hdu1907John (尼姆博弈，存在一种特殊情况全为1时)

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
#include<stdio.h>int main(){    int t,n,c,a[105],k;    scanf("%d",&t);    while(t--)    {        k=0;c=0;        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]); c^=a[i];            if(a[i]>1) k=1;        }        //当c为0时，也就是先手是输家       if(c&&k||!k&&n%2==0)        printf("John\n");        else printf("Brother\n");    }}