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
3
3 5 1
1
1
Sample Output
John
Brother
Brother
Source
Southeastern Europe 2007
依旧是anti-nim
AC记录喜+1
/*by SilverN*/
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int n;
int s;
int main(){
int T;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
bool flag=;
s=;
int i,j;
int a;
for(i=;i<=n;i++){
scanf("%d",&a);
s^=a;
if(a>)flag=;
}
if(flag){
if(s==)printf("Brother\n");
else printf("John\n");
}
else
if(n&)printf("Brother\n");
else printf("John\n"); }
return ;
}