HDU 1907 John Nim博弈

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3012    Accepted Submission(s): 1701

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

/*
1907 Nim博弈 
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
*/
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
  int t,i,n,a[50],ans,f;
  scanf("%d",&t);
  while(t--)
  {
    scanf("%d",&n);
    f=0;
    ans=0;
    for(i=0;i<n;i++)
    {
      scanf("%d",&a[i]);
      ans^=a[i];
      if(a[i]>1)
        f=1;
    }
    if(f==0)//都是1
    {
      if(n%2==0)
        printf("John\n");
      else
        printf("Brother\n");
    } 
    else
    {
      if(ans==0)//奇局势  
        printf("Brother\n");
      else
        printf("John\n"); 
    }
  }
  return 0;
}