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 }