*HDU1907 博弈

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4537    Accepted Submission(s): 2602

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

 

 

Source

Southeastern Europe 2007

题意：
有n堆物品，两个人，每次只能取一堆中的若干个，最后取完者败。

代码

//S0，T2必败，其他情况必胜
#include<iostream>
using namespace std;
int main()
{
    int t,n,x;
    cin>>t;
    while(t--)
    {
        cin>>n;
        int ans=0,flag=0;
        for(int i=0;i<n;i++)
        {
            cin>>x;
            ans=(ans^x);
            if(x>1) flag++;
        }
        if(!flag)
        {
            if(n&1) cout<<"Brother\n";
            else cout<<"John\n";
        }
        else
        {
            if(ans==0&&flag>=2) cout<<"Brother\n";
                else cout<<"John\n";
        }

    }
    return 0;
}