1003 Hangover

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 + 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)

 1 #include < stdio.h >   
 2
 3 int  main()         
 4 { 
 5   double n,s,i;                   //n是要求的板长  
 6   while(scanf("%lf",&n) && n!=0) 
 7   { 
 8      i=0;                         //i是当前第几块板  
 9      s=0;                         //s是当前板的总长  
10      while(s<n) 
11      { 
12         i++; 
13         s+=1/(i+1);               //叠加板后的长  
14      } 
15      printf("%.0lf card(s)\n",i); 
16   } 
17   return 0; 
18}
19

一块板长默认为1
一块板长为1/2
二块板长为1/2+1/3
三块板长为1/2+1/3+1/4
……
n块板长为1/2+1/3+1/4+……+1/(n+1)
循环递推相加，循环条件为输入