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这份源码由MapX5.02所附的C++源码改写而来。其独特之处在于，源码中还包含一份Step by Step 的开发文档。非常有助于初次在VC6下使用MapX开发GIS的入门者。

其中包括西门子S7PLC的60个相关程序，是用Step 7软件进行开发的，对初学者很有好处！

ISE 7.1使用教程，对于初学者进行step-by-step的ISE7.1使用说明

Boost provides free peer-reviewed portable C++ source libraries. We emphasize libraries that work well with the C++ Standard Library. Boost libraries are intended to be widely useful, and usable across a broad spectrum of applications. The Boost license encourages both commercial and non-commercial ...

An adaptive fuzzy integral sliding mode controller for mismatched time-varying linear systems is presented in this paper. The proposed fuzzy integral sliding mode controller is designed to have zero steady state system error under step inputs and alleviate the undesired chattering around the sliding ...

These routines model tropospheric radiowave propagation over variable terrain and calculates propagation loss vs. height and range. Propagation loss is displayed in dB contours on a height vs. range plot. TPEM is based on the split-step Fourier PE method and was originally developed from an early PE ...

symbian C++ 入门经典 初学者必读，step by step 目录 1 概述 5 2 预备知识 5 3 Symbian OS Platform及应用开发5 3.1 内核和用户库.5 3.2 文件和目录5 3.3 目标平台及其变种6 3.4 Symbian OS的目标类型.6 4 应用开发工具包（SDKs）和工具7 4.1 SDK安装后的目录结构.7 4.1.1 根目录.7 4.1.2 Epoc32\.7 4.1.3 Epoc32Ex\8 4.2 ...

Fully revised to cover the latest standards and technologies, XML and Java(TM), Second Edition provides the practical solutions developers need to design powerful and portable Web-based applications. Featuring step-by-step examples, this book focuses on harnessing the power of Java(TM) and XML toget ...

Overview If you have been wanting to learn Java, check out the newly revised fourth edition of the best-seller Sams Teach Yourself Programming with Java in 24 Hours. This step-by-step tutorial will teach you how to create simple Java programs and applets. Comprised of 24 one-hour lessons, this new e ...

void Knight(int i , int j) { // printf("％d ％dn",i,j) if (board[i][j] != 0 || i < 0 || i >= Size || j < 0 || j >= Size ) { return } step++ board[i][j]=step if (step == Size*Size) { showboard() system("PAUSE") return } //DFS Knight(i-2,j-1) //left Knight(i-2,j+1) Knight(i+2,j-1) //right ...