[POJ][1003]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)

这是传说中的水题，题目说的物理原理正确与否我不知道，得出结果还是很简单的，只是简单的模拟，而且仅仅是float类型就能搞定，直接给出AC代码

#include<iostream>
using namespace std;
int main()
{
float c;
cin>>c;
while(c!=0)
{
int n=1;
float num=1/float(n+1);
while(num<c) num+=1/float(++n+1);
cout<<n<<" card(s)"<<endl;
cin>>c;
}
return 0;
}