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John 尼姆博弈

时间:2017-09-15 13:53:57      阅读:202      评论:0      收藏:0      [点我收藏+]

标签:describe   const   imp   inf   always   scanner   rgs   player   main   

John

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. 

InputThe 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 

OutputOutput 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

对于N堆的糖,一种情况下是每堆都是1,那么谁输谁赢看堆数就知道;
对于不都是1的话,若这些堆是奇异局势,或说他们是非奇异局势,但非奇异局势皆可以转换到奇异局势。

经典的尼姆问题是谁哪拿到最后一个则谁赢,本题是拿最后一个的输。
下面分析第二种情况:
1.初始给的是奇异局势的话,则先取者拿到最后一个为输。
2.初始给的是非奇异局势的话,则先取者为赢。
	对于任何奇异局势(a,b,c),都有a^b^c=0(^是代表异或).
非奇异局势(a,b,c)(a<b<c)转换为奇异局势,只需将c变成a^b,即从c中减去c-(a^b)即可

import java.util.*;

public class Main {
    
    static Scanner sc = new Scanner(System.in);
    
    public static void main(String[] args){
        
        int t, n, x, f, ans, i;
        t = sc.nextInt();
        while((t--) != 0){
            n = sc.nextInt();
            ans = 0; f = 0;
            for(i = 1; i <= n; i++){
                x = sc.nextInt();
                ans ^= x;
                if(x > 1) f = 1;
            }
            if(f == 0){
                if((n & 1) != 0) System.out.println("Brother");
                else System.out.println("John");
            }
            else{
                if(ans == 0) System.out.println("Brother");
                else System.out.println("John");
            }
        }
    }
}

 

John 尼姆博弈

标签:describe   const   imp   inf   always   scanner   rgs   player   main   

原文地址:http://www.cnblogs.com/yzm10/p/7525717.html

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