网络系统集成与管理实验,自己做的实验报告。

ssd2 的答案啊，还是很好用的，不过需要密码的，没有的可以找我-ssd2

单点登录实际例子，希望有帮助，OA结合，已经测试ok，验证没问题的，希望多多益善指教

开源Tustena CRM的中文翻译，希望对即将使用Tustena CRM的人有好处.-Tustena CRM open-source Chinese translation, and they hope to soon use Tustena CRM will benefit the people.

Mini2440 esential documents files

BREW SDK2.01文档 BREW SDK 2.01 版向开发者提供环境，供其创建 BREW 2.01设备使用的应用程序。 本 API含有 BREW 2.01 工具及模拟功能介绍。 有关详细信息，请参阅发布声明的“新增功能”部分。-BREW SDK2.01 document 2.01 version of the BREW SDK to developers for the environment, BREW for its creation 2.01 equipment used by applications. The BREW API containing 2.01 simulation tools and features. For more information, please refer to a statement of the "new features" section.

嵌入式数据库sqlite-3.3.6移植到uClinux上，基于s3c44b0x开发板-embedded database sqlite- 3.3.6 transplant uClinux, Based on s3c44b0x development board

电子工程师必备的手册,上有CAN的知识,GPS,EMC,EMI相关资料-electronic engineers required manual on CAN knowledge, GPS, EMC, EMI-related information

st72321.pdf,关于STARM7的芯片资料-st72321.pdf, information on the chip STARM7

卡拉OK点歌系统。VC++ & Access。使用ODBC操作数据库，实现对歌曲的基本管理。-OK karaoke song system. VC++ & Access. Operation of the use of ODBC databases, implementation of the basic management of songs.

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