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eee
- 数据结构实验报告-experimental data structure report
ieee
- this section is include what a ieee and the ieee is consisite-this section include what is a ieee and the i eee is consisite
eee
- 小波变换经典程序
eee
- 这个关于数字识别码的源代码,欢迎大家下载使用,如有问题请联系我.
eee
- 并行AVS实时编解码器设计与实现 介绍了一种并行AVS实时编码器的设计，它包括音视频数据输入、音视频编码、传输流系统复用器、输出和控制部分，其 中重点介绍了视频编码器和传输流系统复用器的设计和实现。实验结果证明，实现标清AVS实时编码器是可行的。
EEE
- 一个数字钟的设计方法及方框图，请指教
eee
- 采用单片机和可编程逻辑器件作为低频数字相位测量仪核心部分
vc21487899244
- 主要源码： void CFDlg::OnChangeEdit1() { UpdateData() m_strExt.MakeUpper() m_list.ResetContent() if(m_strExt.GetLength()==0) return CStdioFile sf if(sf.Open(\"FileExt.txt\",Cfile::modeRead)) { CString strTe
eee
- 某国际贸易中心共有40层，设有载客电梯10部（用E0-E9标识）。要求用面向对象方法和面相对象程序设计语言，实现该高层建筑电梯活动仿真程序。
eee
- 这是宿舍管理系统-This is the hostel management system
eee.rar
- 一种目前最流行的人脸识别算法程序，利用的是特征脸识别算法，程序简单，每行都有注释,A face recognition is currently the most popular algorithm procedures, take advantage of the features of the face recognition algorithm, the procedure is simple, there are notes on each line
EEE.rar
- 飞思卡尔16位单片机MC9S12XE系列的EEPROM操作C语言代码，CodeWarrior环境，绝对能用，本人亲在MC9S12XEQ512上自验证过的,freeescale mcu MC9S12XE EEPROM operation C code
eee
- 一个类似超级玛丽的Java游戏，来自《Developing Games in Java》的一个游戏编程实例，可以全屏运行游戏，卷轴地图。压缩包内附有源码、jar文件与ANT编译文件，学习Java游戏编程的好实例。-A similar super-Mary' s Java Games, from " Developing Games in Java" Programming examples of a game, you can run the game full-scre
EC1363-MICROPROCESSOR-EEE
- EC1363 MICROPROCESSOR--EC1363 MICROPROCESSOR-EEE
cw1_task2
- 西交利物浦大学EEE方向的c语言作业，其中第二种包含有两种解法(there are 4 documents that based on C language, which are the assesment of XJTLU in EEE department)
as3
- 西交利物浦大学EEE方向的c语言作业，其中含有c文件，需要自建项目环境(this document that based on C language, which are the assesment of XJTLU in EEE department)
as4
- 西交利物浦大学EEE方向的c语言作业的Assesment 3，是c文件，需要自建项目环境(this document that based on C language in year 2 assessment 3, you need to build a new project to run)
cw1_task1
- 西交利物浦大学EEE方向的c语言作业，是之前CW1 task1 文件的第二种解法(this document that based on C language, which are the assesment of XJTLU in EEE department. This is the second solution for cw1 task1)