euler1

Problem

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.

#include <stdio.h>


#define NUM    999

int main(int argc, const char *argv[])
{
    int i, sum = 0;
#if 0

    for (i = 0; i <= NUM; i++) {
        if (!(i%3 && i%5)) {
            sum += i;
        }

    }
    printf("sum = %d\n", sum);

#else
    sum = 0;
    sum += (3+(NUM - (NUM%3)))* (NUM/3)/2 ;
    sum += (5+(NUM - (NUM%5)))* (NUM/5)/2 ;
    sum -= (15+(NUM - (NUM%15)))* (NUM/15)/2 ;
    printf("sum = %d\n", sum);
#endif


    return 0;
}