HDOJ 题目2955 kiki's game（博弈）

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kiki's game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/1000 K (Java/Others)
Total Submission(s): 6501 Accepted Submission(s): 3866

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 35 46 60 0

Sample Output

What a pity!Wonderful!Wonderful!

Author

月野兔

Source

HDU 2007-11 Programming Contest

题目大意：一个硬币放在棋盘的最右上角，每一次可以向左或向下或左下移动，kiki和他的一个小伙伴玩，kiki先移，看谁能赢

题目分析：博弈题，我感觉都是有规律可寻的，第一步先找他的必败点,标记为p即右下角，第二部把能通向必败点的标记为必胜点n，第三步，把只能通往必胜点的地方标记为必败点p，否则标记为必胜点n，标完，大概如图所示

很容易看出规律；

ac代码

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

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