一个用于制作Microsoft Word插件的小程序，使用Delphi语言开发

堆排序的free pascal实现代码及其复杂度分析-heap_sort

【分享】超市商品信息管理系统，基于组件技术，CS结构，非常经典！一本教材中的例程。-supermarket goods-- sharing information management system based on component technology, C S, very classic! A Textbook of routines.

进程是程序在一个数据集合上运行的过程，进程是并发进行的，也即系统中的多个进程轮流的占用处理器运行。 我们把若干个进程都能进行访问和修改的那些变量称为公共变量。由于进程是并发执行的，所以，如果对进程访问公共变量不加限制，那么就会产生“与时间有关”的错误，即进程执行后，所得到的结果与访问公共变量的时间有关。为了防止这类错误，系统必须要用同步机构来控制进程对公共变量的访问。一般说，同步机构是由若干条语句――同步原语――所组成。本试验要求学生模拟PV操作同步机构的实现，模拟进程的并发执行，了解进程并发执行时同步机构的作用。 -process is a procedure in the data set to run the process, the process is conducted with the system that is the rotation process more than the occupier runs. We have a number of process can visit and revision of those variables as public variables. As the process is complicated by the implementation, so if the right to visit public process variables without restrictions, it will have a "Time" is wrong, the implementation process, the results obtained with the public to visit the variables of time. To prevent such mistakes, the system must be used to control synchronization process bodies for public variables visit. In general, synchronous bodies of a number of statements-- synchronization primitives-- composed. Th

自己编写的一款小巧的YUV格式视频文件的播放器-I have written a compact of the YUV format video file player

实现pc机和安装无线CF卡的手持设备之间的通信的程序源代码，包含两个部分的代码：pc机端和手持设备端。-achieve PCs and the installation of wireless CF card handheld devices of communication between the program"s source code contains two parts of the code : PCs and handheld devices-end.

一个vc做的语音级文字聊天的服务器和客户端-A client/server used to make sound class written chat which is made in vc.

求集合的各元素的重数，也就是重复次数，和集合的众数，也就是重复次数最多的元素-Seeking re-set the number of each element, that is, repetition, and the collection mode, that is, the highest number of repeated elements

以深度为k的满二叉树(n=2k-1)为例，假设表中每个记录的查找概率相等，即 pi=1/n（1≤i≤n），而树的第i层上有2i-1个结点，因此，折半查找的平均查找长度为： 所以，折半查找的平均时间复杂度为O(log2n)。 -To a depth of k over the binary tree (n = 2k-1) as an example, suppose the table to find the probability of each record the same, that is, pi = 1/n (1 ≤ i ≤ n), and the tree layer of the first i There are 2i-1 nodes, therefore, to find half the average length of search as follows: Therefore, to find half the average time complexity is O (log2n).

枚举例子，可以作为初级读物看看还是可以的吧-Enumerated examples, can be used as primary reading material can take a look at the bar

限制你的鼠标在一个窗口内移动.不能移出这个区域.可以编出网吧管理程序-restrict your mouse in a window within the mobile. Not out of the region. Internet cafes will be made up management procedures