TheAlgorithms-C/project_euler/problem_1/sol3.c

/*
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 N.
'''
'''
This solution is based on the pattern that the successive numbers in the series
follow: 0+3,+2,+1,+3,+1,+2,+3.
*/
#include <stdio.h>

int main()
{
    int n = 0;
    int sum = 0;
    int num = 0;
    scanf("%d", &n);

    while (1)
    {
        num += 3;
        if (num >= n)
            break;
        sum += num;
        num += 2;
        if (num >= n)
            break;
        sum += num;
        num += 1;
        if (num >= n)
            break;
        sum += num;
        num += 3;
        if (num >= n)
            break;
        sum += num;
        num += 1;
        if (num >= n)
            break;
        sum += num;
        num += 2;
        if (num >= n)
            break;
        sum += num;
        num += 3;
        if (num >= n)
            break;
        sum += num;
    }

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