HDU 1907————anti-nim游戏

题目：

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

分析：

    anti-nim游戏：谁最后走最后一步谁输，先手赢的必胜策略是

 1）所有堆石子数都为1而SG值为0；

2）存在某堆石子数不为1且SG值不为0.

#include<cstdio>
int main()
{
    int t,n,a[50],ans;
    int flag;
    scanf("%d",&t);
    while(t--)
    {
        flag=1;
        ans=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]!=1) flag=0;
            ans^=a[i];
        }
        if(flag)
        {
            if(n&1)
                printf("Brother\n");
            else
                printf("John\n");
        }
        else
        {
            if(ans)
            {
                printf("John\n");
            }
            else
                printf("Brother\n");
        }
    }
    return 0;
}