Search - analog

Search - analog - List

[Data structs] sourcecode DL : 0: 以上一共五个在VC环境下编写的程序，分别为串的基本操作，哈夫曼编译码系统，简单词法分析器，进程转换模拟控制，最短路径搜索算法。为节省上传时间，只有源代码，工程文件都删除了，但是上述源码都在本机上编译通过，并且能够正确运行。谢谢！-More than a total of five in the VC environment prepared by the procedure, namely the basic string operations, Huffman encoding and decoding system, a simple lexical analyzer, the process of converting analog control, the shortest path search algorithm. From order to save time, only the source code, project documents are deleted, but the above source are compiled through this machine, and can function correctly. Thanks!
Date : 2026-01-18 Size : 46kb User : zgz
[Data structs] aluo1 DL : 0: 拿“优”的数据结构课程设计：模拟旅馆管理系统的床位的分配与回收功能。里头包括源代码、exe文件、课程设计报告。非常全，值得下载！-Get " excellent" curriculum of the data structure design: Analog hotel management system for the allocation of beds and Recycling function. Inside, including source code, exe files, curriculum design report. Very wide, it is worth to download!
Date : 2026-01-18 Size : 81kb User : aluo
[Data structs] GPSCA DL : 0: GPS数据的通道采集,完成载体姿态的数模建立,并实现定位定向-Access GPS data collection, achieve vector-profile digital-to-analog set up, and implementation-oriented positioning
Date : 2026-01-18 Size : 3.73mb User : lhb
[Data structs] Stack DL : 0: 简单模拟栈的操作：包括栈的初始化，压栈，栈顶元素弹出以及取栈顶元素；采用C实现-Analog stack easy steps: including the stack initialization, push, pop elements as well as the Top-Top-check elements implementation using C
Date : 2026-01-18 Size : 1kb User : yangxh
[Data structs] bank DL : 0: 银行业务模拟编写程序用C++编写数据结构可课程设计-Banking analog programming with C++ to prepare data structure of course design
Date : 2026-01-18 Size : 26kb User : liguifang
[Data structs] WaveData_ DL : 0: 波形数据软件这个东西好啊对控制数模转换芯片至关重要-Waveform data software to control this thing a few good-to-analog converter chip is essential
Date : 2026-01-18 Size : 88kb User : 刘冬
[Data structs] memorymanage DL : 0: 一道北大acm的试题，用双向链表实现模拟内存分配的程序-1 Peking University acm of examination questions, with two-way linked list of memory allocation procedures for analog
Date : 2026-01-18 Size : 450kb User : 尤里
[Data structs] Joseph DL : 0: 数据结构实验：用链表模拟约瑟夫环问题C语言 Joseph-Experimental data structure: the problem with the linked list C-ring analog Joseph Joseph
Date : 2026-01-18 Size : 1kb User : zx
[Data structs] shijianmoni DL : 0: 数据结构队列模拟事件驱动程序。模拟客户到达事件，客户离开事件，事件列表，列队窗口，显示模拟结构-Queue data structure event-driven simulation program. Analog customers to reach the event, the customer to leave the event, event list, lined up window, show the simulated structure
Date : 2026-01-18 Size : 30kb User : teamo38
[Data structs] LinearList DL : 0: 线性表的数组实现、链表实现、模拟指针、间接寻址和箱子排序。-A linear array of the table implementation, linked list implementation, analog pointer, indirect addressing and sorting boxes.
