(ch[ch_i] == R1_ch)
+                {
+                    current = R1;
+                }
+                /* Go to the next Railway station.*/
+                else if(ch[ch_i] == R_ch1 || ch[ch_i] == R_ch2)
+                {
+                    do
+                    {
+                        current = (current + 1) % 40;
+                    } while(current != R1 && current != R2 && current != R3 && current != R4);
+                }
+                /* Go to the next Utility company.*/
+                else if(ch[ch_i] == U_ch)
+                {
+                    do
+                    {
+                        current = (current + 1) % 40;
+                    } while(current != U1 && current != U2);
+                }
+                /* Go back three squares.*/
+                else if(ch[ch_i] == M3)
+                {
+                    current = current - 3;
 
-            ch_i = (ch_i + 1) % 16;
-         }
-      }
+                    /* If you get to Community Chest 3, get a card.*/
+                    if(current == CC3)
+                    {
+                        if(cc[cc_i] == GO_cc)
+                        {
+                            current = GO;
+                        }
+                        else if(cc[cc_i] == JAIL_cc)
+                        {
+                            current=JAIL;
+                        }
 
-      /* Increase the counter for the current square.*/
-      count_squares[current]++;
-   }
+                        cc_i = (cc_i + 1) % 16;
+                    }
+                }
+
+                ch_i = (ch_i + 1) % 16;
+            }
+        }
+
+        /* Increase the counter for the current square.*/
+        count_squares[current]++;
+    }
 }

C/p085.c

@@ -2,6 +2,8 @@
  *
  * Although there exists no rectangular grid that contains exactly two million rectangles, find the area of the grid with the nearest solution.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -10,41 +12,41 @@
 
 int main(int argc, char **argv)
 {
-   int i, j, n, diff, min_diff = INT_MAX, area;
-   double elapsed;
-   struct timespec start, end;
+    int i, j, n, diff, min_diff = INT_MAX, area;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   for(i = 1; i < 100; i++)
-   {
-      for(j = 1; j <= i; j++)
-      {
-         /* In a 2x3 grid, we can take rectangles of height 2 in 3 ways
-          * (two rectangles of height one and one of height 2). For the
-          * width, can take 6 rectangles (3 of width 1, 2 of width 2 and
-          * 1 of width 3). The total is 6x3=18 rectagles. 
-          * Extending to mxn, we can take (m+m-1+m-2+...+1)x(n+n-1+n-2+...+1)= 
-          * m(m+1)/2*n(n+1)/2=m(m+1)*n(n+1)/4 rectangles.*/
-         n = ( i * (i + 1) * j * (j + 1)) / 4;
-         diff = abs(2000000 - n);
+    for(i = 1; i < 100; i++)
+    {
+        for(j = 1; j <= i; j++)
+        {
+            /* In a 2x3 grid, we can take rectangles of height 2 in 3 ways
+             * (two rectangles of height one and one of height 2). For the
+             * width, can take 6 rectangles (3 of width 1, 2 of width 2 and
+             * 1 of width 3). The total is 6x3=18 rectagles.
+             * Extending to mxn, we can take (m+m-1+m-2+...+1)x(n+n-1+n-2+...+1)=
+             * m(m+1)/2*n(n+1)/2=m(m+1)*n(n+1)/4 rectangles.*/
+            n = ( i * (i + 1) * j * (j + 1)) / 4;
+            diff = abs(2000000 - n);
 
-         if(diff < min_diff)
-         {
-            min_diff = diff;
-            area = i * j;
-         }
-      }
-   }
-   
-   clock_gettime(CLOCK_MONOTONIC, &end);
+            if(diff < min_diff)
+            {
+                min_diff = diff;
+                area = i * j;
+            }
+        }
+    }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 85\n");
-   printf("Answer: %d\n", area);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 85\n");
+    printf("Answer: %d\n", area);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }

C/p086.c

@@ -9,6 +9,8 @@
  *
  * Find the least value of M such that the number of solutions first exceeds one million.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -16,41 +18,41 @@
 
 int main(int argc, char **argv)
 {
-   int a, b, c, count = 0;
-   double d, elapsed;
-   struct timespec start, end;
+    int a, b, c, count = 0;
+    double d, elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   a = 0;
+    a = 0;
 
-   while(count <= 1000000)
-   {
-      a++;
+    while(count <= 1000000)
+    {
+        a++;
 
-      for(b = 1; b <= a; b++)
-      {
-         for(c = 1; c <= b; c++)
-         {
-            /* Unfolding the cuboid, it's obvious that the shortest path
-             * is the hypotenuse of a triangle, and the catheti are the
-             * longest side of the cubois and the sum of the other two sides.*/
-            d = sqrt(a*a+(b+c)*(b+c));
+        for(b = 1; b <= a; b++)
+        {
+            for(c = 1; c <= b; c++)
+            {
+                /* Unfolding the cuboid, it's obvious that the shortest path
+                 * is the hypotenuse of a triangle, and the catheti are the
+                 * longest side of the cubois and the sum of the other two sides.*/
+                d = sqrt(a*a+(b+c)*(b+c));
 
-            if(d == (int)d)
-               count++;
-         }
-      }
-   }
+                if(d == (int)d)
+                    count++;
+            }
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 86\n");
-   printf("Answer: %d\n", a);
+    printf("Project Euler, Problem 86\n");
+    printf("Answer: %d\n", a);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p087.c

@@ -8,6 +8,8 @@
  *
  * How many numbers below fifty million can be expressed as the sum of a prime square, prime cube, and prime fourth power?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -23,71 +25,71 @@ int *primes;
 
 int main(int argc, char **argv)
 {
-   int count = 0;
-   int *numbers;
-   long int i, j, k, n;
-   double elapsed;
-   struct timespec start, end;
+    int count = 0;
+    int *numbers;
+    long int i, j, k, n;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Generate primes up to the square root of the limit.*/
-   if((primes = sieve(SQRT_N)) == NULL)
-   {
-      fprintf(stderr, "Error! Sieve function returned NULL\n");
-      return 1;
-   }
+    /* Generate primes up to the square root of the limit.*/
+    if((primes = sieve(SQRT_N)) == NULL)
+    {
+        fprintf(stderr, "Error! Sieve function returned NULL\n");
+        return 1;
+    }
 
-   if((numbers = (int *)calloc(N, sizeof(int))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((numbers = (int *)calloc(N, sizeof(int))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   /* If i>sqrt(n), i*i will be >n, so n won't be a sum of
-    * a square, cube and fourth power.*/
-   for(i = 2; i <= SQRT_N; i++)
-   {
-      /* If i is not prime, try next number.*/
-      if(primes[i])
-      {
-         /* If j>cubic_root(n), j*j*j will be >n.*/
-         for(j = 2; j <= RAD3_N; j++)
-         {
-            if(primes[j])
+    /* If i>sqrt(n), i*i will be >n, so n won't be a sum of
+     * a square, cube and fourth power.*/
+    for(i = 2; i <= SQRT_N; i++)
+    {
+        /* If i is not prime, try next number.*/
+        if(primes[i])
+        {
+            /* If j>cubic_root(n), j*j*j will be >n.*/
+            for(j = 2; j <= RAD3_N; j++)
             {
-               /* If k>fourth_root(n), k*k*k*k will be >n.*/
-               for(k = 2; k <= RAD4_N; k++)
-               {
-                  if(primes[k])
-                  {
-                     n = i * i + j * j * j + k * k * k * k;
+                if(primes[j])
+                {
+                    /* If k>fourth_root(n), k*k*k*k will be >n.*/
+                    for(k = 2; k <= RAD4_N; k++)
+                    {
+                        if(primes[k])
+                        {
+                            n = i * i + j * j * j + k * k * k * k;
 
