matlab中polyfit(x,y,2)函数的c语言的源代码

/**********************************************
*Author :wacs5
*DATE :20090408(YYYMMDD)
*Functtion :多项式拟合polyfit
**********************************************/
#include <stdio.h>
#include <conio.h>
#include <stdlib.h>
#include <math.h>
main()
{
int i,j,m,n=7,poly_n=2;
double x[7]={1,2,3,4,6,7,8},y[7]={2,3,6,7,5,3,2};
double a[3];
void polyfit(int n,double *x,double *y,int poly_n,double a[]);

system("cls");
polyfit(n,x,y,poly_n,a);

for (i=0;i<poly_n+1;i++)/*这里是升序排列，Matlab是降序排列*/
printf("a[%d]=%g\n",i,a[i]);
getch();
}


/*==================polyfit(n,x,y,poly_n,a)===================*/
/*=======拟合y=a0+a1*x+a2*x^2+……+apoly_n*x^poly_n========*/
/*=====n是数据个数 xy是数据值 poly_n是多项式的项数======*/
/*===返回a0,a1,a2,……a[poly_n]，系数比项数多一（常数项）=====*/
void polyfit(int n,double x[],double y[],int poly_n,double a[])
{
int i,j;
double *tempx,*tempy,*sumxx,*sumxy,*ata;
void gauss_solve(int n,double A[],double x[],double b[]);
tempx=calloc(n,sizeof(double));
sumxx=calloc(poly_n*2+1,sizeof(double));
tempy=calloc(n,sizeof(double));
sumxy=calloc(poly_n+1,sizeof(double));
ata=calloc((poly_n+1)*(poly_n+1),sizeof(double));
for (i=0;i<n;i++)
{
tempx[i]=1;
tempy[i]=y[i];
}
for (i=0;i<2*poly_n+1;i++)
for (sumxx[i]=0,j=0;j<n;j++)
{
sumxx[i]+=tempx[j];
tempx[j]*=x[j];
}
for (i=0;i<poly_n+1;i++)
for (sumxy[i]=0,j=0;j<n;j++)
{
sumxy[i]+=tempy[j];
tempy[j]*=x[j];
}
for (i=0;i<poly_n+1;i++)
for (j=0;j<poly_n+1;j++)
ata[i*(poly_n+1)+j]=sumxx[i+j];
gauss_solve(poly_n+1,ata,a,sumxy);

free(tempx);
free(sumxx);
free(tempy);
free(sumxy);
free(ata);
}

void gauss_solve(int n,double A[],double x[],double b[])
{
int i,j,k,r;
double max;
for (k=0;k<n-1;k++)
{
max=fabs(A[k*n+k]); /*find maxmum*/
r=k;
for (i=k+1;i<n-1;i++)
if (max<fabs(A[i*n+i]))
{
max=fabs(A[i*n+i]);
r=i;
}
if (r!=k)
for (i=0;i<n;i++) /*change array:A[k]&A[r] */
{
max=A[k*n+i];
A[k*n+i]=A[r*n+i];
A[r*n+i]=max;
}
max=b[k]; /*change array:b[k]&b[r] */
b[k]=b[r];
b[r]=max;
for (i=k+1;i<n;i++)
{
for (j=k+1;j<n;j++)
A[i*n+j]-=A[i*n+k]*A[k*n+j]/A[k*n+k];
b[i]-=A[i*n+k]*b[k]/A[k*n+k];
}
}

for (i=n-1;i>=0;x[i]/=A[i*n+i],i--)
for (j=i+1,x[i]=b[i];j<n;j++)
x[i]-=A[i*n+j]*x[j];
}