Ｃ＋＋项目在线求解😌😌

线性最小二乘拟合程序
某系统输入输出呈线性关系：Y=aX + b，但参数a、b未知。通过对系统输入（X）和输出（Y）的数据统计，应用最小二乘估计确定a和b的值。现有一次测量活动得到如下10组[X，Y]数据：
[1, 9.762]
[2, 12.719]
[3, 15.497]
[4, 18.143]
[5, 18.982]
[6, 21.008]
[7, 24.596]
[8, 26.285]
[9, 29.369]
[10, 31.17]
请编程实现：输入上述测量数据后，输出参数a和b的值，以及10个测量值的离差（离差：应用拟合函数计算的Y值与测量的Y值之差）。
程序要求：
（1） 一切计算过程全部用程序实现，具备通用性，即可以拟合任意Y=aX + b型线性关系；
（2） 计算完成后，可将数据导入Excel显示测量数据（散点）和拟合后的直线，并将该显示图形加入到结题报告中；
（3） 输出数据可以输出到文本文件，
尽量简单一点，不需要太复杂，太精简

https://blog.csdn.net/yangziluomu/article/details/103503387
这里有现成代码

https://blog.csdn.net/llittleSun/article/details/115045660?ops_request_misc=%257B%2522request%255Fid%2522%253A%2522167022328916800182795866%2522%252C%2522scm%2522%253A%252220140713.130102334..%2522%257D&request_id=167022328916800182795866&biz_id=0&utm_medium=distribute.pc_search_result.none-task-blog-2~all~top_click~default-1-115045660-null-null.142^v67^control,201^v4^add_ask,213^v2^t3_esquery_v1&utm_term=%E6%9C%80%E5%B0%8F%E4%BA%8C%E4%B9%98%E6%B3%95%E6%8B%9F%E5%90%88&spm=1018.2226.3001.4187

如有帮助，望采纳

//Linear Fit
#include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
int main()
{
    int i,j,k,n;
    cout<<"\nEnter the no. of data pairs to be entered:\n";        //To find the size of arrays
    cin>>n;
    double x[n],y[n],a,b;
    cout<<"\nEnter the x-axis values:\n";                //Input x-values
    for (i=0;i<n;i++)
        cin>>x[i];
    cout<<"\nEnter the y-axis values:\n";                //Input y-values
    for (i=0;i<n;i++)
        cin>>y[i];
    double xsum=0,x2sum=0,ysum=0,xysum=0;                //variables for sums/sigma of xi,yi,xi^2,xiyi etc
    for (i=0;i<n;i++)
    {
        xsum=xsum+x[i];                        //calculate sigma(xi)
        ysum=ysum+y[i];                        //calculate sigma(yi)
        x2sum=x2sum+pow(x[i],2);                //calculate sigma(x^2i)
        xysum=xysum+x[i]*y[i];                    //calculate sigma(xi*yi)
    }
    a=(n*xysum-xsum*ysum)/(n*x2sum-xsum*xsum);            //calculate slope
    b=(x2sum*ysum-xsum*xysum)/(x2sum*n-xsum*xsum);            //calculate intercept
    double y_fit[n];                        //an array to store the new fitted values of y    
    for (i=0;i<n;i++)
        y_fit[i]=a*x[i]+b;                    //to calculate y(fitted) at given x points
    cout<<"S.no"<<setw(5)<<"x"<<setw(19)<<"y(observed)"<<setw(19)<<"y(fitted)"<<endl;
    cout<<"-----------------------------------------------------------------\n";
    for (i=0;i<n;i++)
        cout<<i+1<<"."<<setw(8)<<x[i]<<setw(15)<<y[i]<<setw(18)<<y_fit[i]<<endl;//print a table of x,y(obs.) and y(fit.)    
    cout<<"\nThe linear fit line is of the form:\n\n"<<a<<"x + "<<b<<endl;        //print the best fit line
    return 0;
}