北大ACM 1003题—Hangover

Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 89330		Accepted: 43182

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)

Source

Mid-Central USA 2001

#include <iostream>
using namespace std;

double GetLength(int n)
{
	double out=0;
	if(n==1)
	{
		out=0.5;
	}
	else
	{
		for(int i=2;i<n;i++)
			out+=1.0/i;	
	}
	return out;
}

int main()
{
	double input[100],temp;
	int i=0;
	do
	{
		cin>>temp;
		input[i]=temp;
		i++;
	}while(temp!=0);

	for(int j=0;j<i-1;j++)
	{
		for(int m=1;;m++)
		{
			if(input[j]>=GetLength(m)&&input[j]<=GetLength(m+1))
			{
				cout<<m-1<<" card(s)"<<endl;
				break;
			}
		}
	}
	return 0;
}