偶数求和

#include<stdio.h>
int main()
{
 int N,i,j,k,M;
 while(scanf("%d%d",&N,&M)!=EOF)
 {
  int sum=0,k=0,flag=1;
  for(i=2;i<=2*N;i=i+2)
  {
   sum+=i;
   k++;
   if(k==M)
   printf(flag?"%d":" %d",sum/M),sum=0,k=0,flag=0;
   else if(i==2*N&&k!=M)
   printf(flag?"%d":" %d",sum/k);
  }
  puts("");
 }
 return 0;
}

posted on 2011-04-28 15:59  more think, more gains  阅读(148)  评论(0编辑  收藏  举报

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