hdu 2147 kiki's game(博弈论)

最新推荐文章于 2020-08-04 12:27:03 发布

原创最新推荐文章于 2020-08-04 12:27:03 发布 · 916 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#C++ #博弈 #巴士游戏

探讨在特定棋盘游戏中的胜负规律，通过分析游戏规则和步数条件，判断玩家kiki是否能赢得与ZZ的对局。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

kiki's game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/10000 K (Java/Others)
Total Submission(s): 7594 Accepted Submission(s): 4507

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

题目分析：
因为行和列分别要走n-1步和m-1步，所以只要保证剩下的步数有一个是二的倍数即可保证不输，那么也就是当前点必须存在至少一个奇数
所以就是n-1和m-1要至少存在一个奇数

#include"stdio.h"

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