Project Euler, Problem #10, C

/* The task: the sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
 * Find the sum of all the primes below two million. */

/* Don't compile */

/* A more efficient version */

/* This version creates an array of non-prime numbers and then checks
 * all the values in the range of 2 to 2000000 against the array and
 * accumulates primes. */

#include <stdio.h>

int main()
{
     int i, j;
     char non_primes[2000001] = { 0 };
     long int acc;
     for(i = 2; i <= 2000000; i++)
          for(j = 2; j <= 2000000 / i; j++)
               non_primes[i * j] = 1;

     for(i = 2; i <= 2000000; i++)
          if(non_primes[i] == 0)
               acc += i;

     printf("%ld", acc);
     return(0);
}

/* 142913828922
 * real 0m0,460s
 * user 0m0,442s
 * sys  0m0,004s */


/* A less efficient version */

/* The idea behind this solution is that if a number cannot be divided
 * by any smaller prime number it is prime itself. */

#include <stdio.h>

int main()
{
     int primes[1999999] = {2};
     /* indices */
     int i, j;
     /* initialized to 2 because 2 is already in the array */
     long int sum = 2;
     /* current index of primes array */
     int mi = 0;
     /* prime flag */
     char pf;

     for(i = 3; i <= 2000000; i += 2)
     {
          pf = 1;
          for(j = 0; j <= mi; j++)
          {
               if(i % primes[j] == 0)
               {
                    pf = 0;
                    break;
               }
          }

          if(pf == 1)
          {
               mi++;
               primes[mi] = i;
               sum += i;
          }
     }
     printf("%ld\n", sum);
     return(0);
}

/* 142913828922
 *
 * real 1m27,239s
 * user 1m27,011s
 * sys  0m0,060s */