(Problem 1)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.

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <ctype.h>
 4    
 5 void solve()
 6 {
 7   int sum,i;
 8   sum=0;
 9   for(i=3; i<1000; i++)
10   {
11      if(i%3==0 || i%5==0)
12      {
13         sum+=i;
14      }
15   }
16   printf("%d\n",sum);
17      
18 }
19    
20 int main()
21 {
22   solve();
23   return 0;
24 }

 

Answer:
233168

 

posted @ 2013-07-26 10:55  cpoint  阅读(233)  评论(0编辑  收藏  举报
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