[POJ 1003] Hangover C++解題

 

 

Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 95164		Accepted: 46128

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)

翻譯：

翻譯：

　　若將一疊卡片放在一張桌子的邊緣，你能放多遠？如果你有一張卡片，你最遠能達到卡片長度的一半。（我們假定卡片都正放在桌 子上。）如果你有兩張卡片，你能使最上的一張卡片覆蓋下面那張的1/2，底下的那張可以伸出桌面1/3的長度，即最遠能達到 1/2 + 1/3 = 5/6 的卡片長度。一般地，如果你有n張卡片，你可以伸出 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) 的卡片長度，也就是最上的一張卡片覆蓋第二張1/2，第二張超出第三張1/3，第三張超出第四張1/4，依此類推，最底的一張卡片超出桌面1/(n + 1)。下面有個圖形的例子：

　　

　　現在給定伸出長度C（0.00至5.20之間），輸出至少需要多少張卡片。

 

解決思路

這是一道水題，直接按照要求模擬就可以了。

源碼

/*
poj 1000
version:1.0
author:Knight
Email:[e-mail]
*/

#include<cstdio>
using namespace std;
int main(void)
{
    double c;
    int i;
    double Overhangs;
    while(scanf("%lf", &c) == 1)
    {
        if (0.0 == c)
        {
            return 0;
        }
        Overhangs = 0;
        for (i=1; i; i++)
        {
            Overhangs += 1.0 / (i + 1);
            if (Overhangs >= c)
            {
                break;
            }
        }
        printf("%d card(s)\n", i);
    }
    return 0;
}