这是一个学写JAVA的好东西，主要介绍了JAVA在网络上的应用

java的API 当不知道如何使用时，我们可以查询这个API-java" s API when not know how to use, we can query the API

dining philosophers problem

《FrontPage2000教程》是洪恩在线网站上的教材，但通俗易懂，适合初学网页制作者阅读 用Dreamweaver MX 2004设计属于你自己的留言本 原作者：我佛山人 出处：5D多媒体 -"FrontPage2000 Guide" is a Hong Sishan online on the website material, but easily understood and suitable for a novice web reading producers use Dreamweaver MX 2004 to design your own voice the Author : I Foshan people Author : 5D Multimedia

简单描述导弹打飞机的过程，采用类思想，权作参考-A brief description of the process of the missile hit the aircraft, the use of categories of thought, the right for reference

开发dsp的朋友注意了，这本书是有20多年的dsp开发的老手的经验之谈，看了本书收获不是一-Development dsp friends attention, this book is more than 20 years of dsp development of veteran voice of experience, having read this book is not a little bit of harvesting

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.

MODEM调制解调器开发的C语言源程序 MODEM modem developed by C language source-MODEM modem developed by C language source

本文件是C++中的经典著作More Effective C++的中文版本.

数据库开发，通讯录管理系统，VC 6.0+ACCESS，里面有数据表，可直接运行-Database development, address book management system, VC 6.0+ ACCESS, which has data sheets, can be directly run

最小堆头文件(C++)-smallest Live (C)