这是个解压缩软件UNZIP的C源程序代码 ，请站长接收

用M88来直接放出声来，有点效果的,很好玩的-With M88 to direct the sound to come out a little effect, it is fun

建立完整的赫夫曼树，及各种操作，其数据方法，为字母加空格-The establishment of a complete Huffman tree, and a variety of operations, methods of its data for the letter spaces

图像压缩，预测编码，给出了两种预测编码法，及其比较-Image compression, predictive coding, is given two kinds of predictive coding method, and its comparison

Averlogic AL576芯片MPEG4压缩示例源程序，可用于AV576 MPEG4压缩解压程序开发-Averlogic AL576 chip MPEG4 compression sample source code that can be used AV576 MPEG4 compression decompression program development

Daubechies小波变换和嵌入式零树编码(EZW)C语言程序，非常有用-Daubechies wavelet transform and embedded zero-tree coding (EZW) C language program, very useful

随即算法小课设，ｃ＋＋的写得，大二时写的，有不足之处，敬请谅解-immediately algorithm based small classes, c++ write, write when sophomore, is inadequate, locations

在Jpeg中Huffman采用的是范式Huffman编码。经过查相关资料有两种不同说法，请高手指点。谢谢！ 1、构造范式 Huffman 编码的方法大致是： 分别统计从最大编码长度 maxlength 到 1 的每个长度对应了多少个符号。根据这一信息从 maxlength 个 0 开始以递增顺序为每个符号分配编码。例如，编码长度为 5 的符号有 4 个，长度为 3 的有 1 个，长度为 2 的有 3 个，则分配的编码依次为： 00000 00001 00010 00011 001 01 10 11 问题：如果按照这种说法那么我在一付图像中渡到的数据是：编码长度为 6 的符号有 1 个，长度为 5 的有 1 个，长度为 4 的有 3 个，长度为3的个数为4个，长度为2的个数为1个。按照以上的算法得到的编码为：000000；00001；0010；0011；001；010；011；100；01。出现001是0011的前缀。在读取数据流的时候将不能区别。 2、另外一种说法，曾经发帖问过。前面的长度编码加1左移一位得到后面长度的编码。像我这个例子如何确定最短码的起始值？ 谢谢大家指点具体的范式Huffman编码实现-in which Huffman is the paradigm Huffman coding. After investigations related information in two different arguments, please master guiding. Thank you! 1, tectonic paradigm Huffman coding method is generally : statistics separately from the largest maxlength length coding to a length corresponding to each of the number of symbols. According to the information from maxlength-0 progressive order of the distribution of each symbol coding. For example,

mpeg 2 encoder vc++ 6.0实现-mpeg 2 encoder vc++ 6.0 to achieve

一个有用的无损压缩原码用于文件中二进制位操作的函数说明,函数实现,错误处理以及内存中的函数实现-a useful lossless compression code for the original document binary operation of the function that function to achieve, error handling and memory functions to achieve

具有成绩查询功能 适合初学者 表单 连编等功能 做的比较粗糙 刚学半年-Inquiry function with the performance forms suitable for beginners and other functions do not even prepared a rough six months just learning