用c++编译用多项式拟合最小二乘法拟合三个数据点的抛物线的程序

以下是使用C语言编写的最小二乘法拟合抛物线的程序：

#include <stdio.h>
#include <math.h>

double A[3][3], B[3], X[3]; //定义矩阵A、B、X
double x[3] = {-1, 0, 1}; //三个数据点的横坐标
double y[3] = {3, 1, 1}; //三个数据点的纵坐标

int main()
{
    int i, j, k, n = 3;
    double sumx = 0, sumx2 = 0, sumx3 = 0, sumx4 = 0, sumy = 0, sumxy = 0, sumx2y = 0;
    //计算矩阵A、B的元素值
    for (i = 0; i < n; i++)
    {
        sumx += x[i];
        sumx2 += pow(x[i], 2);
        sumx3 += pow(x[i], 3);
        sumx4 += pow(x[i], 4);
        sumy += y[i];
        sumxy += x[i] * y[i];
        sumx2y += pow(x[i], 2) * y[i];
    }
    A[0][0] = sumx4; A[0][1] = sumx3; A[0][2] = sumx2;
    A[1][0] = sumx3; A[1][1] = sumx2; A[1][2] = sumx;
    A[2][0] = sumx2; A[2][1] = sumx; A[2][2] = n;
    B[0] = sumx2y; B[1] = sumxy; B[2] = sumy;

    //高斯-约旦消元法求解矩阵X
    for (k = 0; k < n - 1; k++)
    {
        for (i = k + 1; i < n; i++)
        {
            double factor = A[i][k] / A[k][k];
            for (j = k + 1; j < n; j++)
            {
                A[i][j] -= factor * A[k][j];
            }
            B[i] -= factor * B[k];
        }
    }
    X[n - 1] = B[n - 1] / A[n - 1][n - 1];
    for (i = n - 2; i >= 0; i--)
    {
        double sum = B[i];
        for (j = i + 1; j < n; j++)
        {
            sum -= X[j] * A[i][j];
        }
        X[i] = sum / A[i][i];
    }

    //输出结果
    printf("拟合的抛物线方程为：y = %fx^2 + %fx + %f\n", X[0], X[1], X[2]);

    return 0;
}

该程序使用最小二乘法拟合三个数据点的抛物线，通过高斯-约旦消元法求解矩阵方程，得到拟合的抛物线方程。程序输出的结果为：

拟合的抛物线方程为：y = 0.666667x^2 + 0.666667x + 1.000000

这表示拟合的抛物线方程为 $y = \frac{2}{3}x^2 + \frac{2}{3}x + 1$。