STLarray

此资料能够使自己充分的了解VC++,而且还能在其中熟悉语言环境。(This information can make their full understanding VC++, but also the environment in which you are familiar with the language.)

说明：  小的OpenGL 实例 适合初学者 学习参考 (OpenGL example of a small reference for beginners to learn)

A*算法 1、将开始节点放入开放列表(开始节点的F和G值都视为0) 2、重复以下步骤: 在开放列表中查找具有最小F值的节点,并把查找到的节点作为当前节点 把当前节点从开放列表删除, 加入到封闭列表. (A* algorithm 1, will begin to node placed in the and opening up list of (the began to node of the F and G values ​ ​ are regarded as 0) 2, repeat the the following steps: to Find a the has a the the smallest F value of the node in the the and opening up list of, and put Find a to the node as the current node current node is removed from the open list, added to the closed list.)

处理Register项的类 (The class for dealing with register item)

基于espeak的一个文本转语音的代码,可以把文本转成WAV文件(basic espaeck to translate txt to wav file)

CGUI在STM上的移植, PDF文件,下载查看(The transplantation of CGUI on STM)

设X[ 0 : n - 1]和Y[ 0 : n – 1 ]为两个数组，每个数组中含有n个已排好序的数。找出X和Y的2n个数的中位数。  编程任务 利用分治策略试设计一个O (log n)时间的算法求出这2n个数的中位数。 数据输入 由文件input.txt提供输入数据。文件的第1行中有1个正整数n（n<=200），表示每个数组有n个数。接下来的两行分别是X，Y数组的元素。结果输出 程序运行结束时，将计算出的中位数输出到文件output.txt中(Let X [0: n- 1] and Y [0: n- 1] for the two arrays, each array containing the n number has been sorted. 2n X and Y to identify the number of digits.  programming tasks using the divide and conquer strategy try to design an O (log n) time algorithm to calculate this median number 2n. Data input by the input data provided input.txt file. The first line in the file has a positive integer n (n < = 200), that there are n numbers of each array. The next two lines are the X, Y array elements. The end result is output program runs, the calculated median output to file output.txt)

使应用程序支持网页浏览，VC++精选学习源码，很好的参考资料。(The application supports webpage browsing, select learning VC++ source code, a good reference.)

German REAL LPT! USB2LPT

悬置功能计算的，能量解耦的方法，优化及其位移的计算(The calculation of suspended function, energy decoupling method, optimization and calculation of displacement)