The Art of Assembly Language (Chapter1

一个进程管理的源程序.zip-a process management of the source. Zip

英文版Java 3D API手册，介绍J3D API接口。-The English version of Java 3D API manual introduction J3D API interface.

Java Web 开发：Spring,Hibernate,strut Ajax。PPT-Java Web Development: Spring, Hibernate, strut Ajax. PPT

详细的介绍了堆栈，堆，栈的区别，并给出了实例，解释

实现电梯的升降控制,采用C语言,要求:当用户确定目标楼层,则停靠,当无人操作时回1层等待-Elevator control to achieve take-off and landing, using C language, asked: When a user target floor, then stops, when no one back in operation and waiting for a layer

C语言实现的简单数学算法，简洁，易懂，容易学习。-C language mathematical algorithms to achieve a simple, concise, easy-to-understand, easy to learn.

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.

提出了一种多描述正弦编码器，正弦参数按照奇数位和偶数位交织的方法分解到两个描述中-Proposes a multiple description encoder sinusoidal, sinusoidal parameters in accordance with the odd bit and even bit-cutting method of decomposition to the two description

JAVA BOOK FOR U BROTHER E BOOK DOWNLAD

Java初学者入门教学-Java tutorial for beginners