HDU_2147——组合博弈,转换为P/N图,然后找规律

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!

#include <iostream>
using namespace std;

int a[][];
void init(void)
{
    a[][] = ;    //必败点
    for(int i = ; i < ; i++)   //第一行与第一列
    {
        if(a[][i-])
            a[][i] = ;    //一步能进入必败点标记为必胜点
        else
            a[][i] = ;    //不能标记为必败点

        if(a[i-][])
            a[i][] = ;    //一步能进入必败点标记为必胜点
        else
            a[i][] = ;    //不能标记为必败点
    }
    for(int i = ; i < ; i++)   //中间的位置
    {
        for(int j = ; j < ; j++)
        {
            if(a[i][j-] || a[i-][j] || a[i-][j-])
                a[i][j] = ;    //一步能进入必败点标记为必胜点
            else
                a[i][j] = ;    //不能标记为必败点
        }
    }
}

int main()
{
    /*
    画P/N图得出当n,m为奇数时kiki必败
    init();
    for(int i = 1; i < 50; i++)
    {
        for(int j = 1; j < 50; j++)
        {
            if(a[i][j])
            {
                cout << i%2 << " " << j%2 << endl;
            }
        }
    }
    */
    int n, m;
    while(cin >> n >> m && (n || m))
    {
        if(n% && m%)
            cout << "What a pity!" << endl;
        else
            cout << "Wonderful!" << endl;
    }

    return ;
}