动态装入菜单项

modelsim的教程吧（中文的）. 很好用的.对出学者帮助比较大的哦.-modelsim it (English). Very good use. Right from academics to help big oh.

这是PKI入门，以及理解PKI的非常好的一本书-This is the entry PKI, PKI and understanding a very good book

《C++ 陷阱》内容简介 转自：http://infopower.com.cn/showdetail.php?isbn=7-5083-1494-8 C++ 是一门高效率但是复杂的编程语言，因为它的复杂性，使得程序员有机会去面对形形色色的各种设计和编程错误。如何去规避这些陷井，编写出更具重用性和易维护的C++代码，正是本书的目标所在。作者Stephen C. Dewhurst结合了将近二十年的C++开发经验，罗列了最常见的99个小问题，并详细阐述这些问题的解决之道，让读者在不知觉之间掌握C++的高级编程技巧，充分领悟C++的细微之处。阅读本书，必会对实际工作带来巨大的帮助。- C++ Trap carried Description: http://infopower.com.cn/showdetail.php?isbn=7-5083-1494-8 C++ Are an efficient but complex programming language because of its complexity, programmers have the opportunity to make the face of all sorts of various design and programming errors. How to circumvent these pitfalls, the preparation of a more reusability and easy maintenance of the C++ Code, it is the objective of this book. Author Stephen C. Dewhurst combines nearly two decades of C++ Development experience, listed the 99 most common minor problems, and elaborate on these issues a solution so that the reader can not grasp consciousness between C++ high-level programming skills, to fully comprehend C++ s

一篇介绍JAVA架构基于数据库应用程序的理论文章,其中介绍了目前常用的四种技术 密码:-a framework based on Java Database Application of theoretical articles, which introduced the four usual technical Password :

visual c++ 利用套接字开发网络通信程序-vc development network communication program using sockets

匿名通信代码，用C++编写，特别与大家一起分享，如有不足，还望给出指导意见-Anonymous communication code, using C++ Prepared, especially with the U.S. share, if insufficient, but also gives guidance on Point

统计学习理论和SVM(支持向量机).ppt入门讲义之一-statistical learning theory and SVM (support vector machine). Ppt s one of the portal

ASP.NET数据绑定函数Eval()的机制让你明白如何使用Eval（）-ASP.NET data binding function Eval () allows you to understand the mechanism of how to use Eval ()

Fortran常用算法程序集-徐士良间：第四章所有的源程序，欢迎寻找此书源程序的朋友下载使用　-Fortran algorithm commonly used procedures set- between徐士良: Chapter IV of the source of all, welcome to the book to find friends to download the source code to use

已知：Sn= 1＋1／2＋1／3＋…＋1／n。显然对于任意一个整数K，当n足够大的时候，Sn大于K。 　　现给出一个整数K（1