Java程序练习-Hangover

本文探讨了利用卡片构建堆叠时的最大过挂长度问题,通过数学推导和实例展示了如何实现不同数量卡片的最优化配置,以达到特定的过挂长度。详细分析了每增加一张卡片所能带来的过挂效果,并提供了计算所需卡片数量以实现目标过挂长度的方法。

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描述
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.


输入
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.
输出
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.
样例输入
1.00
3.71
0.04
5.19
0.00
样例输出
3 card(s)
61 card(s)
1 card(s)
273 card(s)

参考代码

import java.util.*; public class Main { public static void main(String[] args) { Scanner cin = new Scanner(System.in); while(cin.hasNextFloat()){ float ft = cin.nextFloat(); if(ft == 0.0){ break; } float temp = 0; int n = 1; while(temp <= ft){ temp += 1.0 / (n + 1); n ++; } System.out.println((n - 1)+" card(s)"); } } }


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