HDU 2147 kiki's game(巴什博弈)

最新推荐文章于 2020-12-09 12:37:38 发布

还是太年轻

最新推荐文章于 2020-12-09 12:37:38 发布

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分类专栏：博弈论

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本文介绍了一款基于巴什博弈原理的棋盘游戏。玩家需要将棋盘上的硬币移动到特定位置，最终无法行动的玩家输掉游戏。文章提供了判断先手玩家胜率的代码实现，并解释了奇偶性在游戏中的关键作用。

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kiki's game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/10000 K (Java/Others)
Total Submission(s): 12354 Accepted Submission(s): 7509

Problem Description

Recently kiki has nothing to do. While she is bored, an idea appears in his mind, she just playes the checkerboard game.The size of the chesserboard is n*m.First of all, a coin is placed in the top right corner(1,m). Each time one people can move the coin into the left, the underneath or the left-underneath blank space.The person who can't make a move will lose the game. kiki plays it with ZZ.The game always starts with kiki. If both play perfectly, who will win the game?

Input

Input contains multiple test cases. Each line contains two integer n, m (0<n,m<=2000). The input is terminated when n=0 and m=0.

Output

If kiki wins the game printf "Wonderful!", else "What a pity!".

Sample Input

Sample Output

What a pity!
Wonderful!
Wonderful!

题解：
有两堆石子，一次可以从其中一堆取出一个，也可以同时从
两堆中各取一个，最后取完者获胜。巴什博弈的原型。
n,m只要有一个为偶数则先手赢。
#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m)&&(n+m))
    {
        if(!(n%2)||!(m%2))printf("Wonderful!\n");
        else printf("What a pity!\n");
    }
    return 0;
}


开始不知道这是巴什博弈就自己打了个表但是样例都不能过，不知道为什么。
如有过路大神还请指教：


#include<bits/stdc++.h>
using namespace std;
bool dp[2002][2002];
int main()
{
    for(int i=0;i<2002;i++)
    {
        if(i%2==1)dp[0][i]=dp[i][0]=1;
        else dp[0][i]=dp[i][0]=0;
    }
    for(int i=1;i<6;i++)
    {
        for(int j=1;j<6;j++)
        {
            if(!dp[i-1][j]||!dp[i][j-1]||!dp[i-1][j-1])
                dp[i][j]=1;
        }
    }
    int n,m;
    while(~scanf("%d%d",&n,&m)&&(n+m))
    {
        if(!dp[n][m])printf("What a pity!\n");
        else printf("Wonderful!\n");
    }
    return 0;
}