【hdu 2147】kiki's game

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 3
5 4
6 6
0 0

Sample Output
What a pity!
Wonderful!
Wonderful!

【题目链接】:http://acm.hdu.edu.cn/submit.php?pid=2147

【题解】

这题如果直接用模拟,n*m从必败往必胜的方向搞；
会超时；
正确做法是
先找规律；
手动模拟一下必败到必胜的转换

(0表示到这个点后必败，1表示必胜);
(只要(i,j-1),(i+1,j-1),(i+1j,j)这3个点中有一个点的状态为必败态，则i,j点的状态为必胜态,因为我们可以接下来一步走到那个点去,这样对方就面临那个必败态了;)

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int n,m;

int main()
{
    //freopen("F:\rush.txt","r",stdin);
    rei(n);rei(m);
    while (n!=0)
    {
        if (n%2==0 || m%2==0)
            puts("Wonderful!");
        else
            puts("What a pity!");
        rei(n);rei(m);
    }
    return 0;
}