CodeBus
www.codebus.net
Search
Sign in
Sign up
Hot Search :
Source
embeded
web
remote control
p2p
game
More...
Location :
Home
Search - max
Main Category
SourceCode
Documents
Books
WEB Code
Develop Tools
Other resource
Sub Category
ASP
.NET/ASPX
PHP
JSP
CGI
VC/MFC
Delphi/CppBuilder
Software Engineering
Network Develop
Server
Database
Homepage tools
Photo software
Other
Search - max - List
[
Books
]
max-rpm.rar
DL : 0
MAX-RPM应用详解 讲述Linux下的加解压打包工具
Date
: 2025-12-28
Size
: 1.48mb
User
:
[
Books
]
max-rpm
DL : 0
MAX-RPM应用详解 讲述Linux下的加解压打包工具-MAX-RPM applies the detailed solution to narrate under the Linux Canadian decompression to pack the tool
Date
: 2025-12-28
Size
: 1.48mb
User
:
吴声
[
Books
]
at3DMax基础教程100
DL : 0
at3DMax基础教程100.rar-at3DMax based tutorial 100.rar
Date
: 2025-12-28
Size
: 1.5mb
User
:
张伟杰
[
Books
]
3D Max基础教程
DL : 0
电脑应用\3D Max基础教程.rar-computer applications \ 3D Max basis Guide. Rar
Date
: 2025-12-28
Size
: 2.27mb
User
:
将将
[
Books
]
Example source codes in C++ using 3d studio max mo
DL : 1
Example source codes in C++ using 3d studio max models
Date
: 2025-12-28
Size
: 938kb
User
:
mskia
[
Books
]
3dmax
DL : 0
3dmax的3个事例,请大家共享-3Dmax the three stories, please share
Date
: 2025-12-28
Size
: 923kb
User
:
潘杰
[
Books
]
MPII_Quickstart_Chinese
DL : 0
Max+Plus II 简易用户使 用入门指南-Max Plus II Summary users Beginner's Guide
Date
: 2025-12-28
Size
: 228kb
User
:
姜力
[
Books
]
max+plus ii快速入门
DL : 0
maxplus2是一款应用于硬件编程的编程软件,本文件教你快速掌握其编程,仿真方法。-maxplus2 hardware is a programming application programming software, this document will teach you grasp its programming and simulation methods.
Date
: 2025-12-28
Size
: 336kb
User
:
刘晓飞
[
Books
]
fpga 和 cpld入门教程
DL : 0
本教程定位于FPGA/CPLD的快速入门。以ALTERA公司的芯片和相应的开发软件为目标载体进行阐述,本教程阐述了ALTERA主要系列芯片PLD芯片的结构和特点以及相应的开发软件MAX和Plusa和Quartus的使用-position in the handbook FPGA/CPLD Quick Start. With Altera's chips and the corresponding development of software for the target vector elaborate, the tutorials explain the main chips Altera PLD chips on the structure and characteristics of the corresponding software development MA Plusa and X and the use Quartus
Date
: 2025-12-28
Size
: 4.13mb
User
:
小易
[
Books
]
3DMaxjiaocheng
DL : 0
3Dmax学习教程,比较简单易懂,适合于初学者-study guides, relatively simple to understand and suitable for beginners
Date
: 2025-12-28
Size
: 8.45mb
User
:
张旭
[
Books
]
sj1
DL : 0
电子售货机的vhdl程序,用max-plus2编译,-electronic vending machines in vhdl procedures used max-plus2 compiler,
Date
: 2025-12-28
Size
: 89kb
User
:
木车
[
Books
]
3dmax
DL : 0
3Dmax建模方式与建模策略,很好的3Dmax教材,对建摸有总体的把握!-3Dmax modeling methods and modeling strategies, good 3Dmax materials, the construction has an overall grasp of touch!
Date
: 2025-12-28
Size
: 3.16mb
User
:
寒萧
[
Books
]
FPGAtaxier
DL : 0
摘要: 本文介绍了基于FPGA 的出租车计价器系统的功能、设计思想和实现, 该设计采用模块化自上而下的层次化设计,顶 层设计有5 个模块,各模块中子模块采用VHDL 或图形法设计。在Max+plusⅡ下实现编译、仿真等,最后成功下载到FPGA 芯 片中。完成了可预置自动计费、自动计程、计时、空车显示等多功能计价器。由于FPGA 具有高密度、可编程及有强大的软件 支持等特点,所以该设计具有功能强、灵活和可靠性高等特点,具有一定的实用价值。-Abstract: This paper introduces the function, design idea and realization of taximeter based on FPGA. The design takes the method of top-down and step by step. The whole system was divided into five modules that were described by VHDL or schematic diagram. By using Max+plus Ⅱ accomplish the compiler, simulator and so on. And then it can be downloaded to the FPGA chip . Achieve the goal to make a taximeter with the function of automatism count the money, the kilometer, the time and show empty car on a screen, ether. On FPGA high density and Programmable ability, you can see it has function better, modify convenient and high dependability. It has certainly practical value.
