用最小二乘算法求过零点的斜率

void Min2Method(float &slope, float &intercept, double X[], double Y[], int nCount)
{
	double   SumX, SumY, SumXY, SumX2;
 
	SumX = 0;
	SumX2 = 0;
	for(int i=0; i<nCount; i++)
	{
		SumX += X[i];
		SumX2 += (X[i] * X[i]);
	}
 
	SumY = 0;
	for(int i=0; i<nCount; i++)
	{
		SumY += Y[i];
	}
 
	SumXY = 0;
	for(int i=0; i<nCount; i++)
	{
		SumXY += ( X[i] * Y[i]);
	}
 
  	intercept = ( (SumX2*SumY - SumX*SumXY) / (nCount*SumX2 - SumX * SumX));
  	slope = ( (nCount*SumXY - SumX*SumY) / (nCount*SumX2 - SumX * SumX));
}