hdu1907(尼姆博弈)

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2348    Accepted Submission(s): 1267

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

 

     尼姆博弈模型。若火柴异或值为0，为利他态；否则为利己态。本题较传统模拟有点小变化，即取到最后一根者败。若为利他态且为1的堆为0，则胜；若为利己态且为1的堆个数不为0，也胜。其他的输。在双方都采取最优策略的情况下。

import java.util.Scanner;public class hdu1907 {public static void main(String[] str){Scanner key=new Scanner(System.in);int n,i,T,num;int [] pile=new int[500];T=key.nextInt();while(T!=0){T--;num=0;n=key.nextInt();for(i=0;i<n;i++)pile[i]=key.nextInt();int ans;for(ans=0,i=0;i<n;i++){if(pile[i]>1)num++;ans^=pile[i];}if((ans==0&&num==0)||(ans!=0&&num!=0))System.out.println("John");else System.out.println("Brother");}}}