-                     /* Check if the number found is lower than the limit,
-                      * and make sure it hasn't been found before.*/
-                     if(n < N && numbers[n] == 0)
-                     {
-                        count++;
-                        numbers[n] = 1;
-                     }
-                  }
-               }
+                            /* Check if the number found is lower than the limit,
+                             * and make sure it hasn't been found before.*/
+                            if(n < N && numbers[n] == 0)
+                            {
+                                count++;
+                                numbers[n] = 1;
+                            }
+                        }
+                    }
+                }
             }
-         }
-      }
-   }
+        }
+    }
 
-   free(primes);
-   free(numbers);
-   
-   clock_gettime(CLOCK_MONOTONIC, &end);
-   
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
-   
-   printf("Project Euler, Problem 87\n");
-   printf("Answer: %d\n", count);
+    free(primes);
+    free(numbers);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
-   
-   return 0;
+    clock_gettime(CLOCK_MONOTONIC, &end);
+
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+
+    printf("Project Euler, Problem 87\n");
+    printf("Answer: %d\n", count);
+
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }

C/p089.c

@@ -19,6 +19,8 @@
  *
  * Note: You can assume that all the Roman numerals in the file contain no more than four consecutive identical units.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
@@ -28,81 +30,81 @@ int reduce(char *numeral, int l);
 
 int main(int argc, char **argv)
 {
-   int i, l, count = 0, count_reduced = 0;
-   char *numerals[1000];
-   double elapsed;
-   struct timespec start, end;
-   FILE *fp;
+    int i, l, count = 0, count_reduced = 0;
+    char *numerals[1000];
+    double elapsed;
+    struct timespec start, end;
+    FILE *fp;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((fp = fopen("roman.txt", "r")) == NULL)
-   {
-      fprintf(stderr, "Error while opening file %s\n", "roman.txt");
-      return 1;
-   }
+    if((fp = fopen("roman.txt", "r")) == NULL)
+    {
+        fprintf(stderr, "Error while opening file %s\n", "roman.txt");
+        return 1;
+    }
 
-   /* Get the roman numerals from file, for each find its length, then
-    * reduce it and find the new length. At the end, find the difference
-    * of the two lengths.*/
-   for(i = 0; i < 1000; i++)
-   {
-      fscanf(fp, "%ms", &numerals[i]);
-      l = strlen(numerals[i]);
-      count += l;
-      l = reduce(numerals[i], l);
-      count_reduced += l;
-      free(numerals[i]);
-   }
+    /* Get the roman numerals from file, for each find its length, then
+     * reduce it and find the new length. At the end, find the difference
+     * of the two lengths.*/
+    for(i = 0; i < 1000; i++)
+    {
+        fscanf(fp, "%ms", &numerals[i]);
+        l = strlen(numerals[i]);
+        count += l;
+        l = reduce(numerals[i], l);
+        count_reduced += l;
+        free(numerals[i]);
+    }
 
-   fclose(fp);
+    fclose(fp);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 89\n");
-   printf("Answer: %d\n", count-count_reduced);
+    printf("Project Euler, Problem 89\n");
+    printf("Answer: %d\n", count-count_reduced);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int reduce(char *numeral, int l)
 {
-   /* DCCCC = CM.*/
-   if(strstr(numeral, "DCCCC") != NULL)
-   {
-      l -= 3;
-   }
-   /* CCCC = CD.*/
-   else if(strstr(numeral, "CCCC") != NULL)
-   {
-      l -= 2;
-   }
+    /* DCCCC = CM.*/
+    if(strstr(numeral, "DCCCC") != NULL)
+    {
+        l -= 3;
+    }
+    /* CCCC = CD.*/
+    else if(strstr(numeral, "CCCC") != NULL)
+    {
+        l -= 2;
+    }
 
-   /* LXXXX = XC.*/
-   if(strstr(numeral, "LXXXX") != NULL)
-   {
-      l -= 3;
-   }
-   /* XXXX = XL.*/
-   else if(strstr(numeral, "XXXX") != NULL)
-   {
-      l -= 2;
-   }
+    /* LXXXX = XC.*/
+    if(strstr(numeral, "LXXXX") != NULL)
+    {
+        l -= 3;
+    }
+    /* XXXX = XL.*/
+    else if(strstr(numeral, "XXXX") != NULL)
+    {
+        l -= 2;
+    }
 
-   /* VIIII = IX.*/
-   if(strstr(numeral, "VIIII") != NULL)
-   {
-      l -= 3;
-   }
-   /* IIII = IV.*/
-   else if(strstr(numeral, "IIII") != NULL)
-   {
-      l -= 2;
-   }
+    /* VIIII = IX.*/
+    if(strstr(numeral, "VIIII") != NULL)
+    {
+        l -= 3;
+    }
+    /* IIII = IV.*/
+    else if(strstr(numeral, "IIII") != NULL)
+    {
+        l -= 2;
+    }
 
-   return l;
+    return l;
 }

C/p092.c

@@ -10,6 +10,8 @@
  *
  * How many starting numbers below ten million will arrive at 89?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -22,63 +24,63 @@ int chain(int n);
 
 int main(int argc, char **argv)
 {
-   int i, count = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i, count = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   for(i = 1; i < N; i++)
-   {
-      /* Simply create a chain for each number and check if it ends at 89, saving
-       * the result so it can be reused.*/
-      if((chains[i] = chain(i)) == 89)
-      {
-         count++;
-      }
-   }
+    for(i = 1; i < N; i++)
+    {
+        /* Simply create a chain for each number and check if it ends at 89, saving
+         * the result so it can be reused.*/
+        if((chains[i] = chain(i)) == 89)
+        {
+            count++;
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 92\n");
-   printf("Answer: %d\n", count);
+    printf("Project Euler, Problem 92\n");
+    printf("Answer: %d\n", count);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 /* Recursively find the chain for n.*/
 int chain(int n)
 {
-   int tmp = 0, digit;
+    int tmp = 0, digit;
 
-   /* If n=1 or n=89, we reached the end of the chain.*/
-   if(n == 1)
-   {
-      return 1;
-   }
+    /* If n=1 or n=89, we reached the end of the chain.*/
+    if(n == 1)
+    {
+        return 1;
+    }
 
-   if(n == 89)
-   {
-      return 89;
-   }
+    if(n == 89)
+    {
+        return 89;
+    }
 
