F - John HDU - 1907(尼姆博弈)

最新推荐文章于 2018-03-29 19:48:19 发布

galesaur_wcy

最新推荐文章于 2018-03-29 19:48:19 发布

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分类专栏：博弈

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本文介绍了一个关于M&M糖果的游戏，玩家轮流吃掉相同颜色的糖果，拿到最后一颗糖果的人视为输家。通过分析糖果数量及颜色分布，利用博弈论中的异或运算，给出了一种确定赢家的方法。

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Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to eat several M&Ms of the same color. Then his opponent has to make a turn. And so on. Please note that each player has to eat at least one M&M during his turn. If John (or his brother) will eat the last M&M from the box he will be considered as a looser and he will have to buy a new candy box.

Both of players are using optimal game strategy. John starts first always. You will be given information about M&Ms and your task is to determine a winner of such a beautiful game.

Input

The first line of input will contain a single integer T – the number of test cases. Next T pairs of lines will describe tests in a following format. The first line of each test will contain an integer N – the amount of different M&M colors in a box. Next line will contain N integers Ai, separated by spaces – amount of M&Ms of i-th color.

Constraints:
1 <= T <= 474,
1 <= N <= 47,
1 <= Ai <= 4747

Output

Output T lines each of them containing information about game winner. Print “John” if John will win the game or “Brother” in other case.

Sample Input

Sample Output

John
Brother

题目的意思就是说谁拿到最后一颗谁就是输

嗯这个要考虑两种情况就是

1 特殊情况如果全部都是1的话是偶数的话那么就是先手赢不然就是后者赢

2 就是尼姆博弈嗯面对奇异局势的话哪一方都没有办法

代码如下

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
	int t;
    cin>>t;
    while(t--)
    {
    	int n;
    	cin>>n;
    	int flag=0;
    	int temp=0,gg;//0和任何东西yi或都是  任何东西 
        for(int i=1;i<=n;i++)
        {
        	cin>>gg;
        	temp=temp^gg; 
        	if(gg>1)
        	flag=1;
        }
		if(flag)
		{
			//面对奇异局势那么哪一方都没有办法 
			if(temp==0)
				puts("Brother");
			else 
				puts("John");
			
			
		}//一下是特殊情况需要具体考虑下
		//如果都是1 并且是偶数 那么是先手赢 不然就是后手赢 
		else 
		{
			if(n%2==1)
				puts("Brother");
			else 
				puts("John");
		}		
    }
	
	return 0;
}