return to do hat won first appears perf art pear

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?

InputInput 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.

OutputIf 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!




最後一個點一定是必敗點，能一步到必敗點的點一定是必勝點，只能到必勝點的點一定是必敗點。
如此，就可以推出來全部的點

 1 #include<iostream>
 2 using namespace std;
 3 #include<cstdio>
 4 #include<cstring>
 5 int n,m;
 6 bool f[2010][2010];
 7 bool v[2010][2010];
 8 int dfs(int x,int y)
 
 9 {
10 //    cout<<x<<" "<<y<<endl;
11     if (v[x][y]) return f[x][y];
12     int a1=1,a2=1,a3=1;
13     if ((y-1)>0&&x) a1=dfs(x,y-1);
14     if ((x+1)<=n&&y) a2=dfs(x+1,y);
15     if ((x+1)<=n&&(y-1)>0) a3=dfs(x+1,y-1);
16     bool s=(a1&&a2&&a3);
 
17     f[x][y]=!s;
18     v[x][y]=1;
19     if (!s) {
20     return 1;}
21     else
22     {
23     return 0;
24     }
25 }
26 int main()
27 {
28     n=2000;
29     dfs(0,2000);
30     while(cin>>n>>m&&n&&m)
31     {
32         bool ans=f[2001-n][m];
33         if (ans)
34         {
35             cout<<"Wonderful!\n";
36         }
37         else
38         {
39             cout<<"What a pity!\n";
40         }
41     }
42     return 0;
43 }

HDU - 2147 kiki's game