C++ Standard Library, The: A Tutorial and Reference Nicolai M. Josuttis...

junit详解， 也是一本介绍非常详细的书-JUnit explain, but also a very detailed introduction of the book

SIP的RFC文档。RFC3263. RFC3263-SIP RFC documents. RFC3263. RFC3263

TMS C6000系列的教程，很好，献给想学ti的DSP的朋友-C6000 tutorials, very good, dedicated to studying the DSP ti friends

超级好的Mfc学习资料，不看后悔死你，学习Mfc的中文文档，好好看看吧-Mfc super good learning materials, you不看后悔death, learning Mfc Chinese documents, a good look at it

约瑟夫（Joeph）问题的一种描述是：编号为1,2,…,n的n个人按顺时针方向围坐一圈，每人持有一个密码（正整数）。一开始任选一个正整数作为报数上限值m，从第一个人开始按顺时针方向自1开始顺序报数，报到m时停止报数。报m的人出列，将他的密码作为新的m值，从他在顺时针方向上的下一个人开始重新从1报数，如此下去，直至所有人全部出列为止。试设计一个程序求出出列顺序。-Joseph (Joeph) a description of the problem is: numbered 1,2, ..., n, n individuals sitting clockwise circle, each holding a password (positive integer). An optional start a positive integer as the number of reported upper limit of m, from the first person to embark on the clockwise direction starting from a reported sequence number, to report the number of m stop reported. Those who reported m out of the column, his password as the new m value, from his home in a clockwise direction on the next person started to re-reported from a few, and so on, until all of the column until all the up. Try a program designed to find out the column order.

CCNA学习指南,非常好-CCNA study guides, very good

linux程序设计，对于linux初学者相当有用！本书为第四版，英文原版！-linux programming, linux for beginners very useful! For the fourth edition of this book, the English original!

侯俊杰的深入浅出MFC-Hou Junjie"s easy MFC

代码阅读方法与实践，希腊Diomidis Spinellis著，赵学良译-Code Reading: The Open Source Perspective，By Diomidis Spinellis

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