John

Time Limit : 5000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 66   Accepted Submission(s) : 31

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<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){    int t, n, a, sum, flag;    scanf("%d", &t);    while (t--)    {        scanf("%d", &n);        sum = 0; flag = 0;        for (int i = 1; i <= n; i++)        {            scanf("%d", &a);            sum = sum^a;            if (a > 1)flag = 1;        }        if (flag == 0)        {            if (n % 2 == 0)  printf("John\n");            else printf("Brother\n");        }        else if (sum == 0) printf("Brother\n");        else printf("John\n");    }    return 0;}