这是我大二时写的一个迷宫程序，可以实现从任意一个地方走到所指定的终点...

java project of controllers .... so many controls are used-java project of controllers .... so many controls are used....

Mobile手机录像，获取源数据是WMV9格式，输出格式为ASF Requirements: Visual Studio 2005, Windows Mobile 5.0 Pocket PC SDK, ActiveSync 4.0.-Feature Area: Video capture Description: This sample provides the basic video capture functionalities: It captures and encodes video in a WMV file, and captures and encodes an image in a jpeg file. Usage: Build and deploy using Visual Studio 2005. Relevant APIs/Associated Help Topics: ICaptureGraphBUilder2 Assumptions: Your device needs to have a video capture driver installed (the emulator ships with a null capture driver). The video will be saved to the root of your device to "video1.asf". The still picture will be saved to "test.jpg" If you wish, you can change these hardcoded values in the code for the files to be saved to a different location. Requirements: Visual Studio 2005, Windows Mobile 5.0 Pocket PC SDK, ActiveSync 4.0.

自己写的电子琴的程序,可以借鉴一下! 希望指点一写!共同进步!-Their own written procedures for organ, can learn from you! Hope pointing to write one! Common progress!

This package contains a collection of geometry filters-- programs that are for manipulating polgonal representations for 3D graphical objects. This source code is undergoing revision, and I expect that a new version of the code, together with many more filters, will be available shortly.

采用清华TPC-1十六位微机实验台上的并行口扫描键盘控制LED显示，以制作一个时间显示系统。-used Qinghua TPC-1 16 Microcomputer Laboratory platform parallel port scanning keyboard control LED display, for the production of a time display system.

An example of 3D textured tunnel, which allows keyboard movement.

还有一个缺陷就是在只知道密文 x 及公钥（n,e）的情况下，只要将 (x^e) mod n 所得余数 s 再不断地循环操作 s = s^e mod n，此运算不断地循环 e 次之后，很多情况下都可以循环出原文，只是计算量过余多一些罢了。不过有不少情况下，根本都无须循环 e 次，不过对于1024位的 n 级别来说，e 也是一个相当大的数值，所以循环密文的余数以解得原文是有些不现实。 以上内容仅供参考，如有不实，请予更正-there is a defect in only know that the secret and public key-x (n, e) the circumstances, as long as (x ^ e) mod n from the remaining s to continuously cycle operation s = s ^ e mod n, this constant cycle of Operational e occasion, the very many circumstances can be recycled from the original, but I calculated the volume more than just. There are, however, many instances, simply do not need e cycle times, but for 1024 the level n, e is a very large figure, so secret circle the remainder of the text was obtained in the original is a bit unrealistic. The above is for reference only, if not true, I corrected

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一个虚拟桌面的源码, Visual C++实现, 包装成类, 方便使用!-A virtual desktop of the source code, Visual C++ Realize, packing into a category, user-friendly!

A Slovienian Python tutor for beginners. A simple explanation of Python basic.