请帮我用C++实现matlab中interp1函数的spline关键词函数

好的，我可以帮你实现一个C++版本的spline插值函数。下面是代码：

#include <iostream>
#include <vector>

using namespace std;

// Spline插值函数
vector<double> spline(vector<double> x, vector<double> y, vector<double> xx) {
    int n = x.size();
    vector<double> h(n - 1), b(n - 1), u(n - 1), v(n - 1), z(n);
    for (int i = 0; i < n - 1; i++) {
        h[i] = x[i + 1] - x[i];
        b[i] = (y[i + 1] - y[i]) / h[i];
    }
    u[1] = 2 * (h[0] + h[1]);
    v[1] = 6 * (b[1] - b[0]);
    for (int i = 2; i < n - 1; i++) {
        u[i] = 2 * (h[i - 1] + h[i]) - h[i - 1] * h[i - 1] / u[i - 1];
        v[i] = 6 * (b[i] - b[i - 1]) - h[i - 1] * v[i - 1] / u[i - 1];
    }
    z[n - 1] = 0;
    for (int i = n - 2; i >= 1; i--) {
        z[i] = (v[i] - h[i] * z[i + 1]) / u[i];
    }
    z[0] = 0;
    vector<double> yy(xx.size());
    int j = 0;
    for (int i = 0; i < xx.size(); i++) {
        while (j < n - 1 && xx[i] > x[j + 1]) j++;
        double t = (xx[i] - x[j]) / h[j];
        double a = z[j + 1] - z[j];
        double b = -h[j] * (z[j + 1] + 2 * z[j]) / 6 + (y[j + 1] - y[j]) / h[j];
        double c = z[j] / 2;
        double d = (z[j + 1] - z[j]) / (6 * h[j]);
        yy[i] = a * t * t * t + b * t * t + c * t + d;
    }
    return yy;
}

// 测试
int main() {
    vector<double> x = {0, 1, 2, 3, 4};
    vector<double> y = {0, 1, 4, 9, 16};
    vector<double> xx = {0.5, 1.5, 2.5, 3.5};
    vector<double> yy = spline(x, y, xx);
    for (int i = 0; i < yy.size(); i++) {
        cout << "f(" << xx[i] << ") = " << yy[i] << endl;
    }
    return 0;
}

使用方法：输入x和y的值，以及需要插值的xx的值，即可得到插值后的yy值。

注意：该函数只适用于具有单峰特性的函数，对于多峰的函数可能会出现不良效果。