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PAT (Advanced Level) Practice 1008 Elevator 分数 20 Python 解法

题目 The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 second

BUAA OO Unit2 —— Multithreading Elevator

BUAA OO Unit2 —— Multithreading Elevator by Monument_Valley 0. 写在正文前本篇博客是对笔者在北航2022年春季《面向对象设计与构造》课程第二单元的三次作业的总结。本单元的主要任务为通过程序模拟一套多线程运行的电梯系统，并为输入的乘梯请求作出合理的响应，即在不违反电

BUAA OO 第二单元总结报告

一、第二单元电梯作业设计思路初次接触多线程的问题，对于作业的设计思路一开始比较迷茫，后来通过阅读上机实验的代码，觉得采用“生产者-消费者”的模式比较合适，遂模仿实验的思路设计了InputHandler、Schedule、RequestQueue来进行输入数据的实时处理和调度，事实证明这一架构模式的设

BUAA_2022_OO_Unit2总结

OO Unit2总结题目概述本单元作业任务为实现一个模拟多线程的电梯调度系统。 HW5为迭代的基础：多部纵向非换乘电梯。具体为基于一个类似北京航空航天大学新主楼的大楼，大楼有 A,B,C,D,E五个座，每个楼座有对应的一台电梯，可以在楼座内 1-10 层之间运行。系统从标准输入中输入请求信息，

北航面向对象2022第二单元总结（Elevator.exe已停止运行...）

您的电梯已停止运行（NO！）这里是BUAAOO第二次单元作业总结博客。第二单元是BUAA的电梯模型，今年的电梯比起往年看起来友好很多。在第五次作业上手时，因为是第一次搭建相关逻辑，并且对于多线程构造和运行状态处于懵懂阶段，可能会遇到一些麻烦。（看到往年第一次的作业是傻瓜电梯的时候，我还在

BUAA_OO_2022_Unit_2_Summary

零、目录程序设计构架第一次作业第二次作业第三次作业协作图调度器设计同步块与锁分析电梯逻辑分析程序BUG分析自测/互测策略心得体会一、程序设计构架第一次作业需求简述：模拟一个多线程实时电梯系统，各楼座有且仅有一台纵向电梯，处理已知起终点的同楼座乘客

PAT Advanced Level 1008 Elevator

原题传送门 1. 问题描述 2. Solution 1、思路分析题目大意：电梯从0层开始向上，给出该电梯依次按顺序停的楼层数，并且已知上升需要6秒/层，下降需要4秒/层，停下来的话需要停5秒，问走完所有需要停的楼层后总共花了多少时间～分析：累加计算输出～now表示现在的层数，a表示将要去的层数，当a > now，

英语题目作业（16）

原题： Elevator 1000ms 32768K 描述： The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the el

1008 Elevator

The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds t

1008Elevator

1008 Elevator (20 分) The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one

python数据分析实战项目—10000条北京二手房多维度可视化分析（附源码）

文章目录开发工具数据内容实现代码运行效果10000条二手房信息下载地址总结开发工具 python版本：Python 3.6.1 python开发工具：JetBrains PyCharm 2018.3.6 x64 第三方库：pandas ;matplotlib ;seaborn 数据内容实现代码 import pandas as pd import seaborn as sns impo

1008 Elevator (20 分)2021-11-01

The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds

pta甲级1008 Elevator（AC）

The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds

1008 Elevator (20 分)

这个题很无语，最后一个楼层也要加上停的时间…… #include<bits/stdc++.h> using namespace std; int main(){ int t,sum=0,pre=0,now; cin>>t; sum+=t*5; while(t--){ cin>>now; if(now>pre) sum+=(now-pre)*6; else sum+=(pre

The Legend of the Elevator(20210720练习赛D题题解)

传送门目录Description题目分析AC代码 Description The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to mo

1008 Elevator (20 分)

The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds

IO调度算法分析--deadline算法分析

这部分是分析特定的IO调度算法--deadline, 这些算法的实现函数都是在通用层中被调用的, 所以应该对照着上面通用层的操作去理解. IO调度算法总的来说实现了两个功能, 1是IO的合并, 包括bio合并到request, 还有两个request的合并, 2是查找最合适的request, 交到分发队头去, 让驱

封装导出excel函数

/** * 导出excel * @param array $data_array 数据 * @param array $export_title 标题 * @param string $file_name 文件名 * @throws Exception */ function export2excel($data_array, $export_title, $file_name) { // 处理数据 $all_cnt = coun

BUAA_OO 第二单元总结

目录（1）同步块的设置和锁的选择，锁与同步块中处理语句直接的关系（2）调度器设计，调度器与程序中的线程交互（3）架构设计与可扩展性。（4）自己程序的bug（5）别人程序的bug（6）心得题会（1）同步块的设置和锁的选择，锁与同步块中处理语句直接的关系选择synchronized关键字封装代码块，在操作电梯列表、每个

OO第二单元总结

同步块的设置和锁的选择在三次的作业中，我都是选择共享变量waitQueue作为锁。对于需要读取或写waitQueue的语句块，我们需要在其外面加锁。以下以第一次作业为例进行分析。例1： synchronized (waitQueue) { if (end && passenger.isEmpty() && waitQueue.isEmpty())

1008 Elevator

The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds

1008 Elevator (20 分)

今天除夕，祝各位除夕快乐！晚上决定再写一题。这题太简单了。。希望考试的时候也都是这么简单的。 #include<iostream> using namespace std; int main(){ int n; cin>>n; int res=0; int floor[101]; floor[0]=0; for(int i =1;i<=n;i++){ c

A1008 Elevator (20分)_C语言_翻译+解析

The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds

2021-01-28

洛谷 The Fair Nut and Elevator 解析（还是看了别人的思路）题目：解析：代码的实现上，要看清题目，输入第二行代表的是每层的人数。然后，从第i层到第一层电梯要走的层数为i-1层。我的代码：

[PAT A1008]Elevator（水题、数学问题）

1008 Elevator Input Specification：Output Specification：Sample Input：Sample Output：Analysis：Code： The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator