这是一本关于最新LED驱动芯片的图书！对LED行也的朋友有用处...

Intel公司出品的内存测试软件--附源代码-Intel producer of memory testing software-- source

数据通讯与网络教程是进行网络编程和数据通信人员的一本圣经，共分为十部分，这是第八部分，尤其对准备从事网络编程、数据通信编程的工程人员具有指导意义。-data communication and network Guide is network programming and data communications staff of a Bible, is divided into 10 parts, This is the eighth part, in particular the right to engage in network programming, data communications programming staff is instructive.

一本关于算法的好书!带你走进计算机的世界!-A good book on algorithms! Take you into the computer world!

实战XML第二版，一本学习XML的好书。-real XML version of the second, a study of XML books.

pcf8583应用.pdf-pcf8583 application. Pdf

动物公司是本混迹职场的一本好书，推荐大家去看。-Animals occupy a place is the workplace of a good book to recommend everyone to see.

ADM6993F/FXFiber to Fast Ethernet Converter (TS1000 CPE Complied) The ADM6993F/FX is a single chip integrating two 10/100 Mbps MDIX TX/FX transceivers, a three-port 10/100M Ethernet L2 switch controller, and one OAM engine to meet demanding applications, including Fiber-to-Ethernet media converters, especially the fiber to the home (FTTH) media converters. The ADM6993F/FX feature set includes link pass through (LPT), TS1000 OAM frame receiving/processing/transmitting, programmable link status LED display, various loop-back modes, and one configurable MII ports for snooping/inserting OAM frame from/to 100Fx. The ADM6993FX is the environmentally friendly “green” package version.-ADM6993F/FXFiber to Fast Ethernet Converter (TS1000 CPE Complied) The ADM6993F/FX is a single chip integrating two 10/100 Mbps MDIX TX/FX transceivers, a three-port 10/100M Ethernet L2 switch controller, and one OAM engine to meet demanding applications, including Fiber-to-Ethernet media converters, especially

开发应用程序的图形界面,可能会用到多于一个的窗体,动态创建窗体.-err

背包问题动态规划算法JAVA 给定n种物品和一背包。物品i的重量是wi，其价值为vi，背包的容量为c。问应如何选择装入背包中的物品，使得装入背包中物品的总价值最大？ 在选择装入背包的物品时，对每种物品i只有两种选择，即装入背包或不装入背包。不能将物品i装入背包多次，也不能只装入部分的物品i。因此，该问题称为0-1背包问题。-Knapsack problem dynamic programming algorithm for a given n species JAVA goods and one backpack. Item i is the weight of wi, the value for vi, knapsack capacity c. Question should be how to choose the items into the backpack, the backpack into the total value of the largest items? In the selection of the items into the backpack, for each of the items i have only two choices, that is loaded or not loaded backpack backpack. Items should not be many times i load backpack can not only part of the goods loaded i. Therefore, the problem known as the 0-1 knapsack problem.

windows画布介绍-windows on canvas