Anti_SG博弈,SJ定理_______John( POJ 3480 )

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 A_i, separated by spaces – amount of M&Ms of i-th color.

Constraints:
1 <= T <= 474,
1 <= N <= 47,
1 <= A_i <= 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

233 5 111

Sample Output

JohnBrother

题意：

有n堆石子，每堆石子的个数已知，两个人轮流从某个石堆取石子，至少去取一个，最多将该堆石子取完。谁取完最后一个石子谁输。

问先手比赢还是后手比赢？

分析：

明显的Anti_SG游戏，而且模型还是使用的nim游戏模型，根据SJ定理可以得到将所有石堆的sg值异或起来为ans,所有石堆的sg值中大于1的石堆个数为cnt.

先手比赢当且仅当  ans != 0 && cnt > 0 ||  ans == 0 && cnt == 0

如果对Anti_SG游戏或者SJ定理不了解的请看这里 ：    虚位以待。

代码：

#include<stdio.h>#include<string.h>int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,step,ans,flag;        flag = 0;ans = 0;        scanf("%d",&n);        for(int i = 0 ; i < n ; i ++)        {            scanf("%d",&step);            if(step > 1) flag ++;            ans ^= step;        }        if(ans == 0 && flag == 0 || ans != 0 && flag != 0)            printf("John\n");        else            printf("Brother\n");    }    return 0;}