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linux下声卡编程，alsa声卡驱动程序集的参考文档
linux下声卡编程，alsa声卡驱动程序集的参考文档-under linux audio programming, alsa sound card driver set of reference documents
- 2023-05-25 14:10:03下载
- 积分：1
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软件工程中的软件设计实例分析是针对目前面相对象软件设计进行的。...
软件工程中的软件设计实例分析是针对目前面相对象软件设计进行的。-software engineering software design example is the view of the current physiognomy target for the software design.
- 2022-01-26 01:42:35下载
- 积分：1
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VC++深入详解教学视频文本编程学习文本编辑
VC++深入详解教学视频文本编程学习文本编辑-VC++ Depth detailed teaching video programming to study the text of a text editor
- 2022-04-26 22:19:14下载
- 积分：1
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This is a set of software development projects is the set of documents, in stric...
这是一套软件项目开发的全套文档资料,严格按照软件开发过程各个阶段而写的文档-This is a set of software development projects is the set of documents, in strict accordance with the software development process at all stages of writing the document
- 2022-03-25 08:43:21下载
- 积分：1
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SDT创建bin文件的步骤,need to configure option
SDT创建bin文件的步骤,need to configure option-SDT steps to create a bin file, need to configure option
- 2022-03-17 16:35:21下载
- 积分：1
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数字视频序列的运动目标检测和跟踪技术的圣..
数字视频序列运动目标检测与跟踪技术研究，西北大学的文章，讨论了多目标跟踪的问题
-Digital video sequences of moving targets detection and tracking of technical studies, Northwestern University, the article discussed the problem of multi-target tracking
- 2023-02-03 21:00:03下载
- 积分：1
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设计模式精解，一本经典的外国的软件工程方面的书籍
设计模式精解，一本经典的外国的软件工程方面的书籍-Design Patterns Explained.rar the explantation of the design patterns!
- 2022-02-04 02:34:53下载
- 积分：1
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software engineering system descriptions, classic books, particularly for the la...
软件工程的系统说明,经典图书,特别针对大型项目管理-software engineering system descriptions, classic books, particularly for the large-scale project management
- 2023-04-30 15:20:04下载
- 积分：1
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还有一个缺陷就是在只知道密文 x 及公钥（n,e）的情况下，只要将 (x^e) mod n 所得余数 s 再不断地循环操作 s = s^e mod n，此运算不...
还有一个缺陷就是在只知道密文 x 及公钥（n,e）的情况下，只要将 (x^e) mod n 所得余数 s 再不断地循环操作 s = s^e mod n，此运算不断地循环 e 次之后，很多情况下都可以循环出原文，只是计算量过余多一些罢了。不过有不少情况下，根本都无须循环 e 次，不过对于1024位的 n 级别来说，e 也是一个相当大的数值，所以循环密文的余数以解得原文是有些不现实。 以上内容仅供参考，如有不实，请予更正-there is a defect in only know that the secret and public key-x (n, e) the circumstances, as long as (x ^ e) mod n from the remaining s to continuously cycle operation s = s ^ e mod n, this constant cycle of Operational e occasion, the very many circumstances can be recycled from the original, but I calculated the volume more than just. There are, however, many instances, simply do not need e cycle times, but for 1024 the level n, e is a very large figure, so secret circle the remainder of the text was obtained in the original is a bit unrealistic. The above is for reference only, if not true, I corrected
- 2022-08-03 02:51:21下载
- 积分：1
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简单的商场优惠摊派金额,能显示具体到每个柜台的应该返还给消费者的金额....
简单的商场优惠摊派金额,能显示具体到每个柜台的应该返还给消费者的金额.-simple malls assessed the amount of concessions that can show specific to each counter should return to consumers the amount.
- 2023-07-06 23:35:03下载
- 积分：1
