hdu1907

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 945    Accepted Submission(s): 495

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

233 5 111

 

Sample Output

JohnBrother

 

Source

Southeastern Europe 2007

 

#include<iostream>
using namespace std;
int main()
{
    int T;
    cin>>T;
    while(T>0)
    {
        T--;
        int n;
        cin>>n;
        int u,k=0,t;
        cin>>t;
        if(t>1)k++;
        for(int i=1;i<n;i++)
        {
           cin>>u;
           t=t^u;
           if(u>1)k++;
        }
        if(k==0)
        {
            if(n%2==0)cout<<"John"<<endl;
            else cout<<"Brother"<<endl;
        }
        else
        {
            if(t!=0)cout<<"John"<<endl;
            else cout<<"Brother"<<endl;
        }
    }
    return 0;
}

这个是一个博弈问题，详细情况看我转载的博弈题目总结。。。。