poj 3537 Crosses and Crosses 博弈论

思路：每次画X之后都会形成2个子游戏，即i-3和n-i-2.

代码如下：

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 #include<vector>
 #include<cstring>
 using namespace std;
 int sg[];
 int getsg(int n)
 {
     if(n<=) return ;
     if(sg[n]>=) return sg[n];
     bool vis[]={};
     for(int i=;i<=n;i++){
         vis[getsg(n-i-)^getsg(i-)]=;
     }
     int i=;
     while(vis[i]) i++;
     return sg[n]=i;
 }
 int main()
 {
     int n;
     memset(sg,-,sizeof(sg));
     while(scanf("%d",&n)!=EOF){
         puts(getsg(n)?"":"");
     }
     return ;
 }