hdu1907Jhon(妮姆博奕)

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3758    Accepted Submission(s): 2124

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

233 5 111

 

Sample Output

JohnBrother

 题目大意：John和其Brother吃巧克力，初始时共有 n盒，每盒有 a[i]个，两人轮流吃，且每次john开始，最后取完者输；

解题思路：其实在整体上是运用尼姆博弈来解题，对于所有盒中都为 1 时需特判，其余情况就是妮姆博奕。

可知，当初始时，所有盒中巧克力数都为1时，若 n 为奇数，则Brother      win;否则，john  win.

当不是上述情况时就按尼姆博弈的性质求解：当所有堆数异或后的值是0（即奇异局势时），John会赢，否则Brother赢；

代码如下：

#include<stdio.h>#include<string.h>int main(){int t;int n,i;int a[50];scanf("%d",&t);while(t--){int cnt=0,s=0;scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&a[i]);s^=a[i];if(a[i]>1) cnt=1;}if(cnt==0){if(n%2==0) printf("John\n");else printf("Brother\n");}else{if(s>0) printf("John\n");else printf("Brother\n");}}return 0;}