最小二乘法（高斯）

# include
# include
# define N 12   ///12个值
# define M 4
# include
# include
using namespace std;
///这是数据初始化。
double x[N]= {0,5,10,15,20,25,30,35,40,45,50,55};
double y[N]= {0,1.27,2.16,2.86,3.44,3.87,4.15,4.37,4.51,4.58,4.02,4.64};
double xx[7];
double yy[4];
///这是要用数组，高斯消元要用
double a[5][5]=
{
    {0,0,0,0,0},
    {0,0,0,0,0},
    {0,0,0,0,0},
    {0,0,0,0,0},
    {0,0,0,0,0},
};
double b[5]={0,0,0,0,0};
double nei(double t)
{
    return b[1]*t+b[2]*t*t+b[3]*t*t*t;
}
int main()
{
    int n=3,k,i,j;
    double sum,s,num;

    ///这里是求x[i]、x[i]平方、、、、x[i]六次方的累和。就是正规表达式中的各项要求的x[i]部分。
    for(i=1; i<=6; i++)
    {
        sum=0;
        for(j=0; j=1; i--)
    {
        s=0;
        for(j=i+1; j<=n; j++)
            s+=a[i][j]*b[j];
        b[i]=(b[i]-s)/a[i][i];
    }
    ///这里的b[1],b[2],b[3]表示的是拟合函数中的a1,a2,a3.
    for(i=1; i<=n; i++)
        printf("b[%2d]=%lf\n",i,b[i]);
    ///这里输出的是12个值的每一个误差。
    for(i=0;i<12;i++)
    {
        printf("y[%d]=%lf   ",i,y[i]);
        printf("%lf  ",nei(x[i]));
        printf("error  %lf  ",y[i]-nei(x[i]));
        printf("\n");
    }
    printf("\n");
    ///这里求得是12个点平方和的总误差
    num=0;
    for(int i=0;i<12;i++)
    {
        num=(y[i]-nei(x[i]))*(y[i]-nei(x[i]));
    }
    printf("%lf ",num);
    return 0;
}