poj1003

Posted 2020-11-09 nowindxdw

Hangover
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 134353 Accepted: 65331
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)

 

 

//Memory Time  
//3136K  172MS
import java.util.*;
public class poj1003 
{
     public static void main(String[] args) throws Exception
     {
       Scanner cin = new Scanner(System.in);
       while(cin.hasNext())
       {
         double f = cin.nextFloat();
        if(f == 0) return;
        if(f<=0.5)
        {
           System.out.println("1 card(s)");
           continue;
        }
        int ans = 1;
        double sum = 0.5;
        int base = 2;
        do
        {
         base++;
         ans ++;
         sum += 1.0/base;
        }while(sum<f);
        System.out.println(ans + " card(s)");
       } 
     }
}