📄 d1r7.cpp
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#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];
//输入已知的方程组的系数矩阵的r
for (i = 1; i<=2 * n - 1; i++)
{
r[i] = 1 / double(i);
}
//输入已知的方程组的右端向量
for (i = 1; i<=n; i++)
{
y[i] = 0.1 * i;
}
cout.setf(ios::fixed|ios::right);
cout.precision(5);
cout<<"已知的方程组的右端向量"<<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);
//输出方程组的解x
cout<<endl;
cout<<"计算出的方程组的解"<<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;
//将计算出的解x乘以系数矩阵,以验证计算结果正确
cout<<endl;
cout<<"将计算出的解乘以系数矩阵,以验证计算结果正确"<<endl;
cout<<endl;
cout<<"解乘以系数矩阵"<<" "<<"方程组的右端向量"<<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;
}
}
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