SIP的RFC文档。RFC3263. RFC3263

《Microphone Array Signal Processing》，2008年新出的一本关于麦克风阵列处理的好书。- Microphone Array Signal Processing , 2008-year out of a good book on the microphone array processing.

GNOME Programming Guide lines, gnome 编程准则。PDF格式。-GNOME Programming Guide lines, pdf format

CSS开发,虽然不太全面,但一定对你有一些帮助的-CSS development, although not comprehensive, but certainly you have some help

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.

The fastest, easiest, most comprehensive way to learn Adobe Photoshop CS4 Fourteen lessons in Adobe Photoshop CS4 Classroom in a Book cover basic and advanced techniques in Adobe Photoshop, the world’s best image-editing software. Learn how to retouch digital photos, work with layers and masks, navigate the workspace, prepare images print, and explore the latest features. Tips, extra-credit exercises, and step-by-step lessons help you become more productive using Photoshop. Learn to correct and enhance digital photos, create image composites, transform images in perspective, and prepare images for print and the web. Combine images for extended depth of field, and try out the new 3D features in Adobe Photoshop CS4 Extended.

的源代码VB.NET版书-第一章。

传教士与野人问题我们分两种情况考虑。先考虑船在左岸的情况。如果不考虑限制条件，也就是说，船一次可以将三人从左岸运到右岸，然后再有一个人将船送回来。这样，船一个来回可以运过河2人，而船仍然在左岸。而最后剩下的三个人，则可以一次将他们全部从左岸运到右岸。所以，在不考虑限制条件的情况下，也至少需要摆渡[(M+N-3)/2]*2+1次。其中分子上的"－3"表示剩下三个留待最后一次运过去。除以"2"是因为一个来回可以运过去2人，需要[(M+N-3)/2]个来回，而"来回"数不能是小数，需要向上取整，这个用符号[ ]表示。而乘以"2"是因为一个来回相当于两次摆渡，所以要乘以2。而最后的"＋1"，则表示将剩下的3个运过去，需要一次摆渡。-M-C products

j2ee开发书籍,weblogic和jbuild开发详解-books, Weblogic and development Elaborates on jbuild

实现C语言中的malloc()函数，以及free（）函数-C language realization of the malloc () function, as well as free () function

主要讲述了线程间如何通信，最后还介绍了不同语言（包括C++、java、C）如何处理线程通信的问题-This is about Acknowledgments to Threads Primer