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 <tt>+</tt> 1/3 <tt>=</tt> 5/6 card lengths. In general you can make n cards overhang by 1/2 <tt>+</tt> 1/3 <tt>+</tt> 1/4 <tt>+</tt> ... <tt>+</tt>1/(n <tt>+</tt> 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 <tt>+</tt> 1). This is illustrated in the figure below.


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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)




题目概述：给出一个浮点数c，求出使得 不等式 1/2 + 1/3 + ... + 1/(n+1) >= c 成立的最小 n。
 
题目中给出了5.2的上限，所以n数组<= 300即可 这里我是打了个表，发现300的。
Ｏ复杂度Ｏ（300ｍ）ｍ为询问数
 
<pre name="code" class="cpp">#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std; 
double a[300] = {0,0.5},c; 
int main()  
{  
    int num = 0;  
    for(int i = 2; i < 300; i ++)  
        a[i] = a[i - 1] + 1.0/(i + 1);   //注意这里1.0否则是错的，他会当成int存成0
    while(scanf("%lf",&c) && c)  
    {  
        for(int i = 1; i < 300; i ++)  
            if(a[i] >= c)  
            {  
                printf("%d card(s)\n",i);  
                break;  
            }  
    }  
    return 0;  
}