关于博弈（OJ kiki'game)

题目描述：

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? 

输入：

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. 

输出：

If kiki wins the game printf "Wonderful!", else "What a pity!". 

样例输入：

5 3

5 4

6 6

0 0

样例输出：

What a pity!

Wonderful!

Wonderful!

代码：

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

分析：本题其实是一道关于博弈的题目，从右下角往左上角推导，当n=m=2即为2*2方阵时，无论kiki或ZZ谁先进入这个2*2方阵就意味着谁输了，根据这一思路往上推得谁先处于n和m都为奇数的方格时，她就输了。