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