-   /* If the chain for the current n has already been found,
-    * return the value.*/
-   if(chains[n] != 0)
-   {
-      return chains[n];
-   }
+    /* If the chain for the current n has already been found,
+     * return the value.*/
+    if(chains[n] != 0)
+    {
+        return chains[n];
+    }
 
-   while(n > 0)
-   {
-      digit = n % 10;
-      tmp += digit * digit;
-      n /= 10;
-   }
+    while(n > 0)
+    {
+        digit = n % 10;
+        tmp += digit * digit;
+        n /= 10;
+    }
 
-   return chain(tmp);
+    return chain(tmp);
 }

C/p095.c

@@ -12,6 +12,8 @@
  *
  * Find the smallest member of the longest amicable chain with no element exceeding one million.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -26,99 +28,99 @@ int divisors[N] = {0};
 
 int main(int argc, char **argv)
 {
-   int i, min = 0, min_tmp, length, l_max = 0;
-   int chain[100];
-   double elapsed;
-   struct timespec start, end;
+    int i, min = 0, min_tmp, length, l_max = 0;
+    int chain[100];
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   for(i = 4; i <= N; i++)
-   {
-      /* Calculate the divisors of i and save it. Ii is equal to the sum of its proper divisors, 
-       * the length of the chain is 1 and we don't need to check it.*/
-      if((divisors[i] = sum_of_divisors(i, 1)) == i)
-      {
-         continue;
-      }
-      else
-      {
-         min_tmp = i;
-         length = sociable_chain(i, chain, 0, &min_tmp);
-      }
-      
-      if(length > l_max)
-      {
-         l_max = length;
-         min = min_tmp;
-      }
-   }
+    for(i = 4; i <= N; i++)
+    {
+        /* Calculate the divisors of i and save it. Ii is equal to the sum of its proper divisors,
+         * the length of the chain is 1 and we don't need to check it.*/
+        if((divisors[i] = sum_of_divisors(i, 1)) == i)
+        {
+            continue;
+        }
+        else
+        {
+            min_tmp = i;
+            length = sociable_chain(i, chain, 0, &min_tmp);
+        }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+        if(length > l_max)
+        {
+            l_max = length;
+            min = min_tmp;
+        }
+    }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 95\n");
-   printf("Answer: %d\n", min);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 95\n");
+    printf("Answer: %d\n", min);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 /* Function to recursively find the length of the chain.*/
 int sociable_chain(int i, int *chain, int l, int *min)
 {
-   int n;
+    int n;
 
-   /* Save current number in the chain.*/
-   chain[l] = i;
+    /* Save current number in the chain.*/
+    chain[l] = i;
 
-   /* If we reached 1, the chain will never go anywhere.*/
-   if(i == 1)
-   {
-      return -1;
-   }
+    /* If we reached 1, the chain will never go anywhere.*/
+    if(i == 1)
+    {
+        return -1;
+    }
 
-   /* Calculate the divisors of i, or retrieve the value if previously calculated.*/
-   if(divisors[i] != 0)
-   {
-      n = divisors[i];
-   }
-   else
-   {
-      n = sum_of_divisors(i, 1);
-      divisors[i] = n;
-   }
-   
-   /* We are looking for chain where no value is greater than 1000000.*/
-   if(n > N)
-   {
-      return -1;
-   }
+    /* Calculate the divisors of i, or retrieve the value if previously calculated.*/
+    if(divisors[i] != 0)
+    {
+        n = divisors[i];
+    }
+    else
+    {
+        n = sum_of_divisors(i, 1);
+        divisors[i] = n;
+    }
 
-   /* If the next number in the chain is equal to the starting one, the chain is finished.*/
-   if(n == chain[0])
-   {
-      return l + 1;
-   }
+    /* We are looking for chain where no value is greater than 1000000.*/
+    if(n > N)
+    {
+        return -1;
+    }
 
-   /* Check if n is equal to another value of the chain different from start. If it is, the
-    * chain is stuck in a loop that will not return to the starting number.*/
-   for(i = l; i > 0; i--)
-   {
-      if(n == chain[i])
-      {
-         return -1;
-      }
-   }
+    /* If the next number in the chain is equal to the starting one, the chain is finished.*/
+    if(n == chain[0])
+    {
+        return l + 1;
+    }
 
-   /* If the next value is smaller than the minimum value of the chain,
-    * update the minimum.*/
-   if(n < *min)
-   {
-      *min = n;
-   }
+    /* Check if n is equal to another value of the chain different from start. If it is, the
+     * chain is stuck in a loop that will not return to the starting number.*/
+    for(i = l; i > 0; i--)
+    {
+        if(n == chain[i])
+        {
+            return -1;
+        }
+    }
 
-   return sociable_chain(n, chain, l+1, min);
+    /* If the next value is smaller than the minimum value of the chain,
+     * update the minimum.*/
+    if(n < *min)
+    {
+        *min = n;
+    }
+
+    return sociable_chain(n, chain, l+1, min);
 }

C/p096.c

@@ -25,6 +25,8 @@
  *    By solving all fifty puzzles find the sum of the 3-digit numbers found in the top left corner of each solution grid;
  *    for example, 483 is the 3-digit number found in the top left corner of the solution grid above.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -35,266 +37,266 @@ int check(int grid[][9], int i, int j, int n);
 
 int main(int argc, char **argv)
 {
-   char dummy[10], dummy_c;
-   int i, j, sum = 0, partial;
-   int grid[9][9];
-   double elapsed;
-   struct timespec start, end;
-   FILE *fp;
+    char dummy[10], dummy_c;
+    int i, j, sum = 0, partial;
+    int grid[9][9];
+    double elapsed;
+    struct timespec start, end;
+    FILE *fp;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((fp = fopen("sudoku.txt", "r")) == NULL)
-   {
-      fprintf(stderr, "Error while opening file %s\n", "sudoku.txt");
-      return 1;
-   }
+    if((fp = fopen("sudoku.txt", "r")) == NULL)
+    {
+        fprintf(stderr, "Error while opening file %s\n", "sudoku.txt");
+        return 1;
+    }
 
-   while(!feof(fp))
-   {
-      fgets(dummy, sizeof(dummy), fp);
+    while(!feof(fp))
+    {
+        fgets(dummy, sizeof(dummy), fp);
 
-      for(i = 0; i < 9; i++)
-      {
-         for(j = 0; j < 9; j++)
-         {
-            fscanf(fp, "%c", &dummy_c); 
-            grid[i][j] = (int)dummy_c - '0';
-         }
-         
-         fscanf(fp, "%c", &dummy_c);
-      }
+        for(i = 0; i < 9; i++)
+        {
+            for(j = 0; j < 9; j++)
+            {
+                fscanf(fp, "%c", &dummy_c);
+                grid[i][j] = (int)dummy_c - '0';
+            }
 
-      /* For each sudoku in the file, solve it and sum the number in the
-       * top left corner to the total.*/
-      if(!solve_sudoku(grid))
-      {
-         printf("Bad sudoku grid\n");
-         return 1;
-      }
+            fscanf(fp, "%c", &dummy_c);
+        }
 
