2013
11-12

# 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.

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 T lines each of them containing information about game winner. Print “John” if John will win the game or “Brother” in other case.

2
3
3 5 1
1
1

John
Brother

//* @author:
import java.util.*;
import java.math.*;
public class Main {
public static void main(String[] args)  throws Exception{
int nn;
Scanner in=new Scanner(System.in);
nn=in.nextInt();
while ((nn--)!=0) {
int n=in.nextInt();
long ans=0,max=0;
for (int i=1;i<=n;i++) {
long j=in.nextLong();
if (j>max) max=j;
ans=ans^j;
}
if ((ans!=0)&&(max>1)||(ans==0)&&(max<=1)) System.out.println("John");
else System.out.println("Brother");
}
}
}

1. 在方法1里面：

//遍历所有的边，计算入度
for(int i=0; i<V; i++)
{
degree = 0;