Date
: 2025-12-28
Size
: 202kb
User
:
lu
[
Books
]
CPLDEPM7128A
DL : 0
一篇关于7128CPLD的英文介绍,里面包含了44脚到100引脚各个型号的MAX系列cpld-An introduction in English about 7128CPLD, which contains a 44 foot to 100-pin MAX Series all models cpld
Date
: 2025-12-28
Size
: 499kb
User
:
李龙
[
Books
]
EDA
DL : 0
介绍使用MAX+plus 2以及部分实验原理介绍,以及EDA开发工具。-On the use of MAX+ plus 2, as well as to introduce the principle part of the experiment, and the EDA development tools.
Date
: 2025-12-28
Size
: 642kb
User
:
徐婷婷
[
Books
]
FPGAtextbook
DL : 0
本教程主要分为以下几个部分,Max+plusⅡ和QuartusⅡ软件介绍;组合逻辑电路实验;时序逻辑电路实验;数字电路系统设计实验(高级实验);实践训练项目。数字电路系统设计实验和实践训练项目可以选作为课程设计或课程实训的项目。课程设计或课程实训也可以利用硬件提供的MCU 单元结合软硬件进行设计。 本教程适用于应用电子技术专业、自动检测与仪表专业、电子信息专业、计算机控制专业、计算机应用专业等专业的电子设计类课程的教学实验及课程设计使用。-This tutorial is divided into the following sections, Max+ plus Ⅱ and software introduced Quartus Ⅱ combinational logic circuit experiment sequential logic circuit experiment digital circuit design experiments (Advanced Experimental) practical training program. Digital circuit design experiments and practical training projects can be chosen as the training curriculum or course design project. Training curriculum or course can also use the combination of hardware, software and hardware to provide the MCU unit design. This tutorial applies to the professional application of electronic technology, automatic detection and Instrument Engineering, Electronic Information, computer control of a professional, computer applications in electronic design professionals and other professional courses in teaching and curriculum design using experimental.
Date
: 2025-12-28
Size
: 1.65mb
User
:
孙
[
Books
]
Crack_QII90
DL : 0
您现在阅读的是 Quartus II 简介手册。 Altera® Quartus® II 设计软件是适合 单芯片可编程系统 (SOPC) 的最全面的设计环境。 如果您以前用过 MAX+PLUS® II 软件、其它设计软件或 ASIC 设计软件,并且准备改用 Quartus II 软件,或如果您对 Quartus II 软件有了一些了解但想进一步了解 它的功能,那么本手册非常适合您。-You are reading the Quartus II brochure. Altera ® Quartus ® II design software is suitable for Single-Chip Programmable System (SOPC) the most comprehensive design environment. If you ve used MAX+ PLUS ® II software, ASIC design software or other design software, and is prepared to use Quartus II software, or if you have some understanding of Quartus II software, but would like to learn more about Its function, then this manual is for you.
Date
: 2025-12-28
Size
: 29kb
User
:
陈冉
[
Books
]
main
DL : 0
求最大度问题,数据结构与算法。-Given a graph which contain N (1 <= N<= 2,500) nodes numbered from 1 to N. This graph has some edges, suppose the graph contain C (1 <= C <= 6,200) edges (described as two nodes ni1 and ni2 (1 <= ni1 <= N 1 <= ni2 <= N)). Now could you tell me which node has the max degree (For each node, the degree is the number of edges connects to the node.)? Output the max degree please.
Date
: 2025-12-28
Size
: 71kb
User
:
team
[
Books
]
3DS-MAX
DL : 0
3Dmax动画制作教程,经典教程,中级进阶-3Dmax animation tutorial
Date
: 2025-12-28
Size
: 16.05mb
User
:
chyzh
CodeBus
is one of the largest source code repositories on the Internet!
Contact us :
1999-2046
CodeBus
All Rights Reserved.