Problem1-Project Euler

Multiples of 3 and 5

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.

问题很简单，只要检索1000以内能被3或5整除的数字，然后累和就OK了。

 1 #include"stdio.h"
 2 
 3 #define MAX 1000
 4 
 5 int main()        /*problem1：Multiples of 3 and 5*/
 6 {
 7     int sum=0,i;
 8     for(i=1;i<MAX;i++)
 9         if(i%3==0||i%5==0)
10             sum+=i;
11     printf("sum is %d
",sum);
12     return(0);
13 }