Date : 2026-01-18 Size : 4kb User : Robin.Wang
[Data structs] code2 DL : 0: 22. 切木头成绩: 10 / 折扣: 0.9 背景人们需要把一跟很长的木头切成几段，有一家名为 Analog Cutting Machinery (ACM) 的公司正在经营这一业务。他们根据切割前木头的长度来收费，木头越长、收费越高，并且每切割一次就收一次费。显而易见，在这里切割木头时，不同的切割顺序就会产生不同的价钱。譬如一跟 10 米长的木头，需要在 2、4、7 米处切开。如果顺序在这三个位置切割，需要的费用是 10 8 6 = 24，因为木头原始长度为 10 米，切掉两米剩 8 米，在四米处切掉剩 6 米。如果按照 4、2、7 的顺序来切割，花费就是 10 4 6 = 20。任务你的老板有很多木材要切割，现在他希望你能够帮他找到最便宜的切割方式。输入一次输入可能包含多组数据。每一组数据的第一行是木材的长度L (L<=1000)，如果为 0 则表示输入结束。每组数据的第二行是要切割的次数 N (N<=50)，第三行则是切割的位置Ci (0<Ci<L)。以上数据均为整数。输出针对每一组输入，输出切割这段木头的最小费用。 -22. Cutting wood Results: 10/Discount: 0.9 Background People need to put a cut into the wood with long paragraphs, there is a company called Analog Cutting Machinery (ACM) of the company is operating the business. Them according to the length of wood before cutting charges, the longer the wood, the higher the fees, and each time to receive a fee cut. Obviously, cutting wood in here, different order will produce different cutting price. For instance, one with 10 meters of wood, you need to cut the 2,4,7 meters. If the sequence of cutting in these three locations, the cost of required 1086 = 24, because the original length of 10 meters of wood, cut two meters 8 meters left, cut off the remaining four meters 6 meters. If the cut in accordance with the order 4,2,7, cost is 1046 = 20. Task Your boss has a lot of wood to be cut, and now he wants you to help him find the cheapest way of cutting. Input One input may contain multiple sets of data. The first line of each set of dat
Date : 2026-01-18 Size : 1kb User : a123
[Data structs] timelyManner DL : 0: 模拟控制正反向运动加减速及时换向和最高速度控制.-Analog control, positive and negative acceleration and deceleration, reversing a timely manner.
Date : 2026-01-18 Size : 26kb User : yuri
[Data structs] E_Josephus DL : 0: 模拟Josephus问题，编写算法实现击鼓传花。-Analog Josephus problem, write algorithm drum transfer flowers.
Date : 2026-01-18 Size : 162kb User : 巫小泉
[Data structs] yinhangyewumoni DL : 0: 该课程设计是通过对到达银行的客户办理的各种存取款业务进行分析，利用队列使办理不同业务的客户排到不同的队列，设置时间函数，利用动态存储结构实现模拟，并计算出顾客的平均逗留时间。 -The course is designed to reach the bank' s customers through a variety of access models for business analysis, the use of queues for different services to different customers routed to queue, set the time function, using dynamic storage structures for analog, and calculate the customer' s Average length of stay.
Date : 2026-01-18 Size : 112kb User : shifeng
[Data structs] fangzhendianti DL : 0: 模拟某校九层教学楼的电梯系统。该楼有一个自动电梯，能在每层停留。九个楼层由下至上依次称为地下层、第一层、第二层、……第八层，其中第一层是大楼的进出层，即是电梯的“本垒层”，电梯“空闲”时，将来到该层候命。乘客可随机地进出于任何层。对每个人来说，他有一个能容忍的最长等待时间，一旦等候电梯时间过长，他将放弃。模拟时钟从0开始，时间单位为0.1秒。人和电梯的各种动作均要消耗一定的时间单位（简记为t），比如：有人进出时，电梯每隔40t测试一次，若无人进出，则关门；关门和开门各需要20t；每个人进出电梯均需要25t；如果电梯在某层静止时间超过300t，则驶回1层侯命。要求：按时序显示系统状态的变化过程，即发生的全部人和电梯的动作序列。 -Simulation of a nine-school teaching building elevator systems. This building has an automatic elevator in each stay. Followed by the bottom nine floors as the basement, first floor, second floor, ... ... the eighth floor, where the first layer is the layer out of the building, that is the elevator "home base layer," Elevator "free" when came to the floor stand. Passengers can randomly access any layer. For everyone, he can tolerate a maximum waiting time, waiting for the elevator when the time is too long, he will give up. Analog clock starts from 0, the time unit is 0.1 seconds. And lift a variety of actions have to consume a certain unit of time (abbreviated as t), for example: it was out of the elevator 40t test every time, without the movement of people, then closed close and open the door of the need to 20t each personal and out of elevators require 25t if the elevator stationary at a level more than 300t, then drove back a layer ready to go. Requirements: Timing display s
Date : 2026-01-18 Size : 223kb User : fairybroken
[Data structs] elevator-simulation DL : 0: 电梯模拟，模拟某校九层教学楼的电梯系统。该楼有一个自动电梯，能在每层停留。九个楼层由下至上依次称为地下层、第一层、第二层、……第八层，其中第一层是大楼的进出层，即是电梯的“本垒层”，电梯“空闲”时，将来到该层候命。乘客可随机地进出于任何层。对每个人来说，他有一个能容忍的最长等待时间，一旦等候电梯时间过长，他将放弃。模拟时钟从0开始，时间单位为0.1秒。人和电梯的各种动作均要消耗一定的时间单位（简记为t），比如：有人进出时，电梯每隔40t测试一次，若无人进出，则关门；关门和开门各需要20t；每个人进出电梯均需要25t；如果电梯在某层静止时间超过300t，则驶回1层侯命。 -elevator simulation,Simulation of a nine-school teaching building elevator systems. This building has an automatic elevator in each stay. Followed by the bottom nine floors as the basement, first floor, second floor, ... ... the eighth floor, where the first layer is the layer out of the building, that is the elevator "home base layer," Elevator "free" when came to the floor stand. Passengers can randomly access any layer. For everyone, he can tolerate a maximum waiting time, waiting for the elevator when the time is too long, he will give up. Analog clock starts from 0, the time unit is 0.1 seconds. And lift a variety of actions have to consume a certain unit of time (abbreviated as t), for example: it was out of the elevator 40t test every time, without the movement of people, then closed close and open the door of the need to 20t each personal and out of elevators require 25t if the elevator stationary at a level more than 300t, then drove back a layer ready to go.
