Hdu 1907 John

John

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

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

 

分为两种情况：

1 每个盒子里都只有一颗巧克力；

2 至少有一个盒子的巧克力数量大于一。

对于1情况，若是奇数个盒子，那么John输，否则John赢。

对于2情况，则是尼姆博弈。

#include<cstdio>int main(){    int T,N,x,tmp,flag;    scanf("%d",&T);    while(T--)    {        tmp=0;        flag=0;        scanf("%d",&N);        while(N--)        {            scanf("%d",&x);            tmp^=x;            if(x>1)                flag=1;        }        if(flag==1)        {            if(tmp==0)                printf("Brother\n");            else                printf("John\n");        }        else        {            if(tmp!=0)                printf("Brother\n");            else                printf("John\n");        }    }    return 0;}