POJ3480John博弈

易水人去，明月如霜。

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. NextT pairs of lines will describe tests in a following format. The first line of each test will contain an integerN – the amount of different M&M colors in a box. Next line will containN integers A_i, separated by spaces – amount of M&Ms ofi-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：在一般的nim博弈上稍作修改，就是最后取完的输，其他一样；

先手胜当且仅当

（1）所有堆石子数都为1且游戏的SG值为0 ，（2）存在某堆石子数大于1且游戏的SG值不为0

   证明：

（1）若所有堆石子数都为1且SG值为0，则共有偶数堆石子，故先手胜。

（2）

i)只有一堆石子数大于1时，我们总可以对该堆石子操作，使操作后石子堆数为奇数且所有堆得石子数均为1

ii)有超过一堆石子数大于1时，先手将SG值变为0即可，且总还存在某堆石子数大于1

代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int n,m,x;int main(){    scanf("%d",&n);    for(int i=1;i<=n;i++)    {            scanf("%d",&m);            bool flag=0;int sg=0;            for(int i=1;i<=m;i++)            {                    scanf("%d",&x);                    sg^=x;                    if(x>1)flag=1;                    }            if(!flag&&!sg)puts("John");            else if(flag&&sg)puts("John");            else puts("Brother");            }    return 0;}