/****************************************
1.2 一維多項式多組求值
****************************************/
#include "math.h"
#include "stdlib.h"
void plys(a,n,x,m,p)
int n,m;
double a[],x[],p[];
{ 
	int i,j,mm,nn,ll,t,s,kk,k;
	double *b,y,z;
	b=malloc(2*n*sizeof(double));
	y=a[n-1];
	for (i=0; i<=n-1; i++) b[i]=a[i]/y;
	k=log(n-0.5)/log(2.0)+1;
	nn=1;
	for(i=0; i<=k-1; i++) nn=2*nn;
	for(i=n; i<nn-1; i++) b[i]=0.0;
	b[nn-1]=1.0;
	t=nn; s=1;
	for(i=1; i<=k-1; i++)
	{ 
		t=t/2; mm=-t;
		for(j=1; j<=s; j++)
		{ 
			mm=mm+t+t; b[mm-1]=b[mm-1]-1.0; 
			for(kk=2; kk<=t; kk++)
				b[mm-kk]=b[mm-kk]-b[mm-1]*b[mm+t-kk];
		}
		s=s+s;
	}
	for (kk=1; kk<=m; kk++)
	{ 
		for (i=0; i<=(nn-2)/2; i++)
		a[i]=x[kk-1]+b[2*i];
		mm=1; 
		z=x[kk-1];
		for(i=1; i<=k-1; i++)
		{ 
			mm=mm+mm; ll=mm+mm; z=z*z;
			for(j=0; j<=nn-1; j=j+ll)
				a[j/2]=a[j/2]+a[(j+mm)/2]*(z+b[j+mm-1]);
		}
		z=z*z/x[kk-1];
		if (nn!=n) a[0]=a[0]-z;
		p[kk-1]=a[0]*y;
	}
	return;
}