HangOver

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

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
	double n,sum,j;
	int i,k;
	while(scanf("%lf",&n)&&n)
	{
		sum=0.50;
		k=1;
		if(n<=0.50)
		printf("1 card(s)\n");
		else
		{
			for(i=3;sum<n;i++)
			{
				j=i*1.0;
				sum=sum+1/j;
				k++;
			}
			printf("%d card(s)\n",k);
		}
	}
	return 0;
}