【POJ】1003 Hangover

Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 73705		Accepted: 34947

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<vector>
#include<algorithm>

using namespace std;

int main()
{
	vector<double> vf;
	double c = 0;
	cin >> c;
	while(c) {
		vf.push_back(c);
		cin >> c;
	}
	for(vector<double>::iterator iter = vf.begin();
			iter != vf.end();
			++iter) {
		double sum = 0.0;
		int i = 1;
		for(i = 1; sum < *iter; i++) {
			sum += 1.0 / (i + 1);
		}
		cout << i - 1 << " card(s)" << endl;
	}
}