#include "iostream.h"
#include "math.h"

void main()
{
    //program d1r7
    //driver program for routine toeplz
    int i,j,n2,n = 5;
	double sum1;
    n2 = 2 * n;
    double x[6], y[6], r[11];
    //輸入已知的方程組的系數矩陣的r
    for (i = 1; i<=2 * n - 1; i++)
	{
		r[i] = 1 / double(i);
    }
    //輸入已知的方程組的右端向量
    for (i = 1; i<=n; i++)
	{
        y[i] = 0.1 * i;
    }
	cout.setf(ios::fixed|ios::right);
	cout.precision(5);
    cout<<"已知的方程組的右端向量"<<endl;
	cout.width(12);    cout<<y[1]<<endl;
	cout.width(12);    cout<<y[2]<<endl;
	cout.width(12);    cout<<y[3]<<endl;
	cout.width(12);    cout<<y[4]<<endl;
	cout.width(12);    cout<<y[5]<<endl;
    toeplz(r, x, y, n);
    //輸出方程組的解x
    cout<<endl;
    cout<<"計算出的方程組的解"<<endl;
	cout.width(15);    cout<<x[1]<<endl;
	cout.width(15);    cout<<x[2]<<endl;
	cout.width(15);    cout<<x[3]<<endl;
	cout.width(15);    cout<<x[4]<<endl;
	cout.width(15);    cout<<x[5]<<endl;
    //將計算出的解x乘以系數矩陣，以驗證計算結果正確
    cout<<endl;
    cout<<"將計算出的解乘以系數矩陣，以驗證計算結果正確"<<endl;
    cout<<endl;
    cout<<"解乘以系數矩陣"<<"        "<<"方程組的右端向量"<<endl;
    for (i = 1; i<=n; i++)
	{
        sum1 = 0;
        for (j = 1; j<=n; j++)
		{
           sum1 = sum1 + r[n + i - j] * x[j];
        }
        cout<<"    "<<sum1<<"              "<<y[i]<<endl;
    }
}