kiki's game

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/46236809

本文介绍了一种在n*m棋盘上的游戏，玩家从右上角开始，每次可向左、下或左下方移动，直至无法移动者为败。通过分析得出，仅当n和m均为奇数时，先手会输。

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Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/10000 K (Java/Others)
Total Submission(s): 7753 Accepted Submission(s): 4596

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!

题意：在一个n*m的棋盘上，从右上角开始，每次可以往左一格，往下一格，或左下一格。最后不能走的人输，问先手胜负。

思路：打表后可以看出只有当n和m都是奇数时先手才会输。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
int T,t,n,m;
int main()
{
    int i,j,k;
    while(~scanf("%d%d",&n,&m) && n+m>0)
    {
        if(n&1 && m&1)
          printf("What a pity!\n");
        else
          printf("Wonderful!\n");
    }
}