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：
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<algorithm> 5 #include<vector> 6 #include<cstring> 7 using namespace std; 8 int sg[2002]; 9 int getsg(int n) 10 { 11 if(n<=0) return 0; 12 if(sg[n]>=0) return sg[n]; 13 bool vis[2002]={0}; 14 for(int i=1;i<=n;i++){ 15 vis[getsg(n-i-2)^getsg(i-3)]=1; 16 } 17 int i=0; 18 while(vis[i]) i++; 19 return sg[n]=i; 20 } 21 int main() 22 { 23 int n; 24 memset(sg,-1,sizeof(sg)); 25 while(scanf("%d",&n)!=EOF){ 26 puts(getsg(n)?"1":"2"); 27 } 28 return 0; 29 }
