windows下，USB程序，VC编译运行通过，实现了USB的一些基本功能...

详细讲述了 ADC 及 DAC的处理方法与原理-A detailed account of the ADC and DAC principles of the treatment methods and

程序的主框架窗口为心型，比较有创意-Tht mainframe of the program is heart-shape, which seems creative.

AppWizard has created this GraphExp application for you. This application not only demonstrates the basics of using the Microsoft Foundation classes but is also a starting point for writing your application.

本文件主要讲ebd公司给的wince开发流程和pb的应用-This document is the main speaker ebd company to wince development of the flow and the application of pb

0-l背包问题是子集选取问题。一般情况下，0-1背包问题是NP难题。0-1背包 问题的解空间可用子集树表示。解0-1背包问题的回溯法与装载问题的回溯法十分类 似。在搜索解空间树时，只要其左儿子结点是一个可行结点，搜索就进入其左子树。当 右子树有可能包含最优解时才进入右子树搜索。否则将右子树剪去。设r是当前剩余 物品价值总和；cp是当前价值；bestp是当前最优价值。当cp+r≤bestp时，可剪去右 子树。计算右子树中解的上界的更好方法是将剩余物品依其单位重量价值排序，然后 依次装入物品，直至装不下时，再装入该物品的一部分而装满背包。由此得到的价值是 右子树中解的上界。-0-l knapsack problem is selected subset of the problem. Under normal circumstances, 0-1 knapsack problem is NP-hard. 0-1 knapsack problem the solution space available subset of the tree said. Xie 0-1 knapsack problem with the law retroactively loading the retroactive law very similar. The search solution space trees, as long as their son left node is a viable nodes, the search entered its left subtree. When the right subtree is the optimal solution may contain only son into the right tree search. It will cut right subtrees. Suppose that r is the total value of the remaining items; Cp is the current value; Bestp is currently the best value. When cp r bestp, they can cut right subtrees. Calculation right

code to click options if clk=1

CBitPointer- Easy Bit Manipulation

个人考勤软件开发示例，很简单的-personal attendance software development paradigm, a very simple

.Net中创建接收服务的组建-. Net to create a service set up to receive

它是用VC＋＋开发的一个小型编译器，它已实现一个编译器所具有的基本功能-it is with VC++ development of a small compiler, it has achieved a compiler all the basic functions