URAL1036[Lucky tickets]

无聊的递推。


方程:


f[i][j]表示j个苹果放到i个容量为9的盒子一共有多少种放法。


那么f[i][j]=sum{f[i-1][k],0<=j-k<=9}。


最后,我被水了,先是没有写高精度,然后是没有判断S是否为偶数。


CODE:


/*


PROGRAM: $PROGRAM


AUTHOR: Su Jiao


DATE: 2010-3-21


DESCRIPTION:


$DESCRIPTION


*/


#include <iostream>


using std::cin;


using std::cout;


#include <fstream>


using std::ifstream;


using std::ofstream;


#include <sstream>


using std::stringstream;


using std::endl;


#include <vector>


using std::vector;


#include <string>


using std::string;


#include <stack>


using std::stack;


#include <queue>


using std::queue;


#include <set>


using std::set;


#include <map>


using std::map;


using std::pair;


using std::make_pair;


#include <algorithm>


using std::sort;


#include <cassert>


//using std::assert;


#include <iomanip>


 


class BigNumber


{


      static const int BASE=10000;


      vector<int> d;


      public:


      BigNumber(int value=0)


                    :d(1,value)


      {


                    //assert(value<BASE);


      }


      friend BigNumber operator+(const BigNumber& a,const BigNumber& b)


      {


             BigNumber c;


             c.d.resize(a.d.size()>b.d.size()?a.d.size():b.d.size());


             int over=0;


             for (int i=0;i<c.d.size();i++)


                 if (i<a.d.size()&&i<b.d.size())


                 {


                    c.d[i]=(a.d[i]+b.d[i]+over)%BASE;


                    over=(a.d[i]+b.d[i]+over)/BASE;


                 }


                 else if (i<a.d.size())


                 {


                    c.d[i]=(a.d[i]+over)%BASE;


                    over=(a.d[i]+over)/BASE;


                 }


                 else if (i<b.d.size())


                 {


                    c.d[i]=(b.d[i]+over)%BASE;


                    over=(b.d[i]+over)/BASE;


                 }


             if (over) c.d.push_back(over);


             return c;


      }


      friend BigNumber operator*(const BigNumber& a,const BigNumber& b)


      {


             BigNumber c;


             c.d.resize(a.d.size()+b.d.size(),0);


             for (int i=0;i<a.d.size();i++)


                 for (int j=0;j<b.d.size();j++)


                 {


                     c.d[i+j]+=a.d[i]*b.d[j];


                     if (c.d[i+j]>=BASE)


                     {


                        c.d[i+j+1]+=c.d[i+j]/BASE;


                        c.d[i+j]=c.d[i+j]%BASE;


                     }


                 }


             if (!c.d.back()) c.d.pop_back();


             return c;


      }


      BigNumber& operator+=(BigNumber& add)


      {


                 *this=operator+(*this,add);


                 return *this;


      }


      BigNumber& operator*=(int mul)


      {


                 int over=0;


                 for (int i=0;i<d.size();i++)


                 {


                     int old=d[i];


                     d[i]=(old*mul+over)%BASE;


                     over=(old*mul+over)/BASE;


                 }


                 if (over) d.push_back(over);


                 return *this;


      }


      BigNumber& operator/=(int div)


      {


                 int rest=0;


                 for (int i=d.size()-1;i>=0;i–)


                 {


                     int old=d[i];


                     d[i]=(rest*BASE+old)/div;


                     rest=(rest*BASE+old)%div;


                 }


                 if (d[d.size()-1]==0) d.resize(d.size()-1);


                 return *this;


      }


      void print()


      {


           cout<<d[d.size()-1];


           for (int i=d.size()-11;i>=0;i–)


               cout<<std::setfill(‘0’)<<std::setw(4)<<d[i]<<std::setw(1)<<std::setfill(‘ ‘);


           cout<<endl;


      }


};


 


class Application


{


      int N,S;


      public:


      Application()


      {


                    cin>>N>>S;


      }


      int run()


      {


          //答案为将S/2个苹果分到N个果盘里(最多分9个,最少不分)的方案数的平方


          //用生成函数 (1+x^2+x^3+…+x^9)^N


          //然后x^(s/2)的系数的平方


          //然后生成函数的程序我不想写了


          if (S%2)


          {


             cout<<0<<endl;


             return 0;


          }


         


          vector<vector<BigNumber> > f(N,vector<BigNumber>(S/2+1));


          for (int i=0;i<N;i++)


          {


              if (i)


                 for (int j=0;j<=S/2;j++)


                 {


                     f[i][j]=0;


                     for (int k=0;k<=j;k++)


                         if (j-k<=9)


                            f[i][j]+=f[i-1][k];


                 }


              else


                  for (int j=0;j<=S/2;j++)


                      if (j<=9)


                         f[i][j]=1;


          }


          BigNumber answer=f[N-1][S/2]*f[N-1][S/2];


          answer.print();


          return 0;


      }


};


 


int main()


{


    Application app;


    return app.run();


}