Improve problem 95 solution

C/p095.c

@@ -16,56 +16,38 @@
 #include <stdlib.h>
 #include <math.h>
 #include <time.h>
-#include "projecteuler.h"
 
 #define N 1000000
 
 int sum_proper_divisors(int i);
 int sociable_chain(int i, int start, int *min, int l);
 
-int chains[N] = {0};
 /* Vector to save the current chain values. I started with a longer vector,
  * but no chain is longer than 100 elements, so this is sufficient.*/
 int c[100];
 int divisors[N] = {0};
-int *primes;
 
 int main(int argc, char **argv)
 {
-   /* l_max starts from 3 because we're interested in chains of at least 3 elements.*/
+   int i, min = 0, min_tmp, length, l_max = 0;
-   int i, min = 0, min_tmp, length, l_max = 2;
    double elapsed;
    struct timespec start, end;
 
    clock_gettime(CLOCK_MONOTONIC, &start);
 
-   if((primes = sieve(N+1)) == NULL)
-   {
-      fprintf(stderr, "Error! Sieve function returned NULL\n");
-      return 1;
-   }
-
    for(i = 4; i <= N; i++)
    {
-      /* Calculate the divisors of i, or retrieve the value if previously calculated.
+      /* Calculate the divisors of i and save it. Ii is equal to the sum of its proper divisors, 
-       * If i is equal to the sum of its proper divisors, the length of the chain is 1
+       * the length of the chain is 1 and we don't need to check it.*/
-       * (i.e. i is a perfect number.*/
+      if((divisors[i] = sum_proper_divisors(i)) == i)
-      if(divisors[i] == i || (divisors[i] = sum_proper_divisors(i)) == i)
       {
-         length = 1;
+         continue;
-         chains[i] = length;
       }
-      else if(!primes[i])
+      else
       {
          min_tmp = i;
          length = sociable_chain(i, i, &min_tmp, 0);
       }
-      /* If i is prime, 1 is its only proper divisor, so no amicable chain is possible.*/
-      else
-      {
-         length = -1;
-         chains[i] = length;
-      }
       
       if(length > l_max)
       {
@@ -74,8 +56,6 @@ int main(int argc, char **argv)
       }
    }
 
-   free(primes);
-
    clock_gettime(CLOCK_MONOTONIC, &end);
 
    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
@@ -111,30 +91,23 @@ int sum_proper_divisors(int n)
    return sum;
 }
 
+/* Function to recursively find the length of the chain.*/
 int sociable_chain(int i, int start, int *min, int l)
 {
    int n;
 
-   /* If we already calculated that the current value can't form a chain,
+   /* Save current number in the chain.*/
-    * or has a chain starting from a different number, it can't form a
+   c[l] = i;
-    * chain starting with the current number, so just return -1.*/
-   if(chains[i] != 0)
-   {
-      chains[start] = -1;
-      return -1;
-   }
 
-   /* If we reached a prime number, the chain will never return to the starting number.*/
+   /* If we reached 1, the chain will never return go anywhere.*/
-   if(primes[i])
+   if(i == 1)
    {
-      chains[start] = -1;
       return -1;
    }
 
    /* Calculate the divisors of i, or retrieve the value if previously calculated.*/
    if(divisors[i] != 0)
    {
-      chains[start] = -1;
       n = divisors[i];
    }
    else
@@ -143,34 +116,28 @@ int sociable_chain(int i, int start, int *min, int l)
       divisors[i] = n;
    }
    
-   /* If the next number in the chain is equal to the starting one, the chain is finished.*/
-   if(n == start)
-   {
-      chains[start] = l + 1;
-      return l + 1;
-   }
-
-   /* Save n, i.e. the next value in the chain, and check if it's 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.*/
-   c[l] = n;
-
-   for(i = 0; i < l; i++)
-   {
-      if(n == c[i])
-      {
-         chains[start] = -1;
-         return -1;
-      }
-   }
-
    /* We are looking for chain where no value is greater than 1000000.*/
    if(n > N)
    {
-      chains[start] = -1;
       return -1;
    }
 
+   /* If the next number in the chain is equal to the starting one, the chain is finished.*/
+   if(n == start)
+   {
+      return l + 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 == c[i])
+      {
+         return -1;
+      }
+   }
+
    /* If the next value is smaller than the minimum value of the chain,
     * update the minimum.*/
    if(n < *min)