题解:

只要输出n/2即可

代码:

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int T,n,x,y;
int main()
{
    scanf("%d",&T);
    while (T--)
     {
         scanf("%d",&n);
         for (int i=1;i<=3*n/2;i++)scanf("%d%d",&x,&y);
         printf("%d\n",n/2);
     } 
    return 0; 
}

 

posted on 2018-01-07 18:41  宣毅鸣  阅读(79)  评论(0编辑  收藏  举报