# 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)

## AC代码

#include <iostream>
#include <stdio.h>

using namespace std;

int main()
{
double n;
while(cin>>n,n)
{
double sum = 0;
int x;
for(double i=2;sum<n;i++)//如果sum>=n时满足题意
{

sum=sum+1.0/i;
x=i;

}
cout<<x-1<<" card(s)"<<endl;
}
}