URAL1047[Simple calculations]

数学题，设a[1]=x，然后由题意得递推式a[i]=(a[i-1]+c[i-1])*2-a[i-2]，然后将a[n+1]表示为x的线性表达式，然后解一元一次方程，这个就不用高斯消元了。可以参见URAL1046的解法，有点像。

CODE:

/*

PROGRAM: $PROGRAM

AUTHOR: Su Jiao

DATE: 2010-3-26

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 <cmath>

#include <iomanip>

 

class Application

{

      typedef double Type;

      struct Data

      {

             typedef double Type;

             Type x,C;

             void set(Type _x,Type _C)

             {

                  x=(_x),C=(_C);

             }

             void print()

             {

                  cout<<floor(x+0.5)<<” “<<floor(C+0.5)<<endl;

             }

      };

      int N;

      Type Astart,Aend;

      vector<Type> c;

      public:

      Application()

      {

                    cin>>N;

                    cin>>Astart>>Aend;

                    c.resize(N);

                    for (int i=0;i<N;i++) cin>>c[i];

      }

      int run()

      {

          //a[i]=(a[i-1]+a[i+1])/2-c[i]

          //a[i+1]=(a[i]+c[i])*2-a[i-1]

          //a[i]=(a[i-1]+c[i-1])*2-a[i-2]

          vector<Data> a(N+2);

          a[0].set(0,Astart);

          a[1].set(1,0);

          for (int i=2;i<=N+1;i++)

          {

              a[i].x=a[i-1].x*2-a[i-2].x;

              a[i].C=(a[i-1].C+c[i-1–1])*2-a[i-2].C;

          }

          cout<<std::setiosflags(std::ios::fixed)<<std::setprecision(2);

          cout<<(Aend-a[N+1].C)/a[N+1].x<<endl;

          return 0;

      }

};

 

int main()

{

    Application app;

    return app.run();

}