(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.

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;
}

 

posted @ 2014-02-11 13:47  acutus  阅读(150)  评论(0编辑  收藏  举报
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