hdu 1907 John 尼姆博奕

尼姆博奕

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3849    Accepted Submission(s): 2182

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

 

/**========================================== *   This is a solution for ACM/ICPC problem * *   @source：hdu 1907 John *   @type: *   @author: wust_ysk *   @blog:  http://blog.csdn.net/yskyskyer123 *   @email: [e-mail] *===========================================*/#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>using namespace std;typedef long long ll;const int INF =0x3f3f3f3f;int n,rich;char type;bool win(){    if(type=='T'&&!rich)  return true;    if(type=='S'&&rich==1)  return true;    if(type=='S'&&rich>=2)  return true;//之前这里写成了rich==2 Wa一发。    return false;}int main(){    int T,x;scanf("%d",&T);    while(T--)    {      rich=0;      int sum=0;      scanf("%d",&n);      for(int i=1;i<=n;i++)      {          scanf("%d",&x);          if(x>1) rich++;          sum^=x;      }      type=sum?'S':'T';      puts( win()?"John":"Brother");    }   return 0;}