COMServices.zip

软件测试与junit实践一书的源代码，需要的请下载-The practice of software testing with junit book source code, need to download

c++Primer chn

Linux设备驱动（第一版）中文-Linux device driver (first edition) Chinese

actionScirpt3_help_cn,actionScript3的帮助手册。个人感觉比较好-actionScirpt3

参数估计.doc 通过本章的学习， 我们应该知道： 1. 统计抽样推断的常用术语 2. 抽样推断基于什么样的原理 3. 点估计与区间估计的具体操作 4. 抽样组织方式及其相应的误差计算 5. 每次抽样需要多大的样本容量 -Parameter estimation. Doc learning through this chapter, we should know: 1. Statistical sampling of terms commonly used inference 2. Sample inference based on what kind of principle 3. Point estimation and interval estimation of the specific operation 4. Sample organization and its corresponding error Calculation 5. each sample require much sample size

这是一个控制台程序，列出给定ASF文件的所有元数据属性。 -This is a console procedures set out to document the ASF will all metadata attributes.

Globus Toolkit 3.0 Quick Start（Globus Toolkit 3.0 快速入门） 这本 IBM 红皮书描述了 ITSO 小组使用 Globus Toolkit 3.0 的早期经验。GT3 是一种重要的网格实现，它能够适应最新的开放网格服务基础设施（Open Grid Services Infrastructure，OGSI） 本红皮书针对那些想要学习 GT3，但具备很少甚至没有 Globus 早期版本或者网格计算经验的人，为他们提供了重要的入门资料。本书向您展示了如何实现 GT3 演示，以及如何为验证概念而构造某些场景。同时，本文还阐述了 GT3 工具箱组件的高级概念和整体架构。 特别提示：这篇材料并不曾提交给任何正式的 IBM 测试，就这样直接出版了。文中的主题并不适合进行严格的评审。IBM 不会为本文的正确性和完整性负任何责任。对文中信息和任何技术实现方法的使用都由用户自己负责，用户要利用自己的能力对其进行评估，然后才能集成到自己的运行环境之中。-Globus Toolkit 3.0 Quick Start (Globus Too lkit 3.0 Quick Start) The Red Book describes the IBM ITSO team used Globus Too lkit 3.0 in the early experience. Globus Toolkits 3 is an important achievement grid, It can adapt to the latest Open Grid Services Infrastructure (Open Grid Services Infra structure, OGSI) against the Red Book for those who want to learn Globus Toolkits 3. but with little or no earlier versions or Globus grid computing experience. they provide important information portal. The book to show you how to achieve

约瑟夫（Joeph）问题的一种描述是：编号为1,2,…,n的n个人按顺时针方向围坐一圈，每人持有一个密码（正整数）。一开始任选一个正整数作为报数上限值m，从第一个人开始按顺时针方向自1开始顺序报数，报到m时停止报数。报m的人出列，将他的密码作为新的m值，从他在顺时针方向上的下一个人开始重新从1报数，如此下去，直至所有人全部出列为止。试设计一个程序求出出列顺序。-Joseph (Joeph) a description of the problem is: numbered 1,2, ..., n, n individuals sitting clockwise circle, each holding a password (positive integer). An optional start a positive integer as the number of reported upper limit of m, from the first person to embark on the clockwise direction starting from a reported sequence number, to report the number of m stop reported. Those who reported m out of the column, his password as the new m value, from his home in a clockwise direction on the next person started to re-reported from a few, and so on, until all of the column until all the up. Try a program designed to find out the column order.

tink in c