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
Comments