HDOJ1056 HangOver

HangOver

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14617    Accepted Submission(s): 6485

Problem 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.



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.

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

1.003.710.045.190.00

 

Sample Output

3 card(s)61 card(s)1 card(s)273 card(s)

 如果只有一张卡片，那它可以伸出自己本身的 1/2.

如果有两张，最上面一张可以伸出自己本身的1/2，下一张可以伸出1/3.

如果有三张，最上面一张可以伸出自己本身的1/2，下一张可以伸出1/3，最下面一张可以伸出1/4

如此类推。

假设每张卡片长度是1，给出伸出的总长度，问最小需要几张卡片。

import java.util.Scanner;public class Main{private static Scanner scanner;public static void main(String[] args) {scanner = new Scanner(System.in);while(scanner.hasNext()){double len = scanner.nextDouble();if(len == 0){break;}double sum = 0;double i = 2;int count = 0;while(sum<len){sum += 1/i;i++;count++;}System.out.println(count+" card(s)");}}}