-      partial = grid[0][0] * 100 + grid[0][1] * 10 + grid[0][2];
-      sum += partial;
-   }
+        /* For each sudoku in the file, solve it and sum the number in the
+         * top left corner to the total.*/
+        if(!solve_sudoku(grid))
+        {
+            printf("Bad sudoku grid\n");
+            return 1;
+        }
 
-   fclose(fp);
+        partial = grid[0][0] * 100 + grid[0][1] * 10 + grid[0][2];
+        sum += partial;
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    fclose(fp);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
-   
-   printf("Project Euler, Problem 96\n");
-   printf("Answer: %d\n", sum);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   return 0;
+    printf("Project Euler, Problem 96\n");
+    printf("Answer: %d\n", sum);
+
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 int check(int grid[][9], int i, int j, int n)
 {
-   int k, w, ii, jj;
+    int k, w, ii, jj;
 
-   for(k = 0; k < 9; k++)
-   {
-      if(k != j && grid[i][k] == n)
-      {
-         return 0;
-      }
-   }
-
-   for(k = 0; k < 9; k++)
-   {
-      if(k != i && grid[k][j] == n)
-      {
-         return 0;
-      }
-   }
-
-   ii = 3 * (i / 3);
-   jj = 3 * (j / 3);
-
-   for(k = 0; k < 3; k++)
-   {
-      for(w = 0; w < 3; w++)
-      {
-         if(ii + k != i && jj + w != j && grid[ii+k][jj+w] == n)
-         {
+    for(k = 0; k < 9; k++)
+    {
+        if(k != j && grid[i][k] == n)
+        {
             return 0;
-         }
-      }
-   }
+        }
+    }
 
-   return 1;
+    for(k = 0; k < 9; k++)
+    {
+        if(k != i && grid[k][j] == n)
+        {
+            return 0;
+        }
+    }
+
+    ii = 3 * (i / 3);
+    jj = 3 * (j / 3);
+
+    for(k = 0; k < 3; k++)
+    {
+        for(w = 0; w < 3; w++)
+        {
+            if(ii + k != i && jj + w != j && grid[ii+k][jj+w] == n)
+            {
+                return 0;
+            }
+        }
+    }
+
+    return 1;
 }
 
 int solve_recursive(int grid[][9], int step)
 {
-   int i, j, k;
+    int i, j, k;
 
-   if(step == 81)
-   {
-      return 1;
-   }
+    if(step == 81)
+    {
+        return 1;
+    }
 
-   i = step / 9;
-   j = step % 9;
+    i = step / 9;
+    j = step % 9;
 
-   if(grid[i][j] != 0)
-   {
-      if(solve_recursive(grid, step+1))
-      {
-         return 1;
-      }
-   }          
-   else
-   {
-      for(k = 1; k <= 9; k++)
-      {
-         grid[i][j] = k;
+    if(grid[i][j] != 0)
+    {
+        if(solve_recursive(grid, step+1))
+        {
+            return 1;
+        }
+    }
+    else
+    {
+        for(k = 1; k <= 9; k++)
+        {
+            grid[i][j] = k;
 
-         if(check(grid, i, j, k))
-         {
-            if(solve_recursive(grid, step+1))
+            if(check(grid, i, j, k))
             {
-               return 1;
+                if(solve_recursive(grid, step+1))
+                {
+                    return 1;
+                }
+
+                grid[i][j] = 0;
             }
+        }
 
-            grid[i][j] = 0;
-         }
-      }
+        grid[i][j] = 0;
+    }
 
-      grid[i][j] = 0;
-   }
-
-   return 0;
+    return 0;
 }
 
 int line_complete(int grid[][9], int i)
 {
-   int n;
+    int n;
 
-   for(n = 0; n < 9; n++)
-   {
-      if(grid[i][n] == 0)
-      {
-         return 0;
-      }
-   }
+    for(n = 0; n < 9; n++)
+    {
+        if(grid[i][n] == 0)
+        {
+            return 0;
+        }
+    }
 
-   return 1;
+    return 1;
 }
 
 int column_complete(int grid[][9], int j)
 {
-   int n;
+    int n;
 
-   for(n = 0; n < 9; n++)
-   {
-      if(grid[n][j] == 0)
-      {
-         return 0;
-      }
-   }
+    for(n = 0; n < 9; n++)
+    {
+        if(grid[n][j] == 0)
+        {
+            return 0;
+        }
+    }
 
-   return 1;
+    return 1;
 }
 
 int solve_sudoku(int grid[][9])
 {
-   int i, j, k, w, count, val;
+    int i, j, k, w, count, val;
 
-   for(w = 0; w < 4; w++)
-   {
-      for(i = 0; i < 9; i++)
-      {
-         for(j = 0; j < 9; j++)
-         {
-            if(grid[i][j] == 0)
+    for(w = 0; w < 4; w++)
+    {
+        for(i = 0; i < 9; i++)
+        {
+            for(j = 0; j < 9; j++)
             {
-               count = 0;
-            
-               for(k = 1; k <= 9 && count < 2; k++)
-               {
-                  grid[i][j] = k;
+                if(grid[i][j] == 0)
+                {
+                    count = 0;
 
-                  if(check(grid, i, j, k) == 1)
-                  {
-                     count++;
-                     val = k;
-                  }
-               }
+                    for(k = 1; k <= 9 && count < 2; k++)
+                    {
+                        grid[i][j] = k;
 
-               if(count == 1)
-               {
-                  grid[i][j] = val;
-               }
-               else
-               {
-                  grid[i][j] = 0;
-               }
+                        if(check(grid, i, j, k) == 1)
+                        {
+                            count++;
+                            val = k;
+                        }
+                    }
+
+                    if(count == 1)
+                    {
+                        grid[i][j] = val;
+                    }
+                    else
+                    {
+                        grid[i][j] = 0;
+                    }
+                }
             }
-         }
-      }
+        }
 
-      for(i = 0; i < 9; i++)
-      {
-         for(k = 1; k <= 9 && !line_complete(grid, i); k++)
-         {
-            count = 0;
-
-            for(j = 0; j < 9 && count < 2; j++)
+        for(i = 0; i < 9; i++)
+        {
+            for(k = 1; k <= 9 && !line_complete(grid, i); k++)
             {
-               if(grid[i][j] == 0)
-               {
-                  grid[i][j] = k;
+                count = 0;
 
-                  if(check(grid, i, j, k))
-                  {
-                     count++;
-                     val = j;
-                  }
+                for(j = 0; j < 9 && count < 2; j++)
+                {
+                    if(grid[i][j] == 0)
+                    {
+                        grid[i][j] = k;
 
-                  grid[i][j] = 0;
-               }
+                        if(check(grid, i, j, k))
+                        {
+                            count++;
+                            val = j;
+                        }
+
+                        grid[i][j] = 0;
+                    }
+                }
+
+                if(count == 1)
+                {
+                    grid[i][val] = k;
+                }
             }
+        }
 
