Anti-SG游戏 与 SJ定理笔记（反Nim博弈）

Anti-SG游戏定义

1、决策集合为空的操作者胜。
2、其余规则与SG游戏一致。

SJ定理

对于任意一个Anti-SG游戏，如果定义所有子游戏的SG值为0时游戏结束，先手必胜的条件：
1、游戏的SG值为0且所有子游戏SG值均不超过1。
2、游戏的SG值不为0且至少一个子游戏SG值超过1。

例：
HDU 1907 John

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

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>

using namespace std;

int main(){

    int N;
    cin>>N;
    while(N--){
        int M;
        cin>>M;
        int mid,a;
        a = 0;
        int temp = 0;
        while(M--){
            scanf("%d",&mid);
            if(mid>1)temp = 1;
            a ^= mid;
        }
        if((a && temp) || (!a && !temp))printf("John
");
        else printf("Brother
");
    }

    return 0;
}