*HDU1907 博弈

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4537    Accepted Submission(s): 2602

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

Southeastern Europe 2007

题意：
有n堆物品，两个人，每次只能取一堆中的若干个，最后取完者败。

代码

 //S0，T2必败，其他情况必胜
 #include<iostream>
 using namespace std;
 int main()
 {
     int t,n,x;
     cin>>t;
     while(t--)
     {
         cin>>n;
         int ans=,flag=;
         for(int i=;i<n;i++)
         {
             cin>>x;
             ans=(ans^x);
             if(x>) flag++;
         }
         if(!flag)
         {
             if(n&) cout<<"Brother\n";
             else cout<<"John\n";
         }
         else
         {
             if(ans==&&flag>=) cout<<"Brother\n";
                 else cout<<"John\n";
         }

     }
     return ;
 }