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VC.netWindeoSystem Window operating system based on the expansion of programming...
VC.netWindeoSystem 基于Window 操作系统的扩展编程。-VC.netWindeoSystem Window operating system based on the expansion of programming.
- 2023-02-11 20:30:03下载
- 积分:1
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简单的IOCP服务器源码, 新手一看很容易就在我它。
简单的IOCP服务器源码, 新手一看很容易就在我它。-Simple IOCP server source code, novice a look at it very easy for me.
- 2023-02-22 22:00:03下载
- 积分:1
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A very nice Flash+ XML image carousel source
很漂亮的Flash+XML图片轮播源码, -A very nice Flash+ XML image carousel source
- 2022-07-19 06:50:36下载
- 积分:1
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一个在VC6.0做的多功能计数器,可以实现各种进制的转换等各种复杂的功能...
一个在VC6.0做的多功能计数器,可以实现各种进制的转换等各种复杂的功能-VC6.0 to do a multi-function counter, can achieve a variety of binary conversion, and other complex functions
- 2023-08-13 00:45:07下载
- 积分:1
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MS (Microsoft Media Server protocol) protocol Overview: used to access and recei...
MS (Microsoft Media Server protocol)协议概述:
用来访问并流式接收 Windows Media 服务器中 .asf 文件的一种协议。MMS 协议用于访问 Windows Media 发布点上的单播内容。MMS 是连接 Windows Media 单播服务的默认方法。若观众在 Windows Media -MS (Microsoft Media Server protocol) protocol Overview: used to access and receive streaming Windows Media server. Asf file an agreement. MMS protocol is used to access the Windows Media Release point unicast content. MMS is to connect Windows Media Unicast service of the default method. If the audience in Windows Media
- 2022-02-10 01:55:49下载
- 积分:1
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Switch universal voice platform development, mainly in support avaya, Siemens, E...
交换机通用语音平台开发,主要支持avaya、西门子、爱立信交换机-Switch universal voice platform development, mainly in support avaya, Siemens, Ericsson switch
- 2022-12-22 06:50:04下载
- 积分:1
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origin软件学习资料,相信大家都知道吧
origin软件学习资料,相信大家都知道吧-origin software, learning materials, I believe we all know
- 2022-02-15 21:04:37下载
- 积分:1
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约瑟夫环(Joseph)问题的一种描述是:编号为1,2,…,n的n个人按顺时针方向围坐一圈,每人持有一个密码(正整数)。一开始人选一个正整数作为报数上限值m,从...
约瑟夫环(Joseph)问题的一种描述是:编号为1,2,…,n的n个人按顺时针方向围坐一圈,每人持有一个密码(正整数)。一开始人选一个正整数作为报数上限值m,从第一个人开始按顺时针方向自1开始顺序报数,报到m时停止报数。报m的人出列,将他的密码作为新的m值,从它在顺时针方向的下一个人开始重新从1报数,如此下去,直至所有人全部出列为止。试设计一个程序求出出列顺序-Joseph Central (Joseph) a description of the problem is: numbered 1,2, ..., n of n individuals according to a clockwise direction around a circle, each holding a password (positive integer). The beginning of a positive integer as a candidate at a number of upper limit of m, from the first individual to embark on the clockwise direction since the beginning of the order of 1 is reported the number of stops required to report back at a few m. Those who reported m out of his password as the new value of m, in a clockwise direction from the next person to start off again from the 1, and so on, until all that out until all. Try to design a procedure to derive the column order
- 2022-02-10 01:58:28下载
- 积分:1
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ntfs 文件恢复 希望需要的朋友有价值
ntfs 文件恢复 希望需要的朋友有价值-ntfs file recovery
- 2022-01-25 23:04:59下载
- 积分:1
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工作分配问题,该程序能够有效的进行工作分配,使得最终合乎要求...
工作分配问题,该程序能够有效的进行工作分配,使得最终合乎要求-work
- 2022-03-06 18:04:48下载
- 积分:1