poj 1003

Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 109037		Accepted: 53142

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)
#include <iostream>
#include <cstring>

using namespace std;

double a[500] = {0,0.5};
int main()
{
    for(int i=2;i<=500;i++)
        a[i] = a[i-1] + 1.0 / (i*1.0 + 1.0);
    double n;
    while(cin>>n && n)
    {
        int pos;
        for(int i=1;i<=500;i++)
        {
            if(n <= a[i])
            {
                pos = i;
                break;
            }
        }
        cout<<pos<<" card(s)"<<endl;
    }
    return 0;
}