poj 3575 Crosses and Crosses（SG函数）

Crosses and Crosses

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 3063		Accepted: 1196
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

【思路】

SG函数，博弈

题目treblecross的简化。

【代码】

 #include<cstdio>
 #include<cstring>
 using namespace std;
 
 const int N =  +;
 
 int sg[N],n;
 
 int dfs(int x) {
     if(x<=) return ;
     if(sg[x]!=-) return sg[x];
     int vis[N];
     memset(vis,,sizeof(vis));
     for(int i=;i<=x;i++)
         vis[dfs(i-)^dfs(x-i-)]=;
     for(int i=;;i++)
         if(!vis[i]) return sg[x]=i;
 }
 
 int main() {
     memset(sg,-,sizeof(sg));
     while(scanf("%d",&n)==) {
         if(dfs(n)) puts("");
         else puts("");
     }
     return ;
 }