标签：accept name rip oss ast order eof 问题 函数
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 3514 | Accepted: 1372 | |
| Case Time Limit: 2000MS | ||
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 file contains one integer number n (3 ≤ n ≤ 2000).
Output ‘1’ if the first player wins, or ‘2’ if the second player does.
| #1 | 3 |
|---|---|
| #2 | 6 |
| #1 | 1 |
|---|---|
| #2 | 2 |
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...
#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;
}
By NeighThorn
标签：accept name rip oss ast order eof 问题 函数
原文地址：http://www.cnblogs.com/neighthorn/p/6441252.html
