hdu-2147-博弈

kiki's game

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 40000/10000 K (Java/Others)
Total Submission(s): 12851    Accepted Submission(s): 7823

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!

 

Author

月野兔

 

Source

HDU 2007-11 Programming Contest

 

　　观察发现可以打表预处理一下整个棋盘，从左下角到右上角处理，每个位置只有1/0两种状态。当所有子状态全都是1的时候为0，否则为1.

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 bool f[2200][2200];
 4 int fx[3][2]={0,-1,1,0,1,-1};
 5 int main(){
 6     int n,m,i,j,k;
 7     n=2000;
 8     m=2000;
 9         f[n][1]=0;
10         for(i=n;i>=1;--i){
11             for(j=1;j<=m;++j){
12                 if(i==n&&j==1) continue;
13                 int tot1=0,tot2=0;
14                 for(k=0;k<3;++k){
15                     int dx=i+fx[k][0];
16                     int dy=j+fx[k][1];
17                     /*if(i==5&&j==3)
18                     cout<<dx<<" -"<<dy<<endl;*/
19                     if(dx>=1&&dx<=n&&dy>=1&&dy<=m){
20                         tot1++;
21                         if(f[dx][dy]) tot2++;
22                     }
23                 }
24                 
25                 if(tot1==tot2) f[i][j]=0;
26                 else f[i][j]=1;
27             }
28         }
29     while(cin>>n>>m&&(n&&m))
30         f[2000-n+1][m]?puts("Wonderful!"):puts("What a pity!");
31     return 0;
32 }

 再仔细观察发现......

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