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POJ 3537 Crosses and Crosses

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

#13
#26

Sample Output

#11
#22

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 NeighThornusing 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;}

  


By NeighThorn

POJ 3537 Crosses and Crosses