Improve solution for problem 47

C/p047.c

@@ -15,88 +15,81 @@
 #include <stdlib.h>
 #include <math.h>
 #include <time.h>
-#include "projecteuler.h"
 
 #define N 150000
 
-int count_distinct_factors(int n);
-   
-int *primes;
+int *count_factors(int n);
 
 int main(int argc, char **argv)
 {
-   int i, found = 0;
+   int i, count, res;
+   int *factors;
    double elapsed;
    struct timespec start, end;
 
    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((primes = sieve(N)) == NULL)
+   if((factors = count_factors(N)) == NULL)
    {
-      fprintf(stderr, "Error! Sieve function returned NULL\n");
+      fprintf(stderr, "Error! Count_factors function returned NULL\n");
       return 1;
    }
 
-   /* Starting from 647, count the distinct prime factors of n, n+1, n+2 and n+3.
-    * If they all have 4, the solution is found.*/
-   for(i = 647; !found && i < N - 3; i++)
-   {
-      if(!primes[i] && !primes[i+1] && !primes[i+2] && !primes[i+3])
-      {
-         if(count_distinct_factors(i) == 4 && count_distinct_factors(i+1) == 4 &&
-               count_distinct_factors(i+2) == 4 && count_distinct_factors(i+3) == 4)
-         {
-            found = 1;
-         }
-      }
-   }     
+   count = 0;
 
-   free(primes);
+   for(i = 0; i < N; i++)
+   {
+      if(factors[i] == 4)
+      {
+         count++;
+      }
+      else
+      {
+         count = 0;
+      }
+
+      if(count == 4)
+      {
+         res = i - 3;
+         break;
+      }
+   }
 
    clock_gettime(CLOCK_MONOTONIC, &end);
 
    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", i-1);
+   printf("Answer: %d\n", res);
 
    printf("Elapsed time: %.9lf seconds\n", elapsed);
 
    return 0;
 }
 
-int count_distinct_factors(int n)
+/* Function using a modified sieve of Eratosthenes to count
+ * the distinct prime factors of each number.*/
+int *count_factors(int n)
 {
-   int i, count=0;
+   int i = 2, j;
+   int *factors;
 
-   /* Start checking if 2 is a prime factor of n. Then remove
-    * all 2s factore.*/
-   if(n % 2 == 0)
+   if((factors = (int *)calloc(n, sizeof(int))) == NULL)
    {
-      count++;
-
-      do
-      {
-         n /= 2;
-      }while(n % 2 == 0);
+      return NULL;
    }
 
-   /* Check all odd numbers i, if they're prime and they're a factor
-    * of n, count them and then divide n for by i until all factors i
-    * are eliminated. Stop the loop when n=1, i.e. all factors have
-    * been found.*/
-   for(i = 3; n > 1; i += 2)
+   while(i < n / 2)
    {
-      if(primes[i] && n % i == 0)
+      if(factors[i] == 0)
       {
-         count++;
-
-         do
+         for(j = i; j < n; j += i)
          {
-            n /= i;
-         }while(n % i == 0);
+            factors[j]++;
+         }
       }
+      i++;
    }
 
-   return count;
+   return factors;
 }