John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 831    Accepted Submission(s): 437


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
 

 

Source
 

 

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代码：
#include<stdio.h>
int a[50];
int main()
{
int i,ans,T,n,temp;
    scanf("%d",&T);
while(T--)
    {
        scanf("%d",&n);
        temp=0;//孤单堆
        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");
continue;
        }         
if(ans==0)
           printf("Brother\n");
else
           printf("John\n");
    }    
return 0;
}