卡拉ok多工能播放器

这是一个用VC实现的，简单的视频播放器！-This is a VC achieved using a simple video player!

Load_PNG_Picture_With_Visual Basics

高级视频解码编码器-advanced video codec devices

本例源代码文件由用户按照书中的操作步骤自己生成，“Example-2-1Project_Navigator_Demo源代码”目录下为源代码的参考文件。“Example-2-1Project_Navigator_Democounter”目录下为完整的工程，包括源代码文件、综合与实现的结果文件、ISE工程文件等，可以使用ISE工程管理器打开工程，供读者参考-In this case the source code files by the user in accordance with the steps the book itself is generated, "Example-2-1 Project_Navigator_Demo source" directory as the source code reference document. "Example-2-1 Project_Navigator_Demo counter" directory for a complete project, including source code files, integrated with the realization of the outcome document, ISE project file, etc. You can use ISE Project Manager, open the project for the reader is referred to

vc多媒体编程不可多得的 初学者入门级教学源码（4）-vc multimedia programming rare entry-level teaching beginners source (4)

C#编写的音频播放器，可以播放mp3、wav，wma等格式-C# prepared by the audio player that can play MP3, WAV, WMA formats.

参数pszSound是标识要播放声音的字符串。 参数hmod是包含声音资源的可执行文件的句柄，当设置了SND_RESOURCE时，它不为NULL,其它情况为NULL。 参数fdwSound作用之一是表明声音资源的来源，它是一系列值的组合，通过或运算连在一起，设置适当的值可以控制声音的播放效果。如：SND_LOOP表示循环播放，SND_SYNC表示同步播放，更具体可以参考M-parameters pszSound logo is to play the voice of the string. Parameters hmod includes voice resources of the executable file handle, SND_RESOURCE when installed, it is not NULL, NULL other circumstances. Parameters fdwSound role is one of the voices and the source, it is a series of portfolio value, or processing units, setting appropriate values can control the voice of playback. Such as : SND_LOOP said cycle broadcast simultaneously broadcast SND_SYNC said, can be more specific reference M

游戏中的窗口切换全屏,再有全屏切换为窗口.按F5即可互换!-game of switching full-screen window, then switch to a full-screen window. F5 can swap!

此可用于诺基亚手机播放影片-this can be used to broadcast films Nokia

设X[0:n-1]和Y[0:n-1]为两个数组，每个数组中都有n个已经排好序的数字，试设计一个O（logn）时间的算法，找出X和Y的2n个数的中位数。-Set X [0: n-1] and Y [0: n-1] for the two arrays, each array has n good order have been ranked the number and try to design a O (logn) time algorithm to find of the X and Y in the 2n-digit number.