HDU 1907 John

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 1786    Accepted Submission(s): 966

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

 

Source

Southeastern Europe 2007

 

Recommend

 

 

题意：转化为一对若干堆的火柴，两人依次从中取，规定每次只能从一堆去若干个，

        可以将其全部取走，但不可取，最后去完的为输。

        因为是最后取完为输的，那么......现在没有推出来自己的方法。

        按照，作者的方法，T2 S0为输。否则是赢的。

代码：

 

#include<stdio.h>
int f[48];
int main()
{
    int i,n,m,t,num1,num2;
    while(scanf("%d",&t)>0)
    {
        while(t--)
        {
            scanf("%d",&n);
            m=0;num1=0;num2=0;
            for(i=1;i<=n;i++)
            {
                scanf("%d",&f[i]);
                m=m^f[i];
                if(f[i]==1) num1++;
                if(f[i]>1) num2++;
            }
            if(m==0)//qi yi
            {
                if(num2>=2)
                    printf("Brother\n");
                else printf("John\n");
            }
            if(m>0) //fei qi yi
            {
                if(num1==n)
                    printf("Brother\n");
                else printf("John\n");
            }
        }
    }
    return 0;
}