已知函数表如下： x 0.4 0.55 0.8 0.9 1 y 0.41075 0.57515 0.88811 1.02652 1.17520 请写出用拉格朗日插值多项式求0.5，0.7,0.85三处的函数值的c语言和c++代码

以下是用拉格朗日插值多项式求解函数值的C语言代码：

#include <stdio.h>

double lagrange_interpolation(double x[], double y[], int n, double xi);

int main()
{
    double x[] = {0.4, 0.55, 0.8, 0.9, 1};
    double y[] = {0.41075, 0.57515, 0.88811, 1.02652, 1.17520};

    double x1 = 0.5;
    double x2 = 0.7;
    double x3 = 0.85;

    double y1 = lagrange_interpolation(x, y, 5, x1);
    double y2 = lagrange_interpolation(x, y, 5, x2);
    double y3 = lagrange_interpolation(x, y, 5, x3);

    printf("f(0.5) = %lf\n", y1);
    printf("f(0.7) = %lf\n", y2);
    printf("f(0.85) = %lf\n", y3);

    return 0;
}

double lagrange_interpolation(double x[], double y[], int n, double xi)
{
    double result = 0;

    for (int i = 0; i < n; i++)
    {
        double term = y[i];
        for (int j = 0; j < n; j++)
        {
            if (j != i)
            {
                term *= (xi - x[j]) / (x[i] - x[j]);
            }
        }
        result += term;
    }

    return result;
}

以下是用拉格朗日插值多项式求解函数值的C++代码：

#include <iostream>

using namespace std;

double lagrange_interpolation(double x[], double y[], int n, double xi);

int main()
{
    double x[] = {0.4, 0.55, 0.8, 0.9, 1};
    double y[] = {0.41075, 0.57515, 0.88811, 1.02652, 1.17520};

    double x1 = 0.5;
    double x2 = 0.7;
    double x3 = 0.85;

    double y1 = lagrange_interpolation(x, y, 5, x1);
    double y2 = lagrange_interpolation(x, y, 5, x2);
    double y3 = lagrange_interpolation(x, y, 5, x3);

    cout << "f(0.5) = " << y1 << endl;
    cout << "f(0.7) = " << y2 << endl;
    cout << "f(0.85) = " << y3 << endl;

    return 0;
}

double lagrange_interpolation(double x[], double y[], int n, double xi)
{
    double result = 0;

    for (int i = 0; i < n; i++)
    {
        double term = y[i];
        for (int j = 0; j < n; j++)
        {
            if (j != i)
            {
                term *= (xi - x[j]) / (x[i] - x[j]);
            }
        }
        result += term;
    }

    return result;
}