acm ural [1066] Garland solution report

original problem: http://acm.timus.ru/problem.aspx?space=1&num=1066

content:

Garland
Time Limit: 2.0 second
Memory Limit: 16 MB

 

The New Year garland consists of N lamps attached to a common wire that hangs down on the ends to which outermost lamps are affixed. The wire sags under the weight of lamp in a particular way: each lamp is hanging at the height that is 1 millimeter lower than the average height of the two adjacent lamps.

The leftmost lamp in hanging at the height of A millimeters above the ground. You have to determine the lowest height B of the rightmost lamp so that no lamp in the garland lies on the ground though some of them may touch the ground.

You shall neglect the lamp's size in this problem. By numbering the lamps with integers from 1 to N and denoting the i^th lamp height in millimeters as H_i we derive the following equations:

• H₁ = A
• H_i = (H_i-1 + H_i+1)/2 - 1, for all 1 < i < N
• H_N = B
• H_i ≥ 0, for all 1 ≤ i ≤ N

The sample garland with 8 lamps that is shown on the picture has A = 15 and B = 9.75.

 

Input

 

The input consists of a single line with two numbers N and A separated by a space. N (3 ≤ N ≤ 1000) is an integer representing the number of lamps in the garland, A (10 ≤ A ≤ 1000) is a real number representing the height of the leftmost lamp above the ground in millimeters.

 

Output

 

Write to the output the single real number B accurate to two digits to the right of the decimal point representing the lowest possible height of the rightmost lamp.

 

Sample Input

692 532.81

Sample Output

446113.34

my report:

Note that there’re totally N points respectively (H1,H2,…,HN)
   

And through the given equation Hi=(Hi-1+Hi+1)/2-1
   

We can derive the third number by the former 2.
   

That means we ought to just assume a H2, using half-dividing method.

Here we know that if any number of the sequence appears below zero at the given H2, 

it shall be adjusted to another value, if c(let’s define it as the every number actually it is) 

is less than zero,

we should increase H2, vice versa. We call it enumerate.

code in JAVA:

import java.io.*;
import java.util.*;

 

public class Application
{
    public static void main(String[] args) throws IOException
    {
        new Application().run();
    }

 

    StreamTokenizer in;
    PrintWriter out;

 

    int nextInt() throws IOException
    {
        in.nextToken();
        return (int) in.nval;
    }

 

    double nextDouble() throws IOException
    {
        in.nextToken();
        return (double) in.nval;
    }

 

    void run() throws IOException
    {
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(
            System.in)));
        out = new PrintWriter(new OutputStreamWriter(System.out));
        solve();
        out.flush();
    }

 

    void solve() throws IOException
    {

 

        double a, b, c, l, r, h1, h2;

        int n = nextInt();
        h1 = nextDouble();

 

        l = 0;
        r = h1;

 

        while (true)
        {
            b = h1;
            h2 = (l + r) / 2;
            c = h2;

 

            for (int i = 3; i <= n; i++)
            {
                a = b;
                b = c;
                c = 2 * (b + 1) - a;
                if (c < 0)
                {
                    break;
                }
            }

 

            if (c < 0)
            {
                l = h2;
            }
            else
            {
                r = h2;
            }

 

            if (r - l < 1E-10 && c >= 0)
            {
                break;
            }
        }

 

        out.println(round(c));
    }

 

    double round(double d)
    {
       int i=(int)(d*100+0.5);
       double k=(double)(i/100.00);
       return k;
    }
}