POJ3537[Crosses and Crosses]

所以每放一次，游戏会被分成左右两块。即左边的点点部分，和右边的点点部分。

然后SG(一个局面)=mex{NIM和(SG(左边子局面,SG右边子局面))}，边界SG(x<0)=0。

CODE:

/*

AUTHOR: Su Jiao

DATE: 2010-6-19

DESCRIPTION:

http://acm.pku.edu.cn/JudgeOnline/problem?id=3537

*/

#include <iostream>

using std::cin;

using std::cout;

using std::endl;

#include <vector>

using std::vector;

 

class Application

{

      int n;

      vector<int> _sg;

      int SG(int x)

      {

          if (x<0) return 0;

          if (_sg[x]!=-1) return _sg[x];

          else

          {

              vector<bool> found(n+1,false);

              for (int i=1;i<=x;i++)

                  found[SG(i-2–1)^SG(x-i-2)]=true;

              for (int i=0;i<=n;i++)

                  if (!found[i]) return _sg[x]=i;

          }

      }

      public:

      Application()

      {

                   std::ios::sync_with_stdio(false);

      }

      int run()

      {

          cin>>n;

          _sg.resize(n+1,-1);

          if (SG(n)) cout<<1<<endl;

          else cout<<2<<endl;

          return 0;

      }

};

 

int main()

{

    Application app;

    return app.run();

}