VC编程的技巧及一些编程方法

动态规划法---如何设置友好城市间的航线使的航线数又最大又不相交-dynamic programming-- how to set up a sister city links are so few routes do not intersect also the largest

VCL消息系统的剖析--四级消息处理-VCL Message System Analysis-- four Message Handling

To split a file and vice versa on win32 platform

本系统基于现成gps接口及原生php，实现了对考场车辆的智能启停，会员充值收费，会员管理等基本功能模块，重点是实现了预约上车服务流程自动化管理，简单流程为：用户预约车辆，时间进入预约时间窗口，用户扫码上车，系统自动计时同时车辆自动开启，时间到车辆自动熄火用户下车，系统自动扣费完结。

Microsoft Windwos NT/2000 提供了一个强大的API集来访问系统事件和性能数据的众多计数器。我们既可以实时地得到计数器的值，也可以从一个日志文件中读取计数器数据。功能可为强大，而且使用简单 。下面我就简单谈谈在VC中如何使用Windows的性能计数器。好，废话少说，我们开始。-Microsoft Windows NT/2000 provides a powerful set of APIs to access the system events and performance data of the many counters. We can get real-time counter to the values, they can log files from a counter reading data. Functions as a powerful, but simple to use. Now, I would simply talk about how to use the VC Windows performance counters. Well, nonsense least, we have begun.

An Improved Switching Median filter for Uniformly Distributed Impulse Noise Removal

模拟Pos收银机 1.收银员功能 1) 新增明细(F2 ) 2) 删除明细(F3) 3) 收银(F4) 4) 挂单(F5) 5) 取单(F6) 6) 撤单(F7) 7) 退货(F8) 8) 退出(F9) 2.管理员功能 1) 商品管理 "1:商品查询", "2:库存查询", "3:添加商品", "4:修改商品信息", "0:返回主菜单" 2) 帐户管理 "1:查询账户", "2:添加账户", "3:修改账户信息", "0:返回主菜单" 3) 销售记录查询 "1:按销售日期统计", "2:按收银员统计", "0:返回主菜单" 0) 退出 -Simulation Pos cash register 1. Cashier function 1) Add Details (F2) 2) delete the details of (F3) 3) cash register (F4) 4)挂单(F5) 5) get a single (F6) 6) cancellation (F7) 7) Return (F8) 8) exit (F9) 2. administrator function 1) commodity management

在 c #all 神韵 c# xnatwo 课上没有很好的游戏： enumtype 和 cars3 stagesinclude 指令

用于关闭Windows信使服务弹出窗口的小程序，使用Delphi6编写，调用相关的API函数。最经典之处在于发现了WindowsNT5系列的OS（XP&2K）的信史服务弹出窗体中，信使服务四个字后面有一个空格。-closed for the Windows Messenger Service pop-up window to the small, the use of Delphi6 preparation, call the API function. The most classic is that WindowsNT5 found a series of OS (XP2K) messenger service pop-up window, the courier services are the words behind a box.

旅行售货员问题的解是一棵排列树．在给出的图中寻找一条费用最优的回路-travel salesman problem is the solution with a tree. The plan is to find an optimal cost of the Loop