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2013
11-08

POJ 1003 Hangover [解题报告] Java

Hangover

问题描述 :

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.



输入:

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.

输出:

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.

样例输入:

1.00
3.71
0.04
5.19
0.00

样例输出:

3 card(s)
61 card(s)
1 card(s)
273 card(s)

解题代码:

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;   
    }   
  
}

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