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一个USB驱动代码,实现了USB枚举,组包发包等操作
一个USB驱动代码,实现了USB枚举,组包发包等操作-A USB driver code, and realized the USB enumeration, group packages contracting and other operations
- 2022-04-12 05:09:11下载
- 积分:1
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采用TCP handoff技术实现服务器集群。减少接入点的负载负担,使得集群的性能大大提高。...
采用TCP handoff技术实现服务器集群。减少接入点的负载负担,使得集群的性能大大提高。-Using TCP handoff technology for server clusters. The burden of reducing the load on the access point, making the performance of the cluster increased substantially.
- 2022-07-07 23:10:52下载
- 积分:1
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根据机关食堂消费流水生成各种业务统计报表
根据机关食堂消费流水生成各种业务统计报表-Water canteen consumption in accordance with authority to generate a variety of business statistics
- 2023-02-23 23:05:03下载
- 积分:1
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MinisQL的源码,是用C++写的,仅供参加
MinisQL的源码,是用C++写的,仅供参加-the Sourse code of MiniSql by C++,which is only for refference.
- 2023-08-17 06:45:03下载
- 积分:1
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简单的翻译功能页面,模仿主流翻印网站制作,美工还需要进一步修改....
简单的翻译功能页面,模仿主流翻印网站制作,美工还需要进一步修改.-Simple translation page reprint site mimic mainstream production, artists need to be further modified.
- 2022-04-18 16:29:35下载
- 积分:1
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fillin small vc procedures used in the preparation of small programs can s
写留言的小程序
用vc++编写的小程序
可以-fillin small vc procedures used in the preparation of small programs can s
- 2022-04-26 21:14:10下载
- 积分:1
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EXCEL formula to introduce the main functions, such as FIND, SEARCH
EXCEL的主要函数公式介绍,如FIND,SEARCH-EXCEL formula to introduce the main functions, such as FIND, SEARCH
- 2022-06-01 23:38:30下载
- 积分:1
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基于Wince4.2操作系统,evc4.0开发环境下ActiveX 控件的创建的例程源码...
基于Wince4.2操作系统,evc4.0开发环境下ActiveX 控件的创建的例程源码-Wince4.2 based operating systems, evc4.0 development environment ActiveX controls created routines FOSS
- 2022-01-28 20:08:16下载
- 积分:1
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provides for a division of silent video detection procedures, and results in the...
提供了一个用于视频分割的静音检测程序,结果以帧的位置的形式给出。-provides for a division of silent video detection procedures, and results in the frame position is given in the form.
- 2022-01-24 14:55:32下载
- 积分:1
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分析:用N进制的方法吧。设一个N个单元的数组,对第一个单元做加一操作,满N进
一。每加一次一就判断一下各位数组单元有无重复,有则再转回去做加一操作,没有则...
分析:用N进制的方法吧。设一个N个单元的数组,对第一个单元做加一操作,满N进
一。每加一次一就判断一下各位数组单元有无重复,有则再转回去做加一操作,没有则说明得到了一个排列方案。
-Analysis: N-band approach with it. A N-unit array, and add a unit to do one operation, over N into one. Add 1 each time you click on an array of modules to determine whether or not duplicate, then go back and have to do is add one operation, then there is no order to be a program.
- 2022-06-20 13:07:53下载
- 积分:1