Cisco E

GUI入门的好教材, 可以配套ARM使用, 内含有一些地层的初级函数和硬件接口.-GUI entry good material, you can use matching ARM, which contains a number of strata of primary function and hardware interface.

Verilog的学习资料，可编程器件fpga的开发语言，有重点介绍Verilog的关键语法-Verilog learning materials, they simply PLD development language, and to highlight the key Verilog syntax

关于硬盘低格的详细方法，对于要低格的朋友来说很更要，网上收集-Low on the hard disk of the detailed methodology for grid, the lower grid is more of a friend is to, on-line collection

通常的冒泡是单向的，而这里是双向的，也就是说还要进行反向的工作。 代码看起来复杂，仔细理一下就明白了，是一个来回震荡的方式。 写这段代码的作者认为这样可以在冒泡的基础上减少一些交换（我不这么认为，也许我错了）。 反正我认为这是一段有趣的代码，值得一看。-The bubble is usually a one-way, and here is a two-way, that is to say the job would also like to reverse. The complexity of the code look carefully you will understand the rationale for it is a shock the way back and forth. Author of Writing the code that this can reduce the bubble on the basis of the exchange (I do not think so, I may be wrong). Anyway, I think this is a fun code, worth a visit.

COM技术内幕-微软组件对象模型，与深入解析atl一样为经典丛书系列。-COM Technology Insider- Microsoft

Huffman编码的讲义，讲述了标准huffman编码的原理，也讲了各种变体形式，包括多媒体编码中常用的规范huffman编码原理。-Huffman coded notes about the standard huffman coding principle, but also talk about a variety of variant forms, including multimedia encoding used in the normative principles of huffman coding.

ALtera FPGA CYCLONE系列的功耗计算工具，相信大家会用的着．-ALtera CYCLONE FPGA series of power calculation tool, I believe we would use to.

山东大学网络教育学院 计算机专业《操作系统原理》模拟题 -Shandong University School of Education computer network professionals "operating system principles" that simulated

假设检验例题 7.1 设总体 ，其中参数 ， 为未知，试指出下面统计假设中哪些是简单假设，哪些是复合假设： （1） ； （2） ； （3） ； （4） ； （5） . 解：（1）是简单假设，其余位复合假设 -Hypothesis testing for Example 7.1 overall, including parameters, is unknown, pointed out that the following statistical hypothesis test which is a simple assumption, which is a compound hypothesis: (1) (2) (3) (4) (5). Solution : (1) is a simple assumption, the remaining bit composite hypothesis