英文版DSP2407资料，对其功能进行了说明，供大家学习

舞伴问题假定在一舞会上，男士排成一队，女士排成一队。跳舞开始时，依次从男队和女队的队头各出一人配成舞伴。若两队初始人数不相同，则较长的那一队中未配对者等待下一轮舞曲。设计要求：模拟上述舞伴系统，并能计算对于任何男士A和女士B在哪一轮舞曲中的k次跳舞？ -Question assumes that a dance partner, the men line up a team, ladies and line up a team. Dancing began, followed by men" s and women" s team from the team dubbed the first one from each partner. If the initial number of teams is not the same, then the longer of that team who is not waiting for the next round of the dance pairs. Design requirements: Analog the partner system, and can calculate for any men" s A and women dance in B in which a k-dance?

VC源码：PE文件格式的判断/PE文件格式的判断-source VC : PE file format judgment/PE file format judgment

这个是Windows程序设计第二版 作者为王艳平 的随书光盘-Win Program Design

HTML参考手册。-HTML reference manual.

basics of operating system

一个著名的java开源分页tag，eXtremeComponents中文教程-a famous java revenue pagination tag, eXtremeComponents Chinese Guide

＜Visual Basic.NET编程培训教程＞源代码- source code

Linux常见问题解答集中版.高质量电子书籍.对学习Linux有大帮助,欢迎下载学习.-Linux FAQ concentrated version. High-quality e-books. For learning Linux has great help, welcome to download the study.

网络安全，详细介绍了网络安全方面的基础知识及一些重点环节和应用。

通常的冒泡是单向的，而这里是双向的，也就是说还要进行反向的工作。 代码看起来复杂，仔细理一下就明白了，是一个来回震荡的方式。 写这段代码的作者认为这样可以在冒泡的基础上减少一些交换（我不这么认为，也许我错了）。 反正我认为这是一段有趣的代码，值得一看。-The bubble is usually a one-way, and here is a two-way, that is to say the job would also like to reverse. The complexity of the code look carefully you will understand the rationale for it is a shock the way back and forth. Author of Writing the code that this can reduce the bubble on the basis of the exchange (I do not think so, I may be wrong). Anyway, I think this is a fun code, worth a visit.