DCT离散余弦变换编码和解码，利用vc对bmp图像实行变换，可以对变换和恢复后的结果进行保存和观看比较结果...

这是一个基于LZARI算法的数据压缩的类.Haruhiko Okumura 于1989年7月4日用c语言写实现了这个算法.但是上面用到了一些全局或静态的变量,在MFC下用起来很不方便.我把它改写成了一个c++类,使它可以方便的压缩和解压缩,更重要的是,我新增加了两个接口,这个类可以压缩/解压缩一段内存缓冲区,而不仅仅是文件. 一共提供了5个对外接口: -LZARI algorithm based on the type of data compression. Haruhiko SAP in 1989, July 4 c language daily wrote realization of this algorithm. But above uses some static or global variables, using the MFC which are very convenient. I rewrite it became a c categories, it can convenience compression, but more importantly, I added two new interface, the class can compression/decompression period memory buffer, and not just paper. A total of five external interface :

aan快速算法

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

采用VB+ACCESS编写的汽车销售管理系统,-VB+ ACCESS to prepare the use of the automobile sales management system,

基于windows mfc的多线程程序设计之赛马进度条的显示-Windows mfc-based multi-threaded program design of the race progress bar display

利用VB开发通信程序主要的方法有两种，一是利用VB本身提供的控件（CONTRALS），另一种方法是利用WINDOWS API应用程序接口，WINDOWS API 主要提供了三个动态连接库KERNEL.EXE、USER.EXE、GDI.EXE供开发人员调用-use VB major communication program two ways. First, the use of VB itself, the controls (CONTRALS) Another method is to use Windows API application programming interface, Windows API provides essentially three KERNEL.EXE Dynamic Link Library, USER.EXE, GDI.EXE call for developers

图像处理高通和低通滤波器及重建-matlab code

0_1背包问题用动态规划解给定n种物品和一背包。物品i的重量是wi，其价值为vi，背包的容量为c。问应如何选择装入背包中的物品，使得装入背包中物品的总价值最大？ 在选择装入背包的物品时，对每种物品i只有两种选择，即装入背包或不装入背包。不能将物品i装入背包多次，也不能只装入部分的物品i。因此，该问题称为0-1背包问题。-0_1 knapsack problem using dynamic programming solution of a given n types of goods and one backpack. Item i is the weight of wi, the value for vi, knapsack capacity c. Question should be how to choose the items into the backpack, the backpack into the total value of the largest items? In the selection of the items into the backpack, for each of the items i have only two choices, that is loaded or not loaded backpack backpack. Items should not be many times i load backpack can not only part of the goods loaded i. Therefore, the problem known as the 0-1 knapsack problem.

地图着色把地图上的每个城市抽象为一个点，并给每个城市编号，，相邻的城市之间用直线连接。据此做出邻接矩阵，若第i个城市与第j个城市相邻，则metro[i][j]=1,否则metro[i][j]=0。 算法：按照编号从小到大的顺序检查每个城市，对每个城市从1到4使用4种颜色着色，若当前颜色可用（即不与相邻城市颜色相同），则着色；否则测试下一种颜色。 -Map coloring to each city on the map as a point of abstraction, and to every city number, adjacent lines to connect between cities. Accordingly to make adjacency matrix, if the section i and section j cities adjacent cities, the metro [i] [j] = 1, otherwise the metro [i] [j] = 0. Algorithm: In accordance with the code sequence from small to large checks every city, every city on 1-4 from the use of four kinds of color shading, color can be used if the current (ie not with the same colors as the neighboring cities), then coloring Otherwise, the next test colors.

AMR-WB编解码源码，附带测试源码与exe可执行文件。-Decoder and Encoder for AMR-WB