-            if(count == 1)
+        for(j = 0; j < 9; j++)
+        {
+            for(k = 1; k <= 9 && !column_complete(grid, j); k++)
             {
-               grid[i][val] = k;
+                count = 0;
+
+                for(i = 0; i < 9 && count < 2; i++)
+                {
+                    if(grid[i][j] == 0)
+                    {
+                        grid[i][j] = k;
+
+                        if(check(grid, i, j, k))
+                        {
+                            count++;
+                            val = i;
+                        }
+
+                        grid[i][j] = 0;
+                    }
+                }
+
+                if(count == 1)
+                {
+                    grid[val][j] = k;
+                }
             }
-         }
-      }
+        }
+    }
 
-      for(j = 0; j < 9; j++)
-      {
-         for(k = 1; k <= 9 && !column_complete(grid, j); k++)
-         {
-            count = 0;
-
-            for(i = 0; i < 9 && count < 2; i++)
-            {
-               if(grid[i][j] == 0)
-               {
-                  grid[i][j] = k;
-
-                  if(check(grid, i, j, k))
-                  {
-                     count++;
-                     val = i;
-                  }
-
-                  grid[i][j] = 0;
-               }
-            }
-
-            if(count == 1)
-            {
-               grid[val][j] = k;
-            }
-         }
-      }
-   }
-
-   return solve_recursive(grid, 0);
+    return solve_recursive(grid, 0);
 }

C/p097.c

@@ -5,44 +5,46 @@
  *
  * Find the last ten digits of this prime number.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
 
 int main(int argc, char **argv)
 {
-   int b = 2, e = 7830457, e_first;
-   long int  m = 10000000000, c, result;
-   double elapsed;
-   struct timespec start, end;
+    int b = 2, e = 7830457, e_first;
+    long int  m = 10000000000, c, result;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Using modular exponentiation algorithm to calculate 2^7830457. The algorithm is:
-    * Set c=1, e'=0
-    * Increment e' by 1
-    * Set c=b*c%m, where b is the base (2) and m is the modulo (10000000000 since
-    * we want the last 10 digits.
-    * If e'<e (where e is the exponent), repeat step 3, otherwise c contains the result.*/
-   c = 1;
-   e_first = 0;
+    /* Using modular exponentiation algorithm to calculate 2^7830457. The algorithm is:
+     * Set c=1, e'=0
+     * Increment e' by 1
+     * Set c=b*c%m, where b is the base (2) and m is the modulo (10000000000 since
+     * we want the last 10 digits.
+     * If e'<e (where e is the exponent), repeat step 3, otherwise c contains the result.*/
+    c = 1;
+    e_first = 0;
 
-   while(e_first < e)
-   {
-      e_first++;
-      c = (b * c) % m;
-   }
+    while(e_first < e)
+    {
+        e_first++;
+        c = (b * c) % m;
+    }
 
-   result = (c * 28433 + 1) % m;
+    result = (c * 28433 + 1) % m;
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 97\n");
-   printf("Answer: %ld\n", result);
+    printf("Project Euler, Problem 97\n");
+    printf("Answer: %ld\n", result);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p099.c

@@ -1,13 +1,15 @@
-/* Comparing two numbers written in index form like 2^11 and 3^7 is not difficult, as any calculator would confirm that 
+/* Comparing two numbers written in index form like 2^11 and 3^7 is not difficult, as any calculator would confirm that
  * 2^11 = 2048 < 3^7 = 2187.
  *
  * However, confirming that 632382^518061 > 519432^525806 would be much more difficult, as both numbers contain over three million digits.
  *
- * Using base_exp.txt, a 22K text file containing one thousand lines with a base/exponent pair on each line, 
+ * Using base_exp.txt, a 22K text file containing one thousand lines with a base/exponent pair on each line,
  * determine which line number has the greatest numerical value.
  *
  * NOTE: The first two lines in the file represent the numbers in the example given above.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -15,46 +17,46 @@
 
 int main(int argc, char **argv)
 {
-   int i, max_i = -1, base, exp;
-   double curr, max = 0, elapsed;
-   struct timespec start, end;
-   FILE *fp;
+    int i, max_i = -1, base, exp;
+    double curr, max = 0, elapsed;
+    struct timespec start, end;
+    FILE *fp;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((fp = fopen("base_exp.txt", "r")) == NULL)
-   {
-      fprintf(stderr, "Error while opening file %s\n", "base_exp.txt");
-      return 1;
-   }
+    if((fp = fopen("base_exp.txt", "r")) == NULL)
+    {
+        fprintf(stderr, "Error while opening file %s\n", "base_exp.txt");
+        return 1;
+    }
 
-   for(i = 1; i <= 1000; i++)
-   {
-      fscanf(fp, "%d,%d", &base, &exp);
-      /* If 
-       * a^x > b^y
-       * log(a^x) > log(b^y)
-       * x*log(a) > y*log(b).
-       * So for each b^e, calculate e*log(b) and compare these numbers instead.*/
-      curr = exp * log(base);
+    for(i = 1; i <= 1000; i++)
+    {
+        fscanf(fp, "%d,%d", &base, &exp);
+        /* If
+         * a^x > b^y
+         * log(a^x) > log(b^y)
+         * x*log(a) > y*log(b).
+         * So for each b^e, calculate e*log(b) and compare these numbers instead.*/
+        curr = exp * log(base);
 
-      if(curr > max)
-      {
-         max = curr;
-         max_i = i;
-      }
-   }
+        if(curr > max)
+        {
+            max = curr;
+            max_i = i;
+        }
+    }
 
-   fclose(fp);
-   
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    fclose(fp);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 99\n");
-   printf("Answer: %d\n", max_i);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 99\n");
+    printf("Answer: %d\n", max_i);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }

C/p102.c

@@ -12,6 +12,8 @@
  *
  * NOTE: The first two examples in the file represent the triangles in the example given above.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -19,51 +21,51 @@
 
 typedef struct point
 {
-   int x;
-   int y;
-}point;
+    int x;
+    int y;
+} point;
 
 int main(int argc, char **argv)
 {
-   int count = 0;
-   FILE *fp;
-   point p1, p2, p3;
-   double a, b, c, elapsed;
-   struct timespec start, end;
+    int count = 0;
+    FILE *fp;
+    point p1, p2, p3;
+    double a, b, c, elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((fp = fopen("triangles.txt", "r")) == NULL)
-   {
-      fprintf(stderr, "Error while opening file %s\n", "triangles.txt");
-      return 1;
-   }
+    if((fp = fopen("triangles.txt", "r")) == NULL)
+    {
+        fprintf(stderr, "Error while opening file %s\n", "triangles.txt");
+        return 1;
+    }
 
