文件系统过滤驱动开发教程 大约两年以前作者在驱动开发网上发表了一组描述如何开发Windows 文件系统过滤驱动的文章。非常庆幸这些文章能 给大家带来帮助...

GPS软件接收机设计的经典英文书籍，《Fundamentals of Global Positioning System Receivers - A Software Approach》对设计人员极具有参考价值。-GPS receiver software design of the classic English books, "Fundamentals of Global Positioning System Re ceivers-A Software Approach "of the design staff is a very valuable reference.

比较适合初学习j2me的人取学习。内含源码，中英文版分开。方便对照。-More suitable for the beginning of learning the person to whom to learn j2me. Containing source code, in separate English version. Convenient control.

微软八十年代老程序员的经验之谈，探讨如何编写编写高效无错代码，也是华为开发人员的必备宝典-old programmer with the experience to explore ways of preparing the efficient code no wrong, Huawei is also developing the necessary treasury

在一个划分成网格的操场上，n个士兵散乱地站在网格点上。网格点由整数坐标(x,y)表示。士兵们可以沿网格边上、下、左、右移动一步，但在同一时刻任一网格点上只能有一名士兵。按照军官的命令，士兵们要整齐地列成一个水平队列，即排列成(x,y),(x+1,y),…,(x+n-1,y)。如何选择x 和y的值才能使士兵们以最少的总移动步数排成一列。 实验任务：计算使所有士兵排成一行需要的最少移动步数。-In a grid divided into the playground, n soldiers stood scattered on the grid points. Grid point by the integer coordinates (x, y) express. Soldiers along the edge of the grid, down, left and right moving step, but at the same time any one grid point has only one soldier. In accordance with the officer s command, the soldiers should be neatly列成level queue, that is arranged in (x, y), (x+ 1, y), ..., (x+ N-1, y). How to choose x and y values to make the soldiers with the minimum of the total moves one line up. Experimental tasks: the calculation so that all the soldiers lined up his required minimum moves.

讲解C++的PPT（主要是关于继承与派生）-Explain the C++, PPT (mainly about the inheritance and derived)

优美的打印C语言源程序-beautiful Print C language source

visual c++ 角色扮演游戏程序：基于visual c++6.0的角色扮演游戏程序设计。-visual c++ RPG-GAME

vc++高级编程源代码-vc senior programming source code

protel的封装库 希望对大家有用，这是和99se相适应的，希望大家能够逐渐丰富该文件-protel library package you want to be useful, it is compatible, and 99se, and I hope that we can progressively enrich the document