POJ 3537 Crosses and Crosses

Posted 2020-09-01 NeighThorn

Crosses and Crosses

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 3514		Accepted: 1372
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

分析：

显然，如果一个棋子放在了棋盘上，那么这个棋子的左边两个位置和右边两个位置都不能再放棋子，问题就转化为了一个棋子的左右两边的四个位置不能放，谁不能放谁输...

对于一个$1*n$的棋盘我们把其看成一个规模为$n$的游戏，这个游戏的后继状态我们可以$O(N)$的枚举计算，所以我们可以算出所有游戏的$SG$函数...

代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
//by NeighThorn
using namespace std;

const int maxn=2000+5;

int n,f[maxn];

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

signed main(void){
	scanf("%d",&n);
	memset(f,-1,sizeof(f));f[0]=0;
	puts(sg(n)?"1":"2");
	return 0;
}

　　

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By NeighThorn