hdu1907

John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 945 Accepted Submission(s): 495



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

#include<iostream>

using namespace std;

int main()

{

int T;

cin>>T;

while(T>0)

{

T--;

int n;

cin>>n;

int u,k=0,t;

cin>>t;

if(t>1)k++;

for(int i=1;i<n;i++)

{

cin>>u;

t=t^u;

if(u>1)k++;

}

if(k==0)

{

if(n%2==0)cout<<"John"<<endl;

else cout<<"Brother"<<endl;

}

else

{

if(t!=0)cout<<"John"<<endl;

else cout<<"Brother"<<endl;

}

}

return 0;

}

这个是一个博弈问题，详细情况看我转载的博弈题目总结。。。。