HangOver ACM HDU1056

#include<iostream>
using namespace std;
int main()
{
    int n;
    double sum;
    double len;
    while(cin>>len)
    {
        if(len==0)break;
        sum=0;
        n=0;
        while(sum<len)
        {
            n++;
            sum+=1.0/(n+1);
        }  
        cout<<n<<" card(s)"<<endl; 
    }
    return 0;    
    
}   

HangOver

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4512    Accepted Submission(s): 1748

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)

 

Source

Mid-Central USA 2001