poj日记（1003）

最新推荐文章于 2021-08-10 15:04:06 发布

Bwnly

最新推荐文章于 2021-08-10 15:04:06 发布

阅读量285

点赞数

CC 4.0 BY-SA版权

分类专栏： poj 文章标签： poj 算法

本文链接：https://blog.youkuaiyun.com/luantai6098/article/details/78639338

poj 专栏收录该内容

17 篇文章

订阅专栏

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.

若将一叠卡片放在一张桌子的边缘，你能放多远？如果你有一张卡片，你最远能达到卡片长度的一半。（我们假定卡片都正放在桌子上。）如果你有两张卡片，你能使最上的一张卡片覆盖下面那张的1/2，底下的那张可以伸出桌面1/3的长度，即最远能达到 1/2 + 1/3 = 5/6 的卡片长度。一般地，如果你有n张卡片，你可以伸出 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) 的卡片长度，也就是最上的一张卡片覆盖第二张1/2，第二张超出第三张1/3，第三张超出第四张1/4，依此类推，最底的一张卡片超出桌面1/(n + 1)。下面有个图形的例子：

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)

import java.util.Scanner;  
  
/** 
 * Created by mxc2011 on 2015/1/25. 
 */  
public class POJ1003 {  
    public static void main(String[] args){  
        Scanner in = new Scanner(System.in);  
        while(true) {  
            double m = in.nextDouble();  
            double sum = 0.0;  
            int i = 1;  
            if (m < 0.00001) {  
                System.exit(0);  
            }  
            if (m <= 0.5) {  
                System.out.println("1 card(s)");  
  
            } else {  
                while (sum < m) {  
                    i++;  
                    sum = sum + 1.0 / i;  
  
                }  
                System.out.println((i - 1) + " card(s)");  
            }  
        }  
    }  
}