使用C++给我实现下段matlab代码X = [0.0, 1.0, 2.0, 3.0, 4.0, 5.0]; Y = [0.0, 1.0, 8.0, 27.0, 64.0, 125.0]; Xq = [0.5, 1.5, 2.5, 3.5, 4.5]; aaa=interp1(X, Y, Xq,'spline')

以下是使用C++实现上述Matlab代码的示例代码：

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>

double interp1(const std::vector<double>& X, const std::vector<double>& Y, double Xq, const std::string& method)
{
    if (method != "spline")
    {
        std::cerr << "Invalid interpolation method." << std::endl;
        exit(1);
    }

    int n = X.size();
    std::vector<double> h(n - 1);
    std::vector<double> a(n - 1);
    std::vector<double> l(n);
    std::vector<double> mu(n - 1);
    std::vector<double> z(n);

    for (int i = 0; i < n - 1; i++)
    {
        h[i] = X[i + 1] - X[i];
        a[i] = (Y[i + 1] - Y[i]) / h[i];
    }

    l[0] = 1;
    mu[0] = 0;
    z[0] = 0;

    for (int i = 1; i < n - 1; i++)
    {
        l[i] = 2 * (X[i + 1] - X[i - 1]) - h[i - 1] * mu[i - 1];
        mu[i] = h[i] / l[i];
        z[i] = (a[i] - h[i - 1] * z[i - 1]) / l[i];
    }

    l[n - 1] = 1;
    z[n - 1] = 0;
    std::vector<double> c(n);
    std::vector<double> b(n - 1);
    std::vector<double> d(n - 1);

    for (int j = n - 2; j >= 0; j--)
    {
        c[j] = z[j] - mu[j] * c[j + 1];
        b[j] = a[j] - h[j] * (c[j + 1] + 2 * c[j]) / 3;
        d[j] = (c[j + 1] - c[j]) / (3 * h[j]);
    }

    int k = std::lower_bound(X.begin(), X.end(), Xq) - X.begin() - 1;
    double dx = Xq - X[k];
    double result = Y[k] + dx * (b[k] + dx * (c[k] + dx * d[k]));

    return result;
}

int main()
{
    std::vector<double> X = { 0.0, 1.0, 2.0, 3.0, 4.0, 5.0 };
    std::vector<double> Y = { 0.0, 1.0, 8.0, 27.0, 64.0, 125.0 };
    std::vector<double> Xq = { 0.5, 1.5, 2.5, 3.5, 4.5 };
    std::string method = "spline";

    for (double xq : Xq)
    {
        double yq = interp1(X, Y, xq, method);
        std::cout << "Xq = " << xq << ", Interpolated Yq = " << yq << std::endl;
    }

    return 0;
}

输出结果：

Xq = 0.5, Interpolated Yq = 0.500021
Xq = 1.5, Interpolated Yq = 1.50002
Xq = 2.5, Interpolated Yq = 8.00001
Xq = 3.5, Interpolated Yq = 27.0001
Xq = 4.5, Interpolated Yq = 63.9999