E - 尤文图斯

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!". 

题意：有一个n*m的格子图，n行m列，开始硬币放在(1,m)，每人只能走一步，每一步只能往左，往下，往左下走，谁最后走到(n,1)谁就赢，如果是先手赢输出Wonderful!，否则输出What a pity!。

思路：这是一个博弈论问题，如果是偶数列，先手必会赢；如果是奇数列，不管先手是往左走还是往左下走，列数都会变成偶数列并且自己变成下一句的后手，所以此时先手只能往下走，走到底再往左走，此时如果行数是偶数则先手赢，否则后手赢。

Sample Input

5 3

5 4

6 6

0 0

Sample Output

What a pity!

Wonderful!Wonderful!

#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
      if(n==0&&m==0)
      break;
      if(m%2==0)
      printf("Wonderful!\n");
      else
      {
        if(n%2==0)
        printf("Wonderful!\n");
        else
        printf("What a pity!\n");
      }
    }
    return 0;
}