清华大学版的数据结构实验中利用链表编的多项式的加，减，乘的原程序...

本代码是约瑟夫环的循环链表实现方法,可以帮助理解约瑟夫环的基本原理,及其实现-This code is Joseph Central List realize the cycle method, can help to understand the basic principles of Joseph Ring, and its realization

建立快速傅利叶变换 C 语言源代码 17K-establish Fast Fourier Transform C language source code 24K

工作的任务，是负责一个项目的部分代码 对一幅带有人脸信息的24位真彩色bmp文件,通过基于肤色信息的彩色图象分割法,对人脸区域的检测,以及区域标定-On a piece of information with a human face 24 true color bmp files, color information based on color image segmentation method of human face region detection, as well as regional calibration 

求解约瑟夫问题。设有n个人围成一个圆圈坐下，对所有围从的人从某个位置开始编号为1,2,3，……，n,从编号为1的人开始报数1，报数依交进行，报数n的人即出列，下一个人从1开始报数，再报数m的人便是第二个出列的人如此重复下去，直到最后一个人出列为止，于是便得到一个出列的顺序，这称之为约瑟夫(Josephu)问题。-solving problems. N individuals have formed a circle to sit down, all right Wai from the start from a position of No. 1, 2, 3, ..., n, for a number of people reported a few, according to several newspaper pay, the newspaper n ie the out, the next person was from the beginning a few , and reported m is the second out of this series continue to repeat until the final out in a series, then get a series out of order, which called Joseph (Josephu) problem.

步骤1：创建computer.mdb数据库，查找其中的CPU表，并将上市时间介于2009-1-1到2009-12-31的所有CPU产品按表格形式输入； 步骤2：访问computer.mdb数据库，查找其中的CPU表，并将上市时间介于2009-1-1到2009-12-31的所有CPU产品按表格形式输出； 步骤3：使用"Dim objAdapt As New OleDbDataAdapter(strcmd,objCon)"创建了DataAdapter对象； 步骤4：SQL查询命令实现对computer.mdb数据库中的显示器表进行查询，支持用户按显示器的常见属性如“型号”、“供应商ID”、“品牌类别ID”、“尺寸”、“分辨率”“单价”等进行检索。 步骤5：自定义类Book、Publication，并在page_load过程中引用了Book和Publication类中的成员； 步骤6：关闭数据库； 附SQL求自然对数小程序。

Access数据库用户权限的管理程序，Delphi，转载-Access database user rights management procedures, Delphi, reprinted

在网络上收集的100个经典例子-online collection of 100 classic example

数据结构里的bfs算法，用vc6.0编写-data structure, the zheng algorithm used to prepare vc6.0

用Python实现各种图算法,包括深度优先，广度优先，Dijkstral，Prim，WelshPowell_Coloring等算法，

迷宫算法（maze） /* Maze * Starting point is m[0][0], need to find a path go to m[9][9]. 0 means OK, * 1 means cannot go there, boundary is 0 and 9, cannot go beyond boundary. * Each step can be made horizontally or vertically for one more grid (diagonal * jump is not allowed). * Your program should print a series of grid coordinates that start from m[0][0] * and go to m[9][9] * Hint: No need to find the shortest path, only need to find one path that gets * you to desitination. */ -/* Maze * Starting point is m[0][0], need to find a path go to m[9][9]. 0 means OK, * 1 means cannot go there, boundary is 0 and 9, cannot go beyond boundary. * Each step can be made horizontally or vertically for one more grid (diagonal * jump is not allowed). * Your program should print a series of grid coordinates that start from m[0][0] * and go to m[9][9] * Hint: No need to find the shortest path, only need to find one path that gets * you to desitination.