[1003]Hangover(水题)

探讨如何通过堆叠卡片使其最大限度地悬空于桌面之外的技术问题。文章介绍了一个具体的数学模型,利用该模型可以计算出实现特定悬空距离所需的最少卡片数量。

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                                                                                                                                   Hangover
Time Limit: 1000MSMemory Limit: 10000K
Total Submissions: 102872Accepted: 50054

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>
using namespace std;
int main()
{
    double a,b;
    int num;//记录卡片的个数
    while(cin>>a,a)
    {
        num=1;
        b=1.0/2;//b记录当前卡片的长度
        while(b<a)
        {
            num++;
            b+=1.0/(1+num);
        }
        cout<<num<<" card(s)"<<endl;
    }
    return 0;
}

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