北大oj--1003

/*
 Hangover
Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 101133		Accepted: 49167
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)

题意：第一张卡最多可以伸出桌面1/2，第一张比桌面多了1/2,第二张比第一张多伸出1/3，第二张就比桌面多了1/2+1/3=5/6,
第三张多伸出1/4，第三张比桌面多了1/2+1/3+1/4=26/24,以此类推，第n张比桌面多了1/2+1/3+,,,,+1/(n-1),
给出一个浮点数k，求，至少需要几张卡可以使超出桌面的长度达到k，0.01<=k<=5.20,k=0.00结束输入 
*/

#include <stdio.h>

int main()
{
	float n;
	int count;
	float sum;
	while(scanf("%f",&n)&&n>0)
	{
		sum=0;
		count=0;
		for(int i=2;sum<n;i++)
		{
			sum+=1.0/i;
			count++;
		}
		printf("%d card(s)\n",count);
	}
	return 0;
}