HDUOJ----John

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2095    Accepted Submission(s): 1133

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

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

Recommend

lcy

简单的nim博弈.....(但是要注意全部为1的情况)....那是就不是nim博弈啦！！

代码：我想说的是，要把名字看清楚，开始将Johe---打错了...打成jone  ，哎！！，尼玛，那叫一个错呀！！

//http://acm.hdu.edu.cn/showproblem.php?pid=1907
#include<iostream>
#include<vector>
#include<cstdio>
using namespace std;

int main()
{    
    int t,n,i,x;
    bool flag ;
    cin>>t;
    while(t--)
    {
        scanf("%d",&n);
        vector<int>num(n+1);
        x=0,flag=false;
        for(i=0;i<n;i++)
        {
            scanf("%d",&num[i]);
             x^=num[i];
        }
        for(i=0;i<n;i++)
        {
             if(num[i]>1)
             {
               flag=true;
               break;
             }
        }
        if(flag)
        {
            if(x)     puts("John");
            else    puts("Brother");
        }
        else
        {
            if(n&1)   
                 puts("Brother");  //奇数 win
            else   
                puts("John");   //偶数 win
        }
    }
    return 0;
}