hdu-2147-kiki's game-博弈论-java

最新推荐文章于 2022-10-02 23:40:50 发布

Li-金玉良言

最新推荐文章于 2022-10-02 23:40:50 发布

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分类专栏： hdu 文章标签： java 博弈 hdu

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kiki's game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/10000 K (Java/Others)
Total Submission(s): 10788 Accepted Submission(s): 6540

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!
   
   

   
   解题思路：题目大意两个人轮流在n*m的方格上移动从右上角看谁先到达左下角
   
            经典的巴什博奕从左下角开始逆推
   
            因为两个人轮流走并且每次只走一步所以输赢状态轮流交替
   
            用图形表示就是
   
            输  赢  输  赢  输
   
            赢             赢
   
            输             输
   
            赢             赢
   
            输  赢  输  赢  输
   
          中间的格子无需考虑不影响解题
   
          四条边上都是一个输一个赢
   
          所以当n和m都为奇数时 右上角的状态为必输
   
          那么这题就解完了
   
   

   
   ac代码：
   
   import java.util.Scanner;




public class Main {


	public static void main(String[] args) {
		// TODO 自动生成的方法存根
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNext()) {
				int a = scanner.nextInt();
				int b = scanner.nextInt();
				if (a==0&&b==0) {
					return;
				}
				if (a%2==1&&b%2==1) {
					System.out.println("What a pity!");
				}else {
					System.out.println("Wonderful!");
				}
			
		}
	}


}