博弈论 尼姆博弈，人生第一道博弈纪念一下

John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 5891 Accepted Submission(s): 3421

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

这道题本应该昨天就补的，可是最近实在学不下去，昨天晚上去打忍村了，愧疚一天。
这个题是尼姆博弈的变形题。
题目大意如下：
若干堆不同颜色的糖果，每次至少从某一堆去一个，取最后一个的人为输家。
话不多说，放代码。

#include <iostream>
#include <cstring>
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        int a[50];


        memset(a,0,sizeof(a)); 
        int res=0;
        int count=0;
        for(int i=1;i<=n;i++){
            cin>>a[i];
            if(a[i]>1){
                count++;
            }
            res=res^a[i];
        }
        res=res^0;

        if(res==0){
            if(count>=2){
                cout<<"Brother"<<endl;//T2
            }
            else if(count==1){//不可能存在T1态 
                cout<<""; 
            }
            else if(count==0){//T0 
                cout<<"John"<<endl;
            }

        }else{

            if(count>=2){
                cout<<"John"<<endl;//s2
            }
            else  if(count==1){
                cout<<"John"<<endl;//s1
            }
            else if(count==0){
                cout<<"Brother"<<endl;//s0
            }
        }

    }
} 

下面的链接是斌神的分析。
https://www.cnblogs.com/kuangbin/archive/2011/08/28/2156426.html