3D Rubik

在屏幕上用控制台库在控制台模式下

核心函数是一个状态切换的函数 CAjaxParserDlg::Route。状态切换函数解根据一个 DFA 来对输入的文本进行分析。也就是说，如果你做一个新的DFA，就能分析新的词法。 DFA 的初始化在 CAjaxParserDlg::OnInitDialog。 -core function is a state of the switching function CAjaxParserDlg : : Route. State switching function according to a DFA solution to the input text for analysis. In other words, if you do a new DFA, we can analyze the new morphology. DFA in the initialization CAjaxParserDlg : : OnInitDialog.

获得tcp连接的方式，值得参考，好好学习，天天向上，-was tcp connections, is worth considering, study hard, make progress day by day, ha ha

DAO多线程的技巧-DAO multithreading techniques

这是关于一些界面编程的一些实例为了和大家分享我可是花了很多心机哦!!希望大家能借此机会有一个共同发展中国软件事业的理想!-This is the programming interface for some of the examples and to share with you but I spent a lot of effort! ! Hope that we can take this opportunity to be a common cause of the development of China"s software ideal!

一个封装的进度条类.可以替换自己的颜色显示百分比-The progress of a package of classes. Can replace their percentage of the color display

可变增益放大器AD603在雷达系统中的应用。有用的朋友请下载-Variable Gain Amplifiers AD603 in radar systems. Friend please download useful

约瑟夫环（Joseph）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）。一开始人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，从它在顺时针方向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 Central (Joseph) a description of the problem is: numbered 1,2, ..., n of n individuals according to a clockwise direction around a circle, each holding a password (positive integer). The beginning of a positive integer as a candidate at a number of upper limit of m, from the first individual to embark on the clockwise direction since the beginning of the order of 1 is reported the number of stops required to report back at a few m. Those who reported m out of his password as the new value of m, in a clockwise direction from the next person to start off again from the 1, and so on, until all that out until all. Try to design a procedure to derive the column order

VC++深入详解code，适合vc++编程-VC++ Depth detailed code, for vc++ Programming