本程序是一本很经典的知识响亮基的源代码　学起来比较方便　请大家使用...

javascrip教程-javascrip Guide

新概念Linux学习电子书籍，给大家一个离线认识Linux与学习Linux的平台。-Linux to learn the new concept of e-books give you an understanding of Linux and off-line learning platform for Linux.

j2ee核心编程，此书可谓经典之作。-j2ee core programming, this book is classic.

基于51单片机的数控直流电源的实现方案,调试等!-51 microcontroller-based CNC DC Power Solution and debugging!

XML宝典 (第二版)(PDG)-XML (2nd edition) (PDG)

c++编程实例详解(PDG 电子书需要超星阅读器）-c++ Detailed programming examples (PDG e-book reader needs Chaoxing)

想了解和开发邮件快递的别忧郁了！-would like to know the development of express mail and the other depressed!

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.

apache2中文使用手册-apache2 Chinese Manuals

用于SoC设计的DFT和BIST，讲解了在SOC设计中需要考虑的可测性设计问题 -SoC design for DFT and BIST, explain in the SOC design need to consider design-for-test issues