简单计算，解方程法，设a[1]为x

#include<iostream>

#include<iomanip>

using namespace std;

class myitem

{

public:

double item_x;

double item_const;

myitem Mul(double a)

{

myitem m;

m.item_const = a * this->item_const;

m.item_x = a * this->item_x;

return m;

}

myitem Add(double a)

{

myitem m;

m.item_const = a + this->item_const;

m.item_x = this->item_x;

return m;

}

myitem Sub(myitem a)

{

myitem m;

m.item_const = this->item_const - a.item_const;

m.item_x = this->item_x - a.item_x;

return m;

}

};

int main()

{

//i=1,2,3,4,5,6

    int n = 0;cin>>n;

    myitem a[n+2];

double input;

cin>>input;

a[0].item_x = 0;

a[0].item_const = input;

cin>>input;//结果 

    a[1].item_x = 1;

    a[1].item_const= 0;

    double c[n+1];

    for(int i = 1;i<n+1;i++)

    cin>>c[i];

    for(int i = 2;i<n+2;i++)

    {

    a[i] = a[i-1].Mul(2.0).Add(2.0*c[i-1]).Sub(a[i-2]);

    }

cout<<fixed<<setprecision(2)<<(input-a[n+1].item_const)/a[n+1].item_x;

    return 0;

}