poj3537 Crosses and Crosses----sg 分解游戏

Crosses and Crosses

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 1866		Accepted: 687
Case Time Limit: 2000MS

Description

The game of Crosses and Crosses is played on the field of 1 × n cells. Two players make moves in turn. Each move the player selects any free cell on the field and puts a cross ‘×’ to it. If after the player’s move there are three crosses in a row, he wins.

You are given n. Find out who wins if both players play optimally.

Input

Input file contains one integer number n (3 ≤ n ≤ 2000).

Output

Output ‘1’ if the first player wins, or ‘2’ if the second player does.

Sample Input

  
  
  
  

   
   
   
   
 
    
 
     
#1
 
     
3
 
    
 
    
 
     
#2
 
     
6
 
    
 
   
  
  
  
  

Sample Output

  
  
  
  

   
   
   
   
 
    
 
     
#1
 
     
1
 
    
 
    
 
     
#2
 
     
2
 
    
 
   
  
  
  
  

Source

Northeastern Europe 2007, Northern Subregion

 

在第I个位置放一个X，即可分为两个子游戏，I-3和n-I-2

 

#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<memory.h>
using namespace std;
int sg[2100];
int dfs(int n)
{
    if(n<0) return 0;//n<0
    if(sg[n]>=0) return sg[n];
    bool g[2001]={0};
    for(int i=1;i<=n;i++)
    {
        int t=dfs(i-3)^dfs(n-i-2);
        g[t]=1;
    }
    for(int i=0;;i++)
    if(g[i]==0) return sg[n]=i;
}
int main()
{
    memset(sg,-1,sizeof(sg));
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(dfs(n)) puts("1");
        else puts("2");
    }
}