Date : 2026-01-18 Size : 68kb User : 陈庆
[Data structs] paixu DL : 0: 输入10组数据，每组数据包括学号，姓名，高等数学成绩（10学分），大学英语成绩（12学分），模拟电路成绩（8学分），数字电路成绩（5学分），信号与系统成绩（6学分），嵌入式系统成绩（5学分）。然后输入“学号”，则按照学号大小依次输出各组数据；输入“高数”，则按照高等数学成绩从高到低依次输出各组数据；输入“高数”、“英语”、“信号”、“嵌入式”、“模电”、 “数电”时依次按照相应成绩从高到低依次输出各组数据；输入“学分绩”，则按照学分绩从高到低依次输出各组数据。当输入错误的时候给出提示。要求使用“类”完成实验，计算学分绩作为“类”的公有成员函数，“类”具有构造函数。-Enter the 10 sets of data, each data including student number, name, Higher mathematics achievement (10 credits), College English results (12 credits), Analog circuit performance (8 credits), digital circuit performance (5 credits), Signal and system performance (6 credits), embedded systems performance (5 credits). Then enter the "Student ID", then turn out according to school size, number of data in each group Enter the "high number", then the higher math scores in descending order according to the output of each set of data Enter the "high number", "English", "signal", "embedded", "-mode", "Digital Power" is in turn results in descending order according to the corresponding output data in each group Enter the "credit performance", the results in descending order according to credit data output of each group. When prompted enter the wrong time. Require the use of "class" to complete the experiment, Calculation of credit performance as a "class" of the public me
Date : 2026-01-18 Size : 2kb User : yangliu
[Data structs] OS_driver-and-ticket DL : 0: 要求完成的主要任务: （包括课程设计工作量及其技术要求，以及说明书撰写等具体要求） 1．模拟公共汽车司机和售票员开关门及行车操作的同步模型。 2．设计报告内容应说明： ⑴ 需求分析； ⑵ 功能设计（数据结构及模块说明）； ⑶ 开发平台及源程序的主要部分； ⑷ 测试用例，运行结果与运行情况分析； ⑸ 自我评价与总结 -Required to complete major tasks: (including course design work and technical requirements, as well as written instructions and other specific requirements) 1. Analog bus driver and conductor switch doors and traffic operations synchronization model. 2. Design report shall state: ⑴ needs analysis ⑵ functional design (data structure and the module description) ⑶ source development platform and the main part ⑷ test, analysis of operating results and operating conditions ⑸ self-evaluation and summary
Date : 2026-01-18 Size : 899kb User : 张亮
[Data structs] SavingAccount DL : 0: 账户存储模拟 FIFO 类的深入剖析设计实现一个用于存储整型数据的队列类-ACCOUNT ANALOG/FIFO
Date : 2026-01-18 Size : 599kb User : ada
[Data structs] analog-parking DL : 0: 一个实现模拟停车场的数据结构代码，适合初学者学习-An analog parking lot of the data structure of the code, suitable for beginners to learn
Date : 2026-01-18 Size : 2kb User : 瑞文戴雨

« 12 3 »