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 }

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