HDU1056：HangOver

Problem 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.

 

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.

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)

 

 

#include <stdio.h>



int main()

{

    double a;

    while(~scanf("%lf",&a),a)

    {

        double b = 0,i;

        for(i = 1;b<a;i++)

        {

            b+=1/(i+1);

        }

        printf("%.0lf card(s)\n",i-1);

    }

    return 0;

}