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

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
2
3
3 5 1
1
1

Sample Output
John
Brother

题意，有n堆糖果，两人轮流从任意里拿至少1个，谁拿到最后一个就输。
尼姆博弈。对于N堆的糖，一种情况下是每堆都是1，那么谁输谁赢看堆数就知道；对于不都是1的话，若这些堆是奇异局势，或说他们是非奇异局势，但非奇异局势皆可以转换到奇异局势。
经典的尼姆问题是谁哪拿到最后一个则谁赢，本题是拿最后一个的输。下面分析第二种情况：
1.初始给的是奇异局势的话，则先取者为输。
2.初始给的是非奇异局势的话，则先取者为赢。辗转转换非奇异、奇异的次数是相对的。
我们用（a，b，c）表示某种局势，首先（0，0，0）显然是奇异局势，无论谁面对奇异局势，都必然失败。第二种奇异局势是（0，n，n），只要与对手拿走一样多的物品，最后都将导致（0，0，0）。仔细分析一下，（1，2，3）也是奇异局势，无论对手如何拿，接下来都可以变为（0，n，n）的情形。

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，我们用符号（+）表示这种运算，先看（1，2，3）的按位模2加的结果：

1 =二进制01

2 =二进制10

3 =二进制11 （+）

———————

0 =二进制00 （注意不进位）

对于奇异局势（0，n，n）也一样，结果也是0。

#include <iostream>

using namespace std;

int main()
{int t,n;
 int a,k,sum;
 cin>>t;
 while(t--)
 {k=0;sum=0;
     cin>>n;
     for(int i=0;i<n;i++)
     {
         cin>>a;
         if(a>1)
         {
            k=1;
         }
         sum^=a;
     }
     if(k==0)
     {
         if(sum%2)
         cout<<"Brother"<<endl;
         else
         cout<<"John"<<endl;
     }
     else
     {
         if(sum)
         cout<<"John"<<endl;
         else
         cout<<"Brother"<<endl;
     }
 }

    return 0;
}