HDU-1907 John

浏览数：25 / 时间：2015年06月09日

http://acm.hdu.edu.cn/showproblem.php?pid=1907

John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 2385 Accepted Submission(s): 1289

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

2

3 3 5

1

1 1

Sample Output

John 

Brother

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
   int t,n,i,a[50];
   int temp,ans;
   scanf("%d",&t);
   while(t--)
   {
       temp=0;
       ans=-1;
       scanf("%d",&n);
       for(i=0;i<n;i++)
       {
           scanf("%d",&a[i]);
           if(i==0)
               ans=a[i];
           else
               ans=ans^a[i];
           if(a[i]>1)
                temp=1;
       }
       if(temp==0)
       {
           if(n%2==1)
             printf("Brother\n");
           else
             printf("John\n");
       }
       else
       {
           if(ans==0)
             printf("Brother\n");
           else
             printf("John\n");
           
       }
   }
   return 0;
}
 