-   /* Using the barycentric coordinates method to determine if (0, 0) is inside the triangles.*/
-   while(fscanf(fp, "%d,%d,%d,%d,%d,%d", &p1.x, &p1.y, &p2.x, &p2.y, &p3.x, &p3.y) != EOF)
-   {
-      a = (double)((p2.y - p3.y) * (-p3.x) + (p3.x - p2.x) * (-p3.y)) /
-         ((p2.y - p3.y) * (p1.x - p3.x) + (p3.x - p2.x) * (p1.y - p3.y));
-      b = (double)((p3.y - p1.y) * (-p3.x) + (p1.x - p3.x) * (-p3.y)) /
-         ((p2.y - p3.y) * (p1.x - p3.x) + (p3.x - p2.x) * (p1.y - p3.y));
-      c = 1 - a - b;
+    /* Using the barycentric coordinates method to determine if (0, 0) is inside the triangles.*/
+    while(fscanf(fp, "%d,%d,%d,%d,%d,%d", &p1.x, &p1.y, &p2.x, &p2.y, &p3.x, &p3.y) != EOF)
+    {
+        a = (double)((p2.y - p3.y) * (-p3.x) + (p3.x - p2.x) * (-p3.y)) /
+            ((p2.y - p3.y) * (p1.x - p3.x) + (p3.x - p2.x) * (p1.y - p3.y));
+        b = (double)((p3.y - p1.y) * (-p3.x) + (p1.x - p3.x) * (-p3.y)) /
+            ((p2.y - p3.y) * (p1.x - p3.x) + (p3.x - p2.x) * (p1.y - p3.y));
+        c = 1 - a - b;
 
-      if(a >= 0 && a <= 1 && b >= 0 && b <= 1 && c >= 0 && c <= 1)
-      {
-         count++;
-      }
-   }
+        if(a >= 0 && a <= 1 && b >= 0 && b <= 1 && c >= 0 && c <= 1)
+        {
+            count++;
+        }
+    }
 
-   fclose(fp);
+    fclose(fp);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 102\n");
-   printf("Answer: %d\n", count);
+    printf("Project Euler, Problem 102\n");
+    printf("Answer: %d\n", count);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }

C/p104.c

@@ -7,6 +7,8 @@
 *
 * Given that F_k is the first Fibonacci number for which the first nine digits AND the last nine digits are 1-9 pandigital, find k.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -17,101 +19,101 @@ void fib_calc(mpz_t fib, int n);
 
 int main(int argc, char **argv)
 {
-   unsigned int i = 2, j, fib1 = 1, fib2 = 1, fibn, fib_int, found = 0;
-   char *fib_str;
-   double elapsed;
-   struct timespec start, end;
-   mpz_t fib;
+    unsigned int i = 2, j, fib1 = 1, fib2 = 1, fibn, fib_int, found = 0;
+    char *fib_str;
+    double elapsed;
+    struct timespec start, end;
+    mpz_t fib;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   mpz_init(fib);
+    mpz_init(fib);
 
-   while(!found)
-   {
-      /* Calculate the next Fibonacci number modulo 10^9 and check if the result is 1-9 pandigital.*/
-      fibn = (fib1 + fib2) % 1000000000;
-      fib1 = fib2;
-      fib2 = fibn;
-      i++;
-      
-      /* If the last 9 digits of fib_n are pandigital, calculate the ith Fibonacci number using
-       * the matrix representation.*/
-      if(is_pandigital(fibn, 9))
-      {
-         fib_calc(fib, i);
-         fib_str = mpz_get_str(NULL, 10, fib);
+    while(!found)
+    {
+        /* Calculate the next Fibonacci number modulo 10^9 and check if the result is 1-9 pandigital.*/
+        fibn = (fib1 + fib2) % 1000000000;
+        fib1 = fib2;
+        fib2 = fibn;
+        i++;
 
-         fib_int = 0;
+        /* If the last 9 digits of fib_n are pandigital, calculate the ith Fibonacci number using
+         * the matrix representation.*/
+        if(is_pandigital(fibn, 9))
+        {
+            fib_calc(fib, i);
+            fib_str = mpz_get_str(NULL, 10, fib);
 
-         for(j = 0; j < 9; j++)
-         {
-            fib_int *= 10;
-            fib_int += fib_str[j] - '0';
-         }
+            fib_int = 0;
 
-         if(is_pandigital(fib_int, 9))
-         {
-            found = 1;
-         }
+            for(j = 0; j < 9; j++)
+            {
+                fib_int *= 10;
+                fib_int += fib_str[j] - '0';
+            }
 
-         free(fib_str);
-      }
-   }
+            if(is_pandigital(fib_int, 9))
+            {
+                found = 1;
+            }
 
-   mpz_clear(fib);
+            free(fib_str);
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    mpz_clear(fib);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Project Euler, Problem 104\n");
-   printf("Answer: %d\n", i);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Project Euler, Problem 104\n");
+    printf("Answer: %d\n", i);
 
-   return 0;
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 void fib_calc(mpz_t fib, int n)
 {
-   int i;
-   mpz_t tmp, tmp00, tmp01, tmp10, tmp11;
-   mpz_t fib_matrix_base[2][2], fib_matrix[2][2];
+    int i;
+    mpz_t tmp, tmp00, tmp01, tmp10, tmp11;
+    mpz_t fib_matrix_base[2][2], fib_matrix[2][2];
 
-   mpz_init_set_ui(fib_matrix_base[0][0], 1);
-   mpz_init_set_ui(fib_matrix_base[0][1], 1);
-   mpz_init_set_ui(fib_matrix_base[1][0], 1);
-   mpz_init_set_ui(fib_matrix_base[1][1], 0);
-   mpz_init_set_ui(fib_matrix[0][0], 1);
-   mpz_init_set_ui(fib_matrix[0][1], 1);
-   mpz_init_set_ui(fib_matrix[1][0], 1);
-   mpz_init_set_ui(fib_matrix[1][1], 0);
-   mpz_inits(tmp, tmp00, tmp01, tmp10, tmp11, NULL);
+    mpz_init_set_ui(fib_matrix_base[0][0], 1);
+    mpz_init_set_ui(fib_matrix_base[0][1], 1);
+    mpz_init_set_ui(fib_matrix_base[1][0], 1);
+    mpz_init_set_ui(fib_matrix_base[1][1], 0);
+    mpz_init_set_ui(fib_matrix[0][0], 1);
+    mpz_init_set_ui(fib_matrix[0][1], 1);
+    mpz_init_set_ui(fib_matrix[1][0], 1);
+    mpz_init_set_ui(fib_matrix[1][1], 0);
+    mpz_inits(tmp, tmp00, tmp01, tmp10, tmp11, NULL);
 
