/* 牛頓法解多項(xiàng)式的根
   輸入：多項(xiàng)式系數(shù)c[]，多項(xiàng)式度數(shù)n，求在[a,b]間的根
   輸出：根
   要求保證[a,b]間有根
*/

double fabs( double x )
{
	return (x<0)? -x : x;
}

double f(int m, double c[], double x)
{
	int i;
	double p = c[m];

	for (i=m; i>0; i--)
		p = p*x + c[i-1];
	return p;
}

int newton(double x0, double *r, 
		   double c[], double cp[], int n, 
		   double a, double b, double eps)
{
	int MAX_ITERATION = 1000;
	int i = 1;
	double x1, x2, fp, eps2 = eps/10.0;

	x1 = x0;
	while (i < MAX_ITERATION) {
		x2 = f(n, c, x1);
		fp = f(n-1, cp, x1);
		if ((fabs(fp)<0.000000001) && (fabs(x2)>1.0))
			return 0;
		x2 = x1 - x2/fp;
		if (fabs(x1-x2)<eps2) {
			if (x2<a || x2>b)
				return 0;
			*r = x2;
			return 1;
		}
		x1 = x2;
		i++;
	}
	return 0;
}

double Polynomial_Root(double c[], int n, double a, double b, double eps)
{
	double *cp;
	int i;
	double root;

	cp = (double *)calloc(n, sizeof(double));
	for (i=n-1; i>=0; i--) {
		cp[i] = (i+1)*c[i+1];
	}
	if (a>b) {
			root = a; a = b; b = root;
	}
	if ((!newton(a, &root, c, cp, n, a, b, eps)) && 
		(!newton(b, &root, c, cp, n, a, b, eps)))
			newton((a+b)*0.5, &root, c, cp, n, a, b, eps);
	free(cp);
	if (fabs(root)<eps)
		return fabs(root);
	else
		return root;
}