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

233 5 111

 

Sample Output

JohnBrother

 

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" );    }}