HDU-1907-John【nimm 博弈】

最新推荐文章于 2020-10-07 16:37:01 发布

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最新推荐文章于 2020-10-07 16:37:01 发布

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本文链接：https://blog.youkuaiyun.com/m0_37345402/article/details/77428884

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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，最后一个吃掉糖果的玩家算输。通过分析糖果数量，利用nim博弈原理确定胜者。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1907

John

Problem Description

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

题目分析：

n堆石子，最后一个取完的人输。
1）如果全是1的时候，特判，奇数则输。
2）否则 nim博弈，求异或值，为0则输。

#include<iostream>
using namespace std;
int main()
{
	int T,n,x,i;
	cin>>T;
	while(T--)
	{
		cin>>n;
		int ans=0; 
		int flag=0;
		for(i=1;i<=n;i++)
		{
			cin>>x;
			ans^=x;
			if(x!=1)
			flag=1;
		}
		if(flag)
		{
			if(!ans)
			cout<<"Brother"<<endl;
			else
			cout<<"John"<<endl;
		}
		else
		{
			if(n%2)
			cout<<"Brother"<<endl;
			else
			cout<<"John"<<endl;
		}
	}
	return 0;
 }