hdu 1907 John （anti—Nim）

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
http://acm.hdu.edu.cn/showproblem.php?pid=1907

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

 

题意：Nim取石子，取到最后一个的输

若局面异或和为不为0，定义其为S态，否则，定义其为T态

若一堆石子只有1个，定义其为孤独堆，否则，定义其为充裕堆

 

S0：无充裕堆，异或和不为0

S1：有1个充裕堆，异或和不为0

S2：有>=2个充裕堆，异或和不为0

T0：无充裕堆，异或和为0

T1不存在

T2：有>=2个充裕堆，异或和为0

 

S0：一定是有奇数个孤独堆，所以必败

T0：一定是有偶数个孤独堆，必胜

S1：若孤独堆个数为奇数，则拿空充裕堆，那么留给对方的是S0态，所以S1必胜

S2：可以转到S1、T2

T2：可以转到S1、S2

若T2转到了S2，则S2有转回了T2

若T2转到了S1，则T2必败

所以S2必胜

 

#include<cstdio>
using namespace std;
int main()
{
    int T,n,x,sum,yh;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        sum=yh=;
        for(int i=;i<=n;i++)
        {
            scanf("%d",&x);
            yh^=x;
            if(x>) sum++;
        }
        if(yh&&!sum) printf("Brother\n");
        else if(!yh&&sum>=) printf("Brother\n");
        else printf("John\n");
    }
}