hdu 1907 John （anti—Nim）

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
http://acm.hdu.edu.cn/showproblem.php?pid=1907

 

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

 

题意：Nim取石子，取到最后一个的输

若局面异或和为不为0，定义其为S态，否则，定义其为T态

若一堆石子只有1个，定义其为孤独堆，否则，定义其为充裕堆

 

S0：无充裕堆，异或和不为0

S1：有1个充裕堆，异或和不为0

S2：有>=2个充裕堆，异或和不为0

T0：无充裕堆，异或和为0

T1不存在

T2：有>=2个充裕堆，异或和为0

 

S0：一定是有奇数个孤独堆，所以必败

T0：一定是有偶数个孤独堆，必胜

S1：若孤独堆个数为奇数，则拿空充裕堆，那么留给对方的是S0态

　　 若孤独堆个数为偶数，将充裕堆拿的只剩下一个，留给对方的还是S0态，

　　 所以S1必胜

 

S2：可以转到S1、T2

T2：可以转到S1、S2

由于S1是必胜态，所以S2、T2肯定都尽可能不留给对方S1

由Nim游戏证明可知，S2一定可以留给对方T2态（将异或和最大的那个1的那一堆的低位变成异或和）

而T2不一定都给对方S2态

所以S2必胜，T2必败

 

#include<cstdio>
using namespace std;
int main()
{
    int T,n,x,sum,yh;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        sum=yh=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            yh^=x;
            if(x>1) sum++;
        }
        if(yh&&!sum) printf("Brother\n");
        else if(!yh&&sum>=2) printf("Brother\n"); 
        else printf("John\n");
    }
}