HDU 1907 John 尼姆博奕

最新推荐文章于 2018-08-30 01:22:00 发布

LZK1997

最新推荐文章于 2018-08-30 01:22:00 发布

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本文介绍了一个有趣的博弈游戏，玩家需轮流吃掉同色M&M巧克力豆，目标是避免吃到最后一个。通过分析不同颜色巧克力豆的数量，可以预测哪位玩家将获胜。文章提供了AC代码实现，展示了如何运用异或运算确定最优策略。

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John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4471 Accepted Submission(s): 2559

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

Source

Southeastern Europe 2007

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尼姆博弈：有三堆各若干个物品，两个人轮流从某一堆取任意多的物品，规定每次至少取一个，多者不限，最后取光者得胜特例是全为1的时候，加了个flag判断

ac代码付下：

#include <stdio.h>

int main()
{
	int t,n;
	int flag,ans,num;
	
	scanf("%d",&num);
	while(num--)
	{
		scanf("%d",&n);
		flag=ans=0;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&t);
			ans=ans^t;
			if(t>1)
				flag=1;
		}
		if(!flag)
		{
			if(n&1)
			{
				printf("Brother\n");
			}
			else
			{
				printf("John\n");
			}
		} 
		else
		{
			if(!ans)
				printf("Brother\n");
			else
				printf("John\n");
		}
	}
	
	
	return 0;
}