kiki's game+hdu+简单巴什博弈

本文探讨了在特定棋盘游戏中,玩家Kiki与ZZ之间的博弈策略,通过分析棋盘大小对胜利者的影响,揭示了当n和m均为奇数时,Kiki将面临失败的局面。

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: 40000/1000 K (Java/Others)
Total Submission(s): 6799    Accepted Submission(s): 4062


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
5 3 5 4 6 6 0 0
 

Sample Output
What a pity! Wonderful! Wonderful!
解决方案:简单巴什博弈,当n,m都为奇数时,kiki必败。
code:
#include <iostream>
#include<cstdio>
using namespace std;

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

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值