ACM-尼姆博弈之John——hdu1907

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2577    Accepted Submission(s): 1400

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

 

题目：http://acm.hdu.edu.cn/showproblem.php?pid=1907

近期在看博弈系列，搞完了巴什博弈、威佐夫博弈，接下来就是这个尼姆博弈。

这一系列博弈类型可以概括为：

巴什博弈：从一堆石子中拿石子，一次拿1~m个。

威佐夫博弈：从两堆石子中拿石子，方法①任选一堆石子拿k个石子（k≥1），方法②从两堆石子中拿相同数量的石子（当然所拿的数要≥1）

尼姆博弈：从三堆石子中拿石子，每次任选一堆拿任意数目（≥1）的石子。

这些当然是谁先那光，谁获胜。

尼姆博弈解法，和二进制有关。

反正我不知道他们怎么推得，只会用。o(╯□╰)o。。。

给出的数 用异或加起来，若等于0，则为奇异态（必胜态）。

这道题有些不一样，如果John吃的是某个盒子最后一颗，那就判定John为败。

所以，这道题分为两种情况讨论：

①若所有堆的数量都为1。则根据奇偶来判断谁胜。

②其他情况，将所有数据异或起来，判断是否为奇异态。

/**************************************
***************************************
*        Author:Tree                  *
*From :http://blog.csdn.net/lttree    *
* Title : John                        *
*Source: hdu 1907                     *
* Hint  : 尼姆博弈                   *
***************************************
**************************************/
#include <iostream>
#include <algorithm>
using namespace std;
int arr[48];
int main()
{
    int t,n,i,temp;
    cin>>t;
    while( t-- )
    {
        cin>>n;
        for(i=0;i<n;++i)
            cin>>arr[i];
        sort(arr,arr+n);
        // 如果全是1，按照奇偶判断谁获胜
        if( arr[n-1]==1 )
        {
            if( n&1 )   cout<<"Brother"<<endl;
            else    cout<<"John"<<endl;
            continue;
        }
        // 异或加起来
        temp=arr[0]^arr[1];
        for(i=2;i<n;++i)
            temp^=arr[i];
        if( temp==0 )   cout<<"Brother"<<endl;
        else    cout<<"John"<<endl;
    }
    return 0;
}