题目链接:hdu5882 Balanced Game 题解:每种手势的攻防数一样,不难想到n为奇数时游戏平衡. #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main(){ int t, x; scanf("%d", &t); while(t--){ scanf("%d", &x); ) puts(&q…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=5882 解法:一个点必须出度和入度相同就满足题意,所以加上本身就是判断奇偶性 #include<stdio.h> #include<math.h> #include<string.h> #include<stack> #include<set> #include<queue> #include<vector> #include<…

Balanced Game Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 115    Accepted Submission(s): 99 Problem Description Rock-paper-scissors is a zero-sum hand game usually played between two people,…