HDOJ 2147 kiki's game (博弈)

kiki's game

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 40000/10000 K (Java/Others)
Total Submission(s): 8351    Accepted Submission(s): 4976


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
 

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!
注 - 此题为: HDOJ 2147 kiki's game (博弈)

题意:     在一个m*n的棋盘内,从(1,m)点出发,每次可以进行的移动是:左移一,下移一,左下移一。然后kiki每次先走,判断kiki时候会赢(对方无路可走的时候)。

思路:

         P点:就是P个石子的时候,对方拿可以赢(自己输的)   (必输点)
         N点:就是N个石子的时候,自己拿可以赢   (必胜点)

                                       

               PN状态的点:

  HDOJ 2147 kiki's game (博弈)_第1张图片

已AC代码:

#include
using namespace std;
int main()
{
	int n,m;
	while(cin>>n>>m,n,m)
	{
		if((n&1)&&(m&1))
			cout<<"What a pity!"<

你可能感兴趣的:(博弈)