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 <tt>+</tt> 1/3 <tt>=</tt> 5/6 card lengths. In general you can make n cards overhang by 1/2 <tt>+</tt> 1/3 <tt>+</tt> 1/4 <tt>+</tt> ... <tt>+</tt> 1/(n <tt>+</tt> 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 <tt>+</tt> 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)
  
  
太水了，不想多说了
   
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
double x;
int main()
{
   // freopen("1003.in", "r", stdin);
    while((scanf("%lf", &x) != EOF)&&(x != (double)0))
    {
        double tmp = 0;
        long i;
        for (i = 2; tmp < x; i++)
        {
            tmp += (1.0 / i);
        }

        printf("%d ", i - 2);
        printf("card(s)\n");

    }
    return 0;
}