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(154) 评论(0) 收藏 举报
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