hdu 1907（尼姆博弈）

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#hdu #尼姆博弈

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本文探讨了游戏开发过程中的核心策略与算法应用，详细解析了游戏开发中的关键技术和实践，包括游戏引擎、图像处理、AI音视频处理、测试与自动化运维等领域的最新进展与实战案例。

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原题：http://acm.hdu.edu.cn/showproblem.php?pid=1907

尼姆博弈（Nimm Game）：简单地说，就是有n堆若干个东西，两个人轮流从某一堆取任意多的东西，规定每次至少取一个，多者不限，最后取光者得胜。

以下内容有转载部分：

例子：

当n=3时，如果是面对（0，0，0）那就是谁输了，这也叫做奇异局势那么再从下往上推也就是说面对（0，a，b）（a，b不等于0）时也是输的，（0，a，b）也是奇异局势，同样（1，a，b）（这里a，b不相等）也是奇异局势。

计算机算法里面有一种叫做按位模2加，也叫做异或的运算，用符号^表示这种运算，先看（1，2，3）的异或的结果：
　　1 =二进制01
　　2 =二进制10
　　3 =二进制11 ^
　　———————
　　0 =二进制00
　　对于奇异局势（0，n，n）也一样，结果也是0。
　　任何奇异局势（a，b，c）都有a^b^c =0。

异或简单口诀（这指的是二进制里的相同位置上的）：相同为0，不同为1

所以对于取光者胜的题目也就是把每堆数量求异或a1^a2^...^ai'^...^an，如果结果为零则先手必输，否则必赢。

John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2582 Accepted Submission(s): 1403

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

题意：有n中不同颜色的糖，每种的数量为a个，John和Brother轮流吃，每次可以吃任意多个（至少一个）相同颜色的糖。最后一个吃完的输。

AC代码如下：

#include <stdio.h>
/*
author:YangSir
time:2014/5/3
*/
int main()
{
	int n,i,t,a,sum,flag;
	scanf("%d",&t);
	while(t--)
	{
		sum=flag=0;
		scanf("%d",&n);
		for(i=0;i<n;i++)
		{
			scanf("%d",&a);
			sum^=a;
			if(a>1)
				flag=1;//判断是否都为1
		}
		if((sum==0&&flag)||(sum&&flag==0))
			//sum==0&&flag是能化成sum&&flag==0这个形式
			//sum&&flag==0相当于有奇数个堆，每堆个数都为1
			printf("Brother\n");
		else
			printf("John\n");
	}
	return 0;
}