C语言--曲线拟合的最小二乘法

/**************运用高斯消去及最小二乘法求最小二乘拟合的经验直线*******/
/***************本例使用6个拟合点，若调节可调节宏定义中N*********/
#include 
#include
#define N 7        
#define n 2
double *gassy(double sum,double sum_x,double sum_y,double sum_x2,double sum_xy )//运用高斯消去法
{
 	int k,i,j,q,p;
 	double b[n][n+1]={sum,sum_x,sum_y,sum_x,sum_x2,sum_xy},m;
 	double *a;
 	a=(double *)malloc(sizeof(int)*n);
 	//printf("消元过程\n");
	for(k=0;k<=n-2;k++)
	{
	 if(b[k][k]==0)
	   	break;
	 for(i=k+1;i<=n-1;i++)
	  {
		   m=b[i][k]/b[k][k];
		   b[i][k]=0;
		   for(j=k+1;j<=n;j++)
		   {
		    b[i][j]=b[i][j]-m*b[k][j];
		   }
	  }
 	}
	for(q=0;q<n;q++)
	{
	  for(p=0;p<=n;p++)
	   printf("%.4lf\t",b[q][p]);
	  printf("\n");
	}
 //printf("回代过程\n");
	a[n-1]=b[n-1][n]/b[n-1][n-1];
	for(i=n-2;i>=0;i--)
	{
	  int  m;
	  double s[n-1]={0};
	  for(m=n-1;m>i;m--)
	  {
	   s[i]=s[i]+a[m]*b[i][m];
	  }
	  a[i]=(b[i][n]-s[i])/b[i][i];
	
	}
    return a;
}
void main()
{
 int i;
 double x[N]={0.6,1.3,1.64,1.8,2.1,2.3,2.44},y[N]={7.05,12.2,14.4,15.2,17.4,19.6,20.2},*p;
 double sum=0,sum_x=0,sum_x2=0,sum_y=0,sum_xy=0;
 for( i=0;i<N;i++)
 {
  printf("请输入x，y");
  scanf_s("%lf%lf",x[i],y[i]);
 }
 for( i=0;i<N;i++)
 {
  sum++;
  sum_x=sum_x+x[i];
  sum_x2=sum_x2+x[i]*x[i];
  sum_y=sum_y+y[i];
  sum_xy=sum_xy+x[i]*y[i];
 }
 p=gassy(sum,sum_x,sum_y,sum_x2,sum_xy);
 printf("最小二乘经验直线为y=%lf+%lfx",p[0],p[1]);

}