HDU1056 HangOver 水題

今天做了1056題目，純粹的水題。用C++，很快做出來。
看到很多大神都不屑於發這個，估計只有我這種渣渣纔來記錄點滴吧哈哈哈

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)

#include<iostream>
using namespace std;
int main()
{
    double input;
    while(cin>>input){
    int number=2;
    int times=0;
    double output=0;
    while(output<input)
    {
        output+=1.0/number*1.0;
        number++;
        times++;
    }
    if(times!=0)
    cout<<times<<" card(s)"<<endl;
    }
    return 0;
}