HDU1907：John(Nim)

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
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    今天看了下博弈的文章，长篇大论啊。。。姑且分享下那篇文章吧。。。
   
   
   
   
   
   
   
   

    
    
    http://blog.csdn.net/acm_cxlove/article/details/7854530
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    这题就是最基础的nim博弈了
   
   
   
   
   
   
   
   

    
    
    
    只要按照那里总结出来的必胜态与必败态来处理即可
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int main()
{
    int t,n,i;
    int sum1,sum2,ans;
    int a[55];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        sum1 = sum2 = ans = 0;
        for(i = 0;i<n;i++)
        {
            scanf("%d",&a[i]);
            ans^=a[i];
            if(a[i]>=2)
            sum2++;
            else
            sum1++;
        }
        if((ans && sum2) || (!ans && !sum2))
        printf("John\n");
        if((ans && sum1%2 && !sum2) || (!ans && sum2>=2))
        printf("Brother\n");
    }

    return 0;
}