Find the sum of all the multiples of 3 or 5 below 1000.

euler1.c

/*
    Copyright 2016 strupo
    
    This program is free software: you can redistribute it and/or modify
    it under the terms of the GNU General Public License as published by
    the Free Software Foundation, either version 3 of the License, or
    (at your option) any later version.

    This program is distributed in the hope that it will be useful,
    but WITHOUT ANY WARRANTY; without even the implied warranty of
    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
    GNU General Public License for more details.

    You should have received a copy of the GNU General Public License
    along with this program.  If not, see <http://www.gnu.org/licenses/>.
*/

#include <stdio.h>

/* This program sums the multiples of 3 and 5 less than 1000.
 */

int main (void)
{
    int i; /* counter */
    int start = 0; /* start summing from here */
    int stop = 1e3; /* stop summing when you reach this */
    int mult[2]; /* this array contains the numbers whose multiples will be summed */
    long sum = 0L; /* will contain the sum; initialized to zero */
    
    mult[0] = 3;
    mult[1] = 5;
    
    for (i = start; i < end; ++i) {
        if (!(i % mult[0] && i % mult[1])) { /* :DDD */
            sum += (long) i;
        }
    }
    
    printf("Sum of multiples of %d and %d from %d to %d = %ld\n",
            mult[0], mult[1], start, end, sum);
    
    return 0;
}