/* If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
 *
 * 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;

    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;
        }
    }

    clock_gettime(CLOCK_MONOTONIC, &end);

    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("Elapsed time: %.9lf seconds\n", elapsed);

    return 0;
}