-   for(i = 0; i < n - 1; i++)
-   {
-      mpz_mul(tmp00, fib_matrix[0][0], fib_matrix_base[0][0]);
-      mpz_mul(tmp, fib_matrix[0][1], fib_matrix_base[1][0]);
-      mpz_add(tmp00, tmp00, tmp);
-      mpz_mul(tmp01, fib_matrix[0][0], fib_matrix_base[0][1]);
-      mpz_mul(tmp, fib_matrix[0][1], fib_matrix_base[1][1]);
-      mpz_add(tmp01, tmp01, tmp);
-      mpz_mul(tmp10, fib_matrix[1][0], fib_matrix_base[0][0]);
-      mpz_mul(tmp, fib_matrix[1][1], fib_matrix_base[1][0]);
-      mpz_add(tmp10, tmp10, tmp);
-      mpz_mul(tmp11, fib_matrix[1][0], fib_matrix_base[0][1]);
-      mpz_mul(tmp, fib_matrix[1][1], fib_matrix_base[1][1]);
-      mpz_add(tmp11, tmp11, tmp);
-      mpz_set(fib_matrix[0][0], tmp00);
-      mpz_set(fib_matrix[0][1], tmp01);
-      mpz_set(fib_matrix[1][0], tmp10);
-      mpz_set(fib_matrix[1][1], tmp11);
-   }
+    for(i = 0; i < n - 1; i++)
+    {
+        mpz_mul(tmp00, fib_matrix[0][0], fib_matrix_base[0][0]);
+        mpz_mul(tmp, fib_matrix[0][1], fib_matrix_base[1][0]);
+        mpz_add(tmp00, tmp00, tmp);
+        mpz_mul(tmp01, fib_matrix[0][0], fib_matrix_base[0][1]);
+        mpz_mul(tmp, fib_matrix[0][1], fib_matrix_base[1][1]);
+        mpz_add(tmp01, tmp01, tmp);
+        mpz_mul(tmp10, fib_matrix[1][0], fib_matrix_base[0][0]);
+        mpz_mul(tmp, fib_matrix[1][1], fib_matrix_base[1][0]);
+        mpz_add(tmp10, tmp10, tmp);
+        mpz_mul(tmp11, fib_matrix[1][0], fib_matrix_base[0][1]);
+        mpz_mul(tmp, fib_matrix[1][1], fib_matrix_base[1][1]);
+        mpz_add(tmp11, tmp11, tmp);
+        mpz_set(fib_matrix[0][0], tmp00);
+        mpz_set(fib_matrix[0][1], tmp01);
+        mpz_set(fib_matrix[1][0], tmp10);
+        mpz_set(fib_matrix[1][1], tmp11);
+    }
 
-   mpz_set(fib, fib_matrix[0][1]);
+    mpz_set(fib, fib_matrix[0][1]);
 
-   mpz_clears(tmp, tmp00, tmp01, tmp10, tmp11, fib_matrix_base[0][0],
-         fib_matrix_base[0][1], fib_matrix_base[1][0], fib_matrix_base[1][1],
-         fib_matrix[0][0], fib_matrix[1][0], fib_matrix[1][1], NULL);
+    mpz_clears(tmp, tmp00, tmp01, tmp10, tmp11, fib_matrix_base[0][0],
+               fib_matrix_base[0][1], fib_matrix_base[1][0], fib_matrix_base[1][1],
+               fib_matrix[0][0], fib_matrix[1][0], fib_matrix[1][1], NULL);
 }

C/p112.c

@@ -11,6 +11,8 @@
  *
  * Find the least number for which the proportion of bouncy numbers is exactly 99%.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
@@ -20,88 +22,88 @@ int is_bouncy(int n);
 
 int main(int argc, char **argv)
 {
-   int i = 100, n_bouncy = 0;
-   double ratio = 0.0, elapsed;
-   struct timespec start, end;
+    int i = 100, n_bouncy = 0;
+    double ratio = 0.0, elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   while(1)
-   {
-      if(is_bouncy(i))
-      {
-         n_bouncy++;
-      }
+    while(1)
+    {
+        if(is_bouncy(i))
+        {
+            n_bouncy++;
+        }
 
-      ratio = (double)n_bouncy / i;
-      
-      if(ratio == 0.99)
-      {
-         break;
-      }
-     
-      i++;
-   }
+        ratio = (double)n_bouncy / i;
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+        if(ratio == 0.99)
+        {
+            break;
+        }
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+        i++;
+    }
 
-   printf("Project Euler, Problem 112\n");
-   printf("Answer: %d\n", i);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   return 0;
+    printf("Project Euler, Problem 112\n");
+    printf("Answer: %d\n", i);
+
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
+
+    return 0;
 }
 
 /* Check if the number is bouncy.*/
 int is_bouncy(int n)
 {
-   int i = 0, l;
-   char n_string[100];
+    int i = 0, l;
+    char n_string[100];
 
-   sprintf(n_string, "%d", n);
+    sprintf(n_string, "%d", n);
 
-   l = strlen(n_string);
+    l = strlen(n_string);
 
-   /* Ignore consecutive equal digits.*/
-   while(i < l - 1 && n_string[i] == n_string[i+1])
-   {
-      i++;
-   }
+    /* Ignore consecutive equal digits.*/
+    while(i < l - 1 && n_string[i] == n_string[i+1])
+    {
+        i++;
+    }
 
-   /* If all the digits are the same, the number is not bouncy.*/
-   if(i == l - 1)
-   {
-      return 0;
-   }
+    /* If all the digits are the same, the number is not bouncy.*/
+    if(i == l - 1)
+    {
+        return 0;
+    }
 
-   /* If the first two different digits are increasing, if a successive
-    * digit smaller than the previous digit is found, the number is bouncy.*/
-   if(n_string[i] < n_string[i+1])
-   {
-      for(; i < l - 1; i++)
-      {
-         if(n_string[i] > n_string[i+1])
-         {
-            return 1;
-         }
-      }
-   }
+    /* If the first two different digits are increasing, if a successive
+     * digit smaller than the previous digit is found, the number is bouncy.*/
+    if(n_string[i] < n_string[i+1])
+    {
+        for(; i < l - 1; i++)
+        {
+            if(n_string[i] > n_string[i+1])
+            {
+                return 1;
+            }
+        }
+    }
 
-   /* If the first two different digits are decreasing, if a successive
-    * digit larger than the previous digit is found, the number is bouncy.*/
-   if(n_string[i] > n_string[i+1])
-   {
-      for(; i < l - 1; i++)
-      {
-         if(n_string[i] < n_string[i+1])
-         {
-            return 1;
-         }
-      }
-   }
+    /* If the first two different digits are decreasing, if a successive
+     * digit larger than the previous digit is found, the number is bouncy.*/
+    if(n_string[i] > n_string[i+1])
+    {
+        for(; i < l - 1; i++)
+        {
+            if(n_string[i] < n_string[i+1])
+            {
+                return 1;
+            }
+        }
+    }
 
-   return 0;
+    return 0;
 }

C/p124.c

@@ -2,7 +2,7 @@
  *
  * If we calculate rad(n) for 1 ≤ n ≤ 10, then sort them on rad(n), and sorting on n if the radical values are equal, we get:
  *
- *    Unsorted             Sorted         
+ *    Unsorted             Sorted
  *    n     rad(n)      n     rad(n)   k
  *    1     1           1     1        1
  *    2     2           2     2        2
@@ -19,6 +19,8 @@
  *
  * If rad(n) is sorted for 1 ≤ n ≤ 100000, find E(10000).*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
@@ -29,77 +31,77 @@
 
 typedef struct n_rad
 {
-   int n;
-   int radn;
-}n_radn;
+    int n;
+    int radn;
+} n_radn;
 
 int compare(void *a, void *b);
 
 int main(int argc, char **argv)
 {
-   int i, *primes;
-   n_radn **rads;
-   double elapsed;
-   struct timespec start, end;
+    int i, *primes;
+    n_radn **rads;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((primes = sieve(N)) == NULL)
-   {
-      fprintf(stderr, "Error! Sieve function returned NULL\n");
-      return 1;
-   }
+    if((primes = sieve(N)) == NULL)
+    {
+        fprintf(stderr, "Error! Sieve function returned NULL\n");
+        return 1;
+    }
 
