John

John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 60 Accepted Submission(s): 41

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

 

Recommend

lcy

 

/*
题意：有一个盒子，里面装着不同颜色的糖，没人每次只能吃相同颜色的糖。并且最少吃一个，谁吃到最后一个糖就会输
    就输。
初步思路：就是裸的尼姆博弈，石头堆换成相同颜色的糖
 
#错误：少了特判，如果都是1的话，就不需要看异或了。
*/
#include<bits/stdc++.h>
using namespace std;
int t,n,a[];
int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d",&t);
    int res=;
    while(t--){
        scanf("%d",&n);
        res=;
        int flag=;
        for(int i=;i<n;i++){
            scanf("%d",&a[i]);
            res^=a[i];
            if(a[i]>) flag=;
        }
        if(flag){
            printf(res?"John\n":"Brother\n");
        }else{
            printf(n&?"Brother\n":"John\n");
        }
    }
    return ;
}