HDU 2147 kiki's game(规律，博弈)

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转载最新推荐文章于 2019-05-29 10:01:17 发布 · 116 阅读

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#java #php #开发工具

本文介绍了一个关于棋盘游戏的博弈问题，通过分析游戏规则，得出了一种简单的判断胜负的方法。只要将棋盘大小相乘后的结果对2取余数，即可快速判断先手玩家是否能获胜。

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kiki's game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/10000 K (Java/Others)
Total Submission(s): 10763 Accepted Submission(s): 6526

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!
  

Author

月野兔

Source

HDU 2007-11 Programming Contest

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2147

分析：推几组数据就可以猜出结论，(a*b)%2==0?"Wonderful!":"What a pity!"

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int a,b;
 6     while(scanf("%d%d",&a,&b)&&(a&&b))
 7     {
 8         if(a*b%2==0)
 9             printf("Wonderful!\n");
10         else printf("What a pity!\n");
11     }
12     return 0;
13 }