hdu 1907 John

最新推荐文章于 2019-07-19 20:32:06 发布

原创最新推荐文章于 2019-07-19 20:32:06 发布 · 267 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#hdu #anti-Nim博弈

ACM算法_博弈专栏收录该内容

4 篇文章

订阅专栏

本文解析了一道基于Nim博弈的变形题目，通过分析不同颜色M&M糖果的数量，两个玩家轮流取糖果的游戏规则，讨论了如何判断哪位玩家会赢得游戏。文章详细解释了Nim博弈的基本原理，并给出了具体的实现代码。

点击打开链接

John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 5305 Accepted Submission(s): 3061

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

题意：有n个不用的糖果盒子，每个盒子里有相对应的糖果的数量，两个人的博弈，两人轮流拿，最少拿一个最多把这个盒子的全部拿走，拿走最后一个的输，基于Nim博弈的变形即anti-Nim。

Nim博弈

任给N堆石子，两人轮流从任意一堆中取，至少一个，至多这堆的全部，取到最后一颗石子的人获胜，先取的人如何获胜？

T=a^b^c...(n个数异或即可)，若开始的时候T=0，则先取者必败，如果开始的时候T>0，那么只要每次取的石子使T=0,即先取者有获胜的方法。

在这个题中要注意盒子中只有1的个数。

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
	int ca;
	int n,x;
	scanf("%d",&ca);
	while(ca--)
	{
		int t=0,ans=0;
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		{
			scanf("%d",&x);
			t^=x;
			if(x>1)
				ans++;
		}
		if((!ans&&n&1)||(!t&&ans>1))
			printf("Brother\n");
		else
			printf("John\n");
	}
	return 0;
}