zoukankan      html  css  js  c++  java
  • Poj1003

    一、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.

    二、问题分析

           这道题就是题目难点,其他的也没什么难的了。题意呢就是,求1/2 + 1/3 + 1/4 + ... + 1/(n + 1)<input的最大的n.起初想着用调和级数来Log(n)+C-1(C为欧拉常数)求这个式子的,后来发现不太精确。所以就这能暴力求解了,反正也是很简单。要注意的就是,除法的精度在久1/n时,要用1.0/n这样编译器才知道是浮点运算。最后将结果减1既得所求。

    版权声明:本文为博主原创文章,未经博主允许不得转载。

  • 相关阅读:
    开源一些C#不常用知识(附上DEMO)
    开源:C# 代码自动生成工具,支持站点前后台
    Xposed 集成 Android 6.0.1环境中,总结
    Android 视频通信,低延时解决方案
    Android studio,第一个生成,调用成功的jni(说多了都是泪)
    C#之文件缓存
    JavaScript 基本常识
    排序算法
    LeetCode:字符串转换整数 (atoi)
    LeetCode:判断回文数
  • 原文地址:https://www.cnblogs.com/AndyDai/p/4734214.html
Copyright © 2011-2022 走看看