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HDU_1907_基础博弈nim游戏

2024-08-14 12:31:10   218人阅读

John

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4440    Accepted Submission(s): 2541


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
 

哎呀卧槽。。。

之前看nim游戏的论文，性子太急，看了一半（拿最后一个胜的情况），然后以为自己弄懂了，然后遇到这个题就有点懵逼，再把论文看完，发现之前自己理解得并不是很透彻。

无论拿最后一个胜还是拿最后一个败，都是在争夺S1状态（只有一个充裕堆）。仔细一想，确实是这样，当到达S1状态，玩家可以决定下一个状态，即把下一个状态变为T0或S0

也就是说谁争夺到S1，谁就必胜。

在拿最后一个败的情况中，当所有堆为孤单堆时，需要特殊判断一下，原因容易想到。

博弈这方面，还需多多练习，多多揣摩！！！

#include<iostream>#include<cstdio>#include<cstring>#include<stdlib.h>#include<algorithm>#include<cmath>using namespace std;int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        int res=0,num;        int flag=0;        for(int i=0; i<n; i++)        {            scanf("%d",&num);            if(num>1)                flag=1;            res^=num;        }        if(flag)        {            if(res>0)                printf("John\n");            else                printf("Brother\n");        }        else        {            if(n%2==0)                printf("John\n");            else                printf("Brother\n");        }    }    return 0;}

 

 

HDU_1907_基础博弈nim游戏


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