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.


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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)

解题思路:水题。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 
 5 using namespace std;
 6 
 7 int main()
 8 {
 9     double a;
10     int countt=0;
11     double now=0;
12     double sum=0;
13     while(scanf("%lf",&a)&&(fabs(a-0)>0.0001)){
14         sum=0;
15         now=0;
16         countt=1;
17         while(sum<a){
18             countt++;
19             sum+=1.0/countt;
20         }
21         printf("%d card(s)\n",countt-1);
22     }
23     return 0;
24 }
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