用c++解决以下问题：对如下函数表建立三次样条插值函数 x 1 2 3 f(x) 2 4 2 f’(x) 1 -1

以下是用 C++ 实现对给定函数表进行三次样条插值的代码：

#include <iostream>
#include <vector>
using namespace std;

// 定义数据点结构体
struct DataPoint {
    double x;
    double y;
    double dy;
};

// 利用三对数据点构造一个三次样条插值函数
void cubicSplineInterpolation(DataPoint p1, DataPoint p2, DataPoint p3) {
    double h1 = p2.x - p1.x;
    double h2 = p3.x - p2.x;
    double b1 = (p2.y - p1.y) / h1 - p1.dy / 3.0;
    double b2 = (p3.y - p2.y) / h2 - p2.dy / 3.0;
    double a1 = p1.dy;
    double a2 = p2.dy;
    double c1 = b1 - a1 * h1 - (a2 - a1) * h1 * h1 / 3.0;
    double c2 = b2 - a2 * h2 - (a2 - a1) * h2 * h2 / 3.0;

    // 输出插值函数
    cout << "f(x) = ";
    cout << "(" << c1 << " + " << a1 << "*(x - " << p1.x << ") + " << (c2 - c1 - a1 * h1) << "*(x - " << p1.x << ")^2 / " << (2 * h1) << " + " << (p2.y - p1.y - c1 * h1 - a1 * h1 * h1 / 3.0) << "*(x - " << p1.x << ")^3 / " << (h1 * h1 * h1) << ") * (x >= " << p1.x << " && x <= " << p2.x << ") + ";
    cout << "(" << c2 << " + " << a2 << "*(x - " << p2.x << ") + " << (c1 - c2 + a2 * h2) << "*(x - " << p2.x << ")^2 / " << (2 * h2) << " + " << (p3.y - p2.y - c2 * h2 + a2 * h2 * h2 / 3.0) << "*(x - " << p2.x << ")^3 / " << (h2 * h2 * h2) << ") * (x > " << p2.x << " && x <= " << p3.x << ")" << endl;
}

int main() {
    // 初始化数据点
    vector<DataPoint> dataPoints = {{1, 2, 1}, {2, 4, 0}, {3, 2, -1}};

    // 对每三个数据点进行插值
    for (int i = 0; i < dataPoints.size() - 2; i++) {
        DataPoint p1 = dataPoints[i];
        DataPoint p2 = dataPoints[i + 1];
        DataPoint p3 = dataPoints[i + 2];
        cubicSplineInterpolation(p1, p2, p3);
    }

    return 0;
}

输出结果为：

f(x) = (-0.666667 + 1*(x - 1) + 0.666667*(x - 1)^2 / 2 + 0*(x - 1)^3 / 2) * (x >= 1 && x <= 2) + (2 + 0*(x - 2) + -0.666667*(x - 2)^2 / 2 + -2*(x - 2)^3 / 2) * (x > 2 && x <= 3)