HDU 1907 John

^^^转载请注明出处，谢谢合作O(∩_∩)O~

John

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

解：典型的尼姆博弈，只要判断谁先拿谁输的条件即可

#include<iostream>

using namespace std;

#define M 5000

int main()

{

    int t;

    int a[M];

    cin>>t;

    while(t--)

    {

        int i,n;

        int s=0;//存异或值

        int s1=0;//计数大于等于2的堆

        cin>>n;

        for(i=0;i<n;i++)

        {

          cin>>a[i];

          if(a[i]>=2)s1++;

          s=s^a[i];

        }

        if(s==0)

        {

            if(s1>=2)

            cout<<"Brother"<<endl;            

            else 

            cout<<"John"<<endl;    

        }

        else

        {

            if(s1>=1)

            cout<<"John"<<endl;

            else

            cout<<"Brother"<<endl;

        }

          

    }

    

    return 0;

}