poj1003 Hangover

Description
How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.

Input

The input consists of one or more test cases, followed by a line containing the number 0.00 that signals the end of the input. Each test case is a single line containing a positive floating-point number c whose value is at least 0.01 and at most 5.20; c will contain exactly three digits.

Output

For each test case, output the minimum number of cards necessary to achieve an overhang of at least c card lengths. Use the exact output format shown in the examples.

Sample Input

1.00

3.71

0.04

5.19

0.00

          

Sample Output

3 card(s)

61 card(s)

1 card(s)

273 card(s)

import java.util.*;   

public class Main {   

    public static void main(String[] args) {   

        Scanner cin = new Scanner(System.in);   

        String str;   

        int maxCardNum = 0;   

        while(cin.hasNext())   

        {   

            str = cin.nextLine();   

            if(str.equals("0.00"))   

                break;   

            float len = Float.valueOf(str).floatValue();   

            maxCardNum = getCardNum(len);   

            System.out.println(maxCardNum + " card(s)");   

        }   

  

    }   

       

    private static int getCardNum(float len)   

    {   

        float value = 0;   

        int index = 2;     

        while (value < len)   

        {   

            value += 1.0/index;   

            index++;   

        }   

               

        return index-2;   

    }   

  

}