[POJ] #1003# Hangover : 浮点数运算
一. 题目
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 116593 | Accepted: 56886 |
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
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
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)
Source
- 按照固定的累加方式:从高层到低层,相对于其相邻下层可以伸出的长度为(1/2),(1/3),...,(1/n)
- 给定一个指定长度 S,问最少需要累加多上块板,使其相对于桌面的伸出长度大于或等于 S
三. 分析
- 算法核心: 此题比较简单,无需考虑任何算法
- 实现细节: 简单的浮点数累加运算即可
四. 题解
#include <stdio.h> int main()
{
int i;
float length, sum; while () {
sum = ;
scanf("%f\n", &length);
if (!length) break; for (i = ; ; i++) {
sum += ( / (float)i); if (sum >= length) {
printf("%d card(s)\n", i - );
break;
}
}
} return ;
}
[POJ] #1003# Hangover : 浮点数运算的更多相关文章
- POJ.1003 Hangover ( 水 )
POJ.1003 Hangover ( 水 ) 代码总览 #include <cstdio> #include <cstring> #include <algorithm ...
- OpenJudge / Poj 1003 Hangover
链接地址: Poj:http://poj.org/problem?id=1003 OpenJudge:http://bailian.openjudge.cn/practice/1003 题目: Han ...
- [POJ 1003] Hangover C++解题
Hangover Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 95164 Accepted: 46128 De ...
- poj 1003:Hangover(水题,数学模拟)
Hangover Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 99450 Accepted: 48213 Descri ...
- 快速切题 poj 1003 hangover 数学观察 难度:0
Hangover Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 103896 Accepted: 50542 Descr ...
- poj 1003 Hangover
#include <iostream> using namespace std; int main() { double len; while(cin >> len & ...
- [POJ] #1004# Financial Management : 浮点数运算
一. 题目 Financial Management Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 173910 Acc ...
- js,java,浮点数运算错误及应对方法
js,java浮点数运算错误及应对方法 一,浮点数为什么会有运算错误 IEEE 754 标准规定了计算机程序设计环境中的二进制和十进制的浮点数自述的交换.算术格式以及方法. 现有存储介质都是2进制.2 ...
- 深入理解计算机系统(2.8)---浮点数的舍入,Java中的舍入例子以及浮点数运算(重要)
前言 上一章我们简单介绍了IEEE浮点标准,本次我们主要讲解一下浮点运算舍入的问题,以及简单的介绍浮点数的运算. 之前我们已经提到过,有很多小数是二进制浮点数无法准确表示的,因此就难免会遇到舍入的问题 ...
随机推荐
- NDK(6)eclipse下断点调试ndk代码
Using the NDK Plugin 1. First set the path to SDK and NDK: Eclipse -> Window -> Preferences -& ...
- BZOJ 1502 月下柠檬树(simpson积分)
题目链接:http://61.187.179.132/JudgeOnline/problem.php?id=1502 题意:给出如下一棵分层的树,给出每层的高度和每个面的半径.光线是平行的,与地面夹角 ...
- computer使用快捷小技巧
1.快速打开运行 win+R 几个运行实用命令 notepad--------打开记事本 mspaint--------画图板 calc-----------启动计算器 compmgmt.msc-- ...
- CVS数据的导入和导出
2.CSV导入/导出测试 package junit.test; import java.io.File; import java.util.ArrayList; import java.util.L ...
- WCF常见异常-The maximum string content length quota (8192) has been exceeded while reading XML data
异常信息:The maximum string content length quota (8192) has been exceeded while reading XML data 问题:调用第三 ...
- 基于EasyUi的快速开发框架
先看图,下边这个简单的增.删.改.查,如果自己写代码实现,这两个页需要多少行代码? 如果再有类似的增.删.改.查,又需要多少行代码? 我最近搞的这个快速开发框架中,代码行数不超过100. 两页的代码如 ...
- hdu 4614 pieces 状态DP
题意:给你一个长度小于等于16的字符串,每次可以删除一个回文传,问你最少删除干净的字数. 状态+dp dp[i] = min(dp[i],dp[j]+dp[j^i]);(j是i的字串): 连接:htt ...
- UVa 10048 Audiophobia【Floyd】
题意:给出一个c个点,s条边组成的无向图,求一点到另一点的路径上最大权值最小的路径,输出这个值 可以将这个 d[i][j]=min(d[i][j],d[i][k]+d[k][j]) 改成 d[i][j ...
- [转] POJ数学问题
转自:http://blog.sina.com.cn/s/blog_6635898a0100magq.html 1.burnside定理,polya计数法 这个大家可以看brudildi的<组合 ...
- Android程序架构基本内容概述
在Android操作系统中开发的应用程序都有一个结构缜密的架构.我们今天就来对这一Android程序架构做一个详细的分析.帮助大家了解程序开发的特点,以方便将来在应用程序开中明确自己的程序架构. An ...