Problem1-Project Euler

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。

 #include"stdio.h"

 #define MAX 1000

 int main()        /*problem1：Multiples of 3 and 5*/
 {
     int sum=,i;
     for(i=;i<MAX;i++)
         if(i%==||i%==)
             sum+=i;
     printf("sum is %d\n",sum);
     return();
 }