/*  prog3a.c -- Tabulate a Polynomial Using Nested Multiplication
    Written Feb 20, 2001 by Eric Olson for Mathematics 483

    For more information see Step 5 in First Steps in Numerical
    Analysis, 2nd Edition, by Hosking, Joe, Joyce and Turner. */

#include <stdio.h>
#include <math.h>

#define MAXn 10
static double a[MAXn+1];
static int n;

void outpoly(){
    int i;
    printf("\n\n#\tP(x)=%lg+",a[0]);
    for(i=1;i<n;i++){
        printf("%lgx^%d+",a[i],i);
    }
    printf("%lgx^%d\n",a[n],n);
}

double R(){
    int i;
    double m=0.0;
    for(i=0;i<n;i++){
        double r=fabs(a[i]);
        if(r>m) m=r;
    }
    return m/fabs(a[n])+1.0;
}

double P(double x,double *DP){
    int i;
    double p=a[n],q=0.0;
    for(i=1;i<=n;i++){
        q=q*x+p;
        p=p*x+a[n-i];
    }
    if(DP) *DP=q;
    return p;
}

main(){
    for(;;){
        if(scanf("%d",&n)!=1) exit(0);
        if(n<0) exit(0);
        if(n>MAXn) {
            puts("Capacity exceeded!");
            exit(1);
        }
        {
            int i;
            for(i=0;i<=n;i++){
                if(scanf("%lg",&a[i])!=1){
                    puts("Incomplete data!");
                    exit(2);
                }
            }
        }
        outpoly();
        {
            int i,imax=20*n-1;
            double r=R();
            printf("#\n#%20s %20s %20s\n","x","P(x)","P'(x)");
            for(i=0;i<=imax;i++){
                double x=r*i*(2.0/imax)-r;
                double p,dp;
                p=P(x,&dp);
                printf(" %20.16lg %20.16lg %20.16lg\n",x,p,dp);
            }
        }
    }
}