POJ 1003 Hangover

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

 

    

    题目大致意思：就是你有很多卡片，然后重叠的放在一起，重叠的规则是：第一张卡片伸出桌面的1 / 2，第二张1 / 3，第三张1 / 4……依次类推第n张1 / （n + 1）。
现在给出伸出桌面的长度，问你最少需要几张卡片才能达到要求。
    其实很简单，就是一个循环，每次sum加1 / （i + 1），当sum >= 给出的长度的时候就跳出循环，然后输出i就好。

 

    代码如下：

#include<iostream>

using namespace std;

int main()
{
    int i;
    double sum, n;
    while (cin >> n&& n)
    {
        sum = 0;
        for (i = 1;; i++)
        {
            sum += 1.0 / (i + 1);
            if (sum >= n)  break;
        }
        cout << i << "card(s)" << endl;
    }
}