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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.

题目大意：

10以下的自然数中，属于3和5的倍数的有3,5,6和9，它们之和是23.

找出1000以下的自然数中，属于3和5的倍数的数字之和。

#include <stdio.h>
#include <string.h>
#include <ctype.h>

void solve()
{
	int sum, i;
	sum = 0;
	for (i = 3; i < 1000; i++) {
		if (i % 3 == 0 || i % 5 == 0) {
			sum += i;
		}
	}
	printf("%d\n", sum);

}

int main()
{
	solve();
	return 0;
}

Answer:233168
Completed on Sun, 31 Mar 2013, 14:35