anti-Nim题目标准代码

John

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2990		Accepted: 1451

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

思路：

情况1、

若所有堆里的石头数都是1，则堆的个数为偶数时，必定是John赢；奇数个时，必定是Brother赢。这都没得选，想不赢都不行。

情况2、

有一堆的个数大于1，其他堆的个数等于1。假如这堆个数为n（n>1），那么John可以选n-1和n个来控制这一步取完之后石头数量为1的堆的奇偶性，转换成情况1。如果其他石头数为1的个数为奇数，那John这里就取n-1个；如果如果其他石头数为1的个数为偶数，John这里就取n个。所以John能够让自己必赢。

情况3、

代码: