hdu 1907(Nim博弈)

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4407    Accepted Submission(s): 2520

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:在n堆中取糖吃,每次可以在其中的一堆中取至少一个,谁取完所有的糖果就输了,问最后谁会赢?

此游戏可以简化为nim博弈模型:

今有若干堆火柴，两人依次从中拿取，规定每次只能从一堆中取若干根， 可将一堆全取走，但不可不取，最后取完者为负，求必胜的方法。

解法:定义：若所有火柴数异或为0，则该状态被称为利他态，用字母T表示；否则， 为利己态，用S表示。

定义：T态中，若充裕堆的堆数大于等于2，则称为完全利他态，用T2表示；若充裕堆的堆数等于0，则称为部分利他态，用T0表示。

[定理5]：S0态，即仅有奇数个孤单堆，必败。T0态必胜。

[定理6]：S1态，只要方法正确，必胜。

中间过程不再赘述:

结论:

必输态有：  T2,S0 
必胜态：    S2,S1,T0.

 

所以这题求出John的必输态即可.

#include <iostream>
#include <cstring>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <map>
using namespace std;

int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--){
        int n,sum = 0,v,_cnt=0,_cnt1=0; ///_cnt 代表孤单堆的数量,_cnt1 代表充裕堆的数量
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%d",&v);
            if(v==1) _cnt++;
            if(v>1) _cnt1++;
            sum^=v;
        }
        if(_cnt%2&&_cnt1==0||_cnt1>=2&&sum==0){
            printf("Brother
");
        }else{
            printf("John
");
        }
    }
    return 0;
}