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 =  2000+10;

int sg[N],n;

int dfs(int x) {
    if(x<=0) return 0;
    if(sg[x]!=-1) return sg[x];
    int vis[N];
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=x;i++)
        vis[dfs(i-3)^dfs(x-i-2)]=1;
    for(int i=0;;i++)
        if(!vis[i]) return sg[x]=i;
}

int main() {
    memset(sg,-1,sizeof(sg));
    while(scanf("%d",&n)==1) {
        if(dfs(n)) puts("1");
        else puts("2");
    }
    return 0;
}