hdu 1907 John(anti-nim)

最新推荐文章于 2019-07-03 12:49:16 发布

原创最新推荐文章于 2019-07-03 12:49:16 发布 · 489 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#C++ #博弈

本文介绍了一种基于M&M糖果的两人博弈游戏，通过分析游戏规则及策略，得出赢家预测的方法。利用亦或运算判断不同颜色糖果堆的状态，进而确定先手玩家（John）与后手玩家（弟弟）的胜负。

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

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" );
    }
}