hdu_1056_HangOver_201311071354

HangOver

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8258 Accepted Submission(s): 3414

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)

Source

Mid-Central USA 2001

 1 #include <stdio.h>
 2 
 3 int main()
 4 {
 5     double n;
 6     while(scanf("%lf",&n),n)
 7     {
 8         int i;
 9         double sum=0;
10         i=0;
11         while(sum<n)
12         {
13             i++;
14             sum+=1.0/(i+1);
15         }
16         printf("%d card(s)
",i);
17     }
18     return 0;
19 }

简单题，按题中给的公式来就行