2147 kiki's game

最新推荐文章于 2021-05-11 13:29:50 发布

WNNSCNS

最新推荐文章于 2021-05-11 13:29:50 发布

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分类专栏：题解

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本文介绍了一种简单但有效的公式法来解决一个关于棋盘游戏的博弈问题。该问题是这样的：一枚棋子从棋盘右上角开始，两名玩家轮流将棋子向左、向下或对角线下移一格，直至无法移动者输。通过分析得出，只要判断初始位置的坐标奇偶性即可快速确定胜者。

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kiki's game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/10000 K (Java/Others)
Total Submission(s): 8453 Accepted Submission(s): 5036

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!

以为可以用递推的传统方式解决这题。。。没想到数据量太大，会MLE、、、然后发现其实就是简单的公式法

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