hdu 2147 kiki's game 博弈

最新推荐文章于 2020-12-09 12:37:38 发布

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最新推荐文章于 2020-12-09 12:37:38 发布

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分类专栏： acm_博弈文章标签：博弈 acm

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本文探讨了Kiki's Game的策略玩法，这是一种基于n*m棋盘的游戏，玩家需将棋子通过特定移动规则从右上角移至左下角。文章提供了判断初始玩家胜败的算法，并附带源代码实现。

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kiki's game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/1000 K (Java/Others)
Total Submission(s): 6933 Accepted Submission(s): 4141

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!

Author

月野兔

Source

HDU 2007-11 Programming Contest

给出 n*m的方格能往下往左或者左下三个方向走从(n,m)开始

走到(1,1)的胜利问谁能赢

#include<cstdio>
#include<cstring>
int main()
{
    int n,m;
    while(scanf("%d %d",&n,&m)!=EOF&&(n||m))
    {
        if(n%2&&m%2)
            printf("What a pity!\n");
        else
            printf("Wonderful!\n");
    }
    return 0;
}