hdu 1907 John(anti-nim)

John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 3308 Accepted Submission(s): 1866



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

Source
Southeastern Europe 2007

题目分析：

反尼姆博弈，就是必胜态是S2,S1,T0,必败态是T2,S0

S代表堆的亦或和不为０，T代表堆的亦或和为０

只有一个石头的堆叫做孤单堆，有多个石头的堆叫做正常堆，那么当正常堆只有一个 时，因为高位一定之存在一个，所以只有Ｓ１而没有Ｔ１,那么根据实际情况的，T2能够转移到的只有S2,S1，因为无论T堆中取出任何数，剩下的堆都不能保证亦或和为０，只可能通过是否拿掉一整堆来判断是S2还是Ｓ１,如果是S1，一定能够通过取石子得到S0，而S0只能得到T0,对于S2，一定能够通过取数到达T2（nim游戏中可以理解到这一点），所以必胜态是S2,S1,T0，根据这个结论就可以得到最终结果

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

int t,n,a,ans,sum;

int main ( )
{
    scanf ( "%d" , &t );
    while ( t-- )
    {
        scanf ( "%d" , &n );
        sum = ans = 0;
        for ( int i = 0 ; i < n ; i++ )
        {
            scanf ( "%d" , &a );
            ans ^= a;
            if ( a > 1 ) sum++;
        }
        if ( ( ans == 0 && sum == 0 ) || ( ans != 0 && sum != 0 ) )
           printf ( "John\n" ); 
        else printf ( "Brother\n" );
    }
}