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TAO学习工具，可以增删命名服务，及开发实例。可以在做tao开发中参考。可以直接编译。...
TAO学习工具，可以增删命名服务，及开发实例。可以在做tao开发中参考。可以直接编译。-TAO learning tools, you can add to, delete from naming services, and development examples. Can make the development of reference tao. Can be directly compiled.
- 2022-01-22 12:46:04下载
- 积分：1
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M3355 Software Architecture
M3355 Software Architecture
- 2023-05-23 04:45:03下载
- 积分：1
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As a software company in Shenzhen, director of research and development center a...
本人作为深圳一家软件公司研发中心总监时的一些工作文档。
此文为项目管理的相关制度。-As a software company in Shenzhen, director of research and development center at some of the work document. This article related to project management system.
- 2022-04-09 23:17:20下载
- 积分：1
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RS485的串口通信
本程序实现了基于API函数的16位的RS485的串口通信！程序能够以XMA5000的通讯协议，采用主机广播方式通讯，串口半双工，帧11位，实现了主机与从机设备的良好通信。
- 2022-09-12 06:45:03下载
- 积分：1
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样式表 样式表 样式表 样式表
样式表 样式表 样式表 样式表-Stylesheet style sheet style sheet style sheet style sheet style sheet style sheet
- 2022-03-19 23:54:44下载
- 积分：1
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软件工程开发文档，包含了软件开发各阶段所需要输出各类文档模板，非常实用。...
软件工程开发文档，包含了软件开发各阶段所需要输出各类文档模板，非常实用。-Software engineering development documentation, including the various stages of software development required for the output types of document templates, very useful.
- 2022-06-02 20:18:17下载
- 积分：1
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如何在C函数中调用C++函数一个简单的例子但很实用
如何在C函数中调用C++函数一个简单的例子但很实用-How to call C function C++ Function a simple example but it is practical
- 2022-05-08 01:57:29下载
- 积分：1
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bank accounts processing system, download unpacked Accounts can be achieved out...
银行帐务处理系统,下载解压后可以实现帐务的出入帐-bank accounts processing system, download unpacked Accounts can be achieved out of the recording
- 2022-03-20 23:34:03下载
- 积分：1
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ClearQuest是一种对缺陷与记录的变化进行跟踪管理的工具。它体现了一个BUG的完整的生命周期，从提交到关闭记录了BUG的改变历史。我读了两遍，很受启发，推...
ClearQuest是一种对缺陷与记录的变化进行跟踪管理的工具。它体现了一个BUG的完整的生命周期，从提交到关闭记录了BUG的改变历史。我读了两遍，很受启发，推荐给大家～文档有两份有点不同，大家随意选择。-ClearQuest is a record of defects and tracking changes in management tools. It is the expression of an BUG complete life cycle, from the records submitted to the closure of the BUG change history. I read it twice and enlightening to recommend to you two documents- a bit different, we choose arbitrarily.
- 2022-03-22 15:21:37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
