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Added comments to the C code for problems from 16 to 25

C/p016.c

@@ -1,3 +1,7 @@
+/* 2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
+ *
+ * What is the sum of the digits of the number 2^1000?*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -11,6 +15,8 @@ int main(int argc, char **argv)
 
    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);
@@ -19,6 +25,7 @@ int main(int argc, char **argv)
 
    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);
    }

C/p017.c

@@ -1,12 +1,24 @@
+/* If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
+ *
+ * If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?
+ *
+ * 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.*/
+
 #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;
@@ -14,14 +26,19 @@ int main(int argc, char **argv)
 
    clock_gettime(CLOCK_MONOTONIC, &start);
 
+   /* 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];
@@ -29,20 +46,27 @@ int main(int argc, char **argv)
       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 "one thousand".*/
    sum += n1000;
 
    clock_gettime(CLOCK_MONOTONIC, &end);

C/p019.c

@@ -1,3 +1,16 @@
+/* You are given the following information, but you may prefer to do some research for yourself.
+ *
+ * 1 Jan 1900 was a Monday.
+ * Thirty days has September,
+ * April, June and November.
+ * All the rest have thirty-one,
+ * Saving February alone,
+ * Which has twenty-eight, rain or shine.
+ * And on leap years, twenty-nine.
+ * A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
+ *
+ * How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -21,6 +34,9 @@ int main(int argc, char **argv)
 
    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))
@@ -32,24 +48,33 @@ int main(int argc, char **argv)
             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++;

C/p020.c

@@ -1,3 +1,10 @@
+/* n! means n × (n − 1) × ... × 3 × 2 × 1
+ *
+ * For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
+ * and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
+ *
+ * Find the sum of the digits in the number 100!*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <time.h>
@@ -11,6 +18,7 @@ int main(int argc, char **argv)
 
    clock_gettime(CLOCK_MONOTONIC, &start);
 
+   /* Calculate the factorial using the GMP Library and sum the digits.*/
    mpz_inits(fact, r, sum, NULL);
 
    mpz_fac_ui(fact, 100);

C/p021.c

@@ -1,9 +1,16 @@
+/* Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
+ * If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.
+ *
+ * For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284.
+ * The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
+
+Evaluate the sum of all the amicable numbers under 10000.*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
 #include <time.h>
-
+#include "projecteuler.h"
-int sum_of_d(int n);
 
 int main(int argc, char **argv)
 {
@@ -15,8 +22,12 @@ int main(int argc, char **argv)
 
    for(i = 2; i < 10000; i++)
    {
-      n = sum_of_d(i);
+      /* Calculate the sum of proper divisors with the function
-      if(i != n && sum_of_d(n) == i)
+       * implemented in projecteuler.c.*/
+      n = sum_of_divisors(i);
+      /* 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) == i)
       {
          sum += i + n;
       }
@@ -35,24 +46,3 @@ int main(int argc, char **argv)
 
    return 0;
 }
-
-int sum_of_d(int n)
-{
-   int i, sum = 1, limit;
-
-   limit = floor(sqrt(n));
-
-   for(i = 2; i <= limit; i++)
-   {
-      if(n % i == 0)
-      {
-         sum += i;
-         if(n != i * i)
-         {
-            sum += (n / i);
-         }
-      }
-   }
-
-   return sum;
-}

C/p022.c

@@ -1,3 +1,11 @@
+/* Using names.txt, a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order.
+ * Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.
+ *
+ * For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list.
+ * So, COLIN would obtain a score of 938 × 53 = 49714.
+ *
+ * What is the total of all the name scores in the file?*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
@@ -29,6 +37,7 @@ int main(int argc, char **argv)
    len = strlen(line);
    n = 1;
 
+   /* Count the names in the file.*/
    for(i = 0; i < len; i++)
    {
       if(line[i] == ',')
@@ -43,6 +52,7 @@ int main(int argc, char **argv)
       return 1;
    }
 
+   /* Save each name in a string.*/
    names[0] = strtok(line, ",\"");
 
    for(i = 1; i < n; i++)
@@ -50,8 +60,10 @@ int main(int argc, char **argv)
       names[i] = strtok(NULL, ",\"");
    }
 
+   /* 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]);
@@ -77,6 +89,7 @@ int main(int argc, char **argv)
    return 0;
 }
 
+/* Function to compare two strings, to pass to the quick_sort function.*/
 int compare(void *string1, void *string2)
 {
    char *s1, *s2;

C/p023.c

@@ -1,7 +1,20 @@
+/* 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.
+ *
+ * As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24.
+ * By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers.
+ * However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed
+ * as the sum of two abundant numbers is less than this limit.
+ *
+ * Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.*/
+
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
 #include <time.h>
+#include "projecteuler.h"
 
 int is_abundant(int n);
 
@@ -16,20 +29,13 @@ int main(int argc, char **argv)
 
    for(i = 0; i < 28123; i++)
    {
-      if(i < 11)
+      /* Find all abundant numbers smaller than 28123.*/
-      {
+      ab_nums[i] = is_abundant(i+1);
-         ab_nums[i] = 0;
-      }
-      else if(is_abundant(i+1))
-      {
-         ab_nums[i] = 1;
-      }
-      else
-      {
-         ab_nums[i] = 0;
-      }
    }
 
+   /* 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])
@@ -44,6 +50,10 @@ int main(int argc, char **argv)
                {
                   sums[sum-1] = 1;
                }
+               else
+               {
+                  break;
+               }
             }
          }
       }
@@ -51,6 +61,7 @@ int main(int argc, char **argv)
 
    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])
@@ -73,22 +84,5 @@ int main(int argc, char **argv)
 
 int is_abundant(int n)
 {
-   int i, sum = 1, limit;
+   return sum_of_divisors(n) > n;
-
-   limit = floor(sqrt(n));
-
-   for(i = 2; i <= limit; i++)
-   {
-      if(n % i == 0)
-      {
-         sum += i;
-
-         if(n != i*i)
-         {
-            sum += (n / i);
-         }
-      }
-   }
-
-   return sum > n;
 }