-   if((rads = (n_radn **)malloc(N*sizeof(n_radn *))) == NULL)
-   {
-      fprintf(stderr, "Error while allocating memory\n");
-      return 1;
-   }
+    if((rads = (n_radn **)malloc(N*sizeof(n_radn *))) == NULL)
+    {
+        fprintf(stderr, "Error while allocating memory\n");
+        return 1;
+    }
 
-   for(i = 0; i < N; i++)
-   {
-      if((rads[i] = (n_radn *)malloc(sizeof(n_radn))) == NULL)
-      {
-         fprintf(stderr, "Error while allocating memory\n");
-         return 1;
-      }
-   }
+    for(i = 0; i < N; i++)
+    {
+        if((rads[i] = (n_radn *)malloc(sizeof(n_radn))) == NULL)
+        {
+            fprintf(stderr, "Error while allocating memory\n");
+            return 1;
+        }
+    }
 
-   for(i = 0; i < N; i++)
-   {
-      rads[i]->n = i + 1;
-      rads[i]->radn = radical(i+1, primes);
-   }
+    for(i = 0; i < N; i++)
+    {
+        rads[i]->n = i + 1;
+        rads[i]->radn = radical(i+1, primes);
+    }
 
-   free(primes);
+    free(primes);
 
-   quick_sort((void **)rads, 0, N-1, compare);
+    quick_sort((void **)rads, 0, N-1, compare);
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 124\n");
-   printf("Answer: %d\n", rads[9999]->n);
+    printf("Project Euler, Problem 124\n");
+    printf("Answer: %d\n", rads[9999]->n);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int compare(void *a, void *b)
 {
-   n_radn *rad1, *rad2;
+    n_radn *rad1, *rad2;
 
-   rad1 = (n_radn *)a;
-   rad2 = (n_radn *)b; 
+    rad1 = (n_radn *)a;
+    rad2 = (n_radn *)b;
 
-   if(rad1->radn != rad2->radn)
-   {
-      return rad1->radn - rad2->radn;
-   }
-   else
-   {
-      return rad1->n - rad2->n;
-   }
+    if(rad1->radn != rad2->radn)
+    {
+        return rad1->radn - rad2->radn;
+    }
+    else
+    {
+        return rad1->n - rad2->n;
+    }
 }

C/p145.c

@@ -6,6 +6,8 @@
  *
  * How many reversible numbers are there below one-billion (109)?*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -16,58 +18,58 @@ int reverse(int n);
 
 int main(int argc, char **argv)
 {
-   int i, s, count = 0;
-   double elapsed;
-   struct timespec start, end;
+    int i, s, count = 0;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   /* Brute force approach, sum each number and their reverse and
-    * check if there are only odd digits.*/
-   for(i = 11; i < N; i++)
-   {
-      if(i % 10 != 0)
-      {
-         s = i + reverse(i);
+    /* Brute force approach, sum each number and their reverse and
+     * check if there are only odd digits.*/
+    for(i = 11; i < N; i++)
+    {
+        if(i % 10 != 0)
+        {
+            s = i + reverse(i);
 
-         while(s > 0)
-         {
-            if((s % 10) % 2 == 0)
+            while(s > 0)
             {
-               break;
+                if((s % 10) % 2 == 0)
+                {
+                    break;
+                }
+                s /= 10;
             }
-            s /= 10;
-         }
 
-         if(s == 0)
-         {
-            count++;
-         }
-      }
-   }
+            if(s == 0)
+            {
+                count++;
+            }
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 145\n");
-   printf("Answer: %d\n", count);
+    printf("Project Euler, Problem 145\n");
+    printf("Answer: %d\n", count);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }
 
 int reverse(int n)
 {
-   int reverse = 0;
+    int reverse = 0;
 
-   while(n > 0)
-   {
-      reverse *= 10;
-      reverse += n % 10;
-      n /= 10;
-   }
+    while(n > 0)
+    {
+        reverse *= 10;
+        reverse += n % 10;
+        n /= 10;
+    }
 
-   return reverse;
+    return reverse;
 }

C/p206.c

@@ -1,60 +1,62 @@
 /* Find the unique positive integer whose square has the form 1_2_3_4_5_6_7_8_9_0,
  * where each “_” is a single digit.*/
 
+#define _POSIX_C_SOURCE 199309L
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
 
 int main(int argc, char **argv)
 {
-   int i, found = 0;
-   /* Since the square of n has 19 digits, n must be at least 10^9.*/
-   long int n = 1e9, p;
-   double elapsed;
-   struct timespec start, end;
+    int i, found = 0;
+    /* Since the square of n has 19 digits, n must be at least 10^9.*/
+    long int n = 1e9, p;
+    double elapsed;
+    struct timespec start, end;
 
-   clock_gettime(CLOCK_MONOTONIC, &start);
+    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   while(!found)
-   {
-      /* If the square on n ends with 10, n must be divisible by 10.*/
-      n += 10;
-      p = n * n;
+    while(!found)
+    {
+        /* If the square on n ends with 10, n must be divisible by 10.*/
+        n += 10;
+        p = n * n;
 
-      /* A square divisible by 10 is also divisible by 100.*/
-      if(p % 100 != 0)
-      {
-         continue;
-      }
+        /* A square divisible by 10 is also divisible by 100.*/
+        if(p % 100 != 0)
+        {
+            continue;
+        }
 
-      /* Check if the digits of the square correspond to the given pattern.*/
-      i = 9;
-      p /= 100;
+        /* Check if the digits of the square correspond to the given pattern.*/
+        i = 9;
+        p /= 100;
 
-      while(p > 0)
-      {
-         if(p % 10 != i)
-         {
-            break;
-         }
-         p /= 100;
-         i--;
-      }
+        while(p > 0)
+        {
+            if(p % 10 != i)
+            {
+                break;
+            }
+            p /= 100;
+            i--;
+        }
 
-      if(p == 0)
-      {
-         found = 1;
-      }
-   }
+        if(p == 0)
+        {
+            found = 1;
+        }
+    }
 
-   clock_gettime(CLOCK_MONOTONIC, &end);
+    clock_gettime(CLOCK_MONOTONIC, &end);
 
-   elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
+    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
 
-   printf("Project Euler, Problem 206\n");
-   printf("Answer: %ld\n", n);
+    printf("Project Euler, Problem 206\n");
+    printf("Answer: %ld\n", n);
 
-   printf("Elapsed time: %.9lf seconds\n", elapsed);
+    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
-   return 0;
+    return 0;
 }