myself from the oracle oracle made mention 805 client streamline their procedure...

人力资源管理系统（附源码），用VFP实现的，已经连编了，比较完善！-Human Resource Management System (with source code), with VFP implementation has not even compiled, and more perfect!

数据结构及算法 对于参加ACM的学生有一定帮助 里面有详细的解答-Data structure and algorithm for ACM students to participate in some help there are detailed answers

几个C的小程序，有汉诺塔，数环打印，链表的创建并自动排序 -several small programs, Tower of Hanoi, several Central Print, Chain automatically create and order

数据隐私保护K-匿名问题的动态多次发布类似m-invarience算法-Data Privacy Protection K-anonymity problem on many occasions issued a similar dynamic m-invarience algorithm

假设一个文件中出现了8种符号S0，SQ，S2，S3，S4，S5，S6，S7，那么每种符号要编码，至少需要3bit。假设编码成000，001， 010，011，100，101，110，111。那么符号序列S0S1S7S0S1S6S2S2S3S4S5S0S0S1编码后变成 000001111000001110010010011100101000000001，共用了42bit。我们发现S0，S1，S2这3个符号出现的频率比较大，其它符号出现的频率比较小，我们采用这样的编码方案：S0到S7的码辽分别01，11，101，0000，0001，0010，0011， 100，那么上述符号序列变成011110001110011101101000000010010010111，共用了39bit。尽管有些码字如 S3，S4，S5，S6变长了(由3位变成4位)，但使用频繁的几个码字如S0，S1变短了，所以实现了压缩。对于上述的编码可能导致解码出现非单值性：比如说，如果S0的码字为01，S2的码字为011，那么当序列中出现011时，你不知道是S0的码字后面跟了个1，还是完整的一个S2的码字。因此，编码必须保证较短的编码决不能是较长编码的前缀。符合这种要求的编码称之为前缀编码。要构造符合这样的二进制编码体系，可以通过二叉树来实现。-Suppose a file appears in eight kinds of symbols S0, SQ, S2, S3, S4, S5, S6, S7, then each symbol to be encoded, at least 3bit. Suppose encoding 000,001, 010,011,100,101,110,111. Then the symbolic sequence S0S1S7S0S1S6S2S2S3S4S5S0S0S1 encoded into 000001111000001110010010011100101000000001, sharing a 42bit. We found that S0, S1, S2 these three symbols the frequency of relatively large, the other symbols the frequency is relatively smal

数据结构课程的实验之一，有关于数组的循环位移，简单的小程序-Experimental data structure course one of the cycle of displacement on the array, a simple applet

算法，钱币组合。可设定钱币种类数量，找出组合种数。-Algorithm, coin combinations. Can set the number of coin types, identify the combination of species.

题目:参加运动会的N个学校编号为1~N.比赛分成M个男子项目和W个女子项目,项目编号分别为1~M和M+1~M+W.由于个项目参加人数差别较大,有些项目取前5名,得分顺序为7,5,3,2,1 还有些项目只取前3名,得分顺序为5,3,2.写一个统计程序产生各种成绩单和得分报表.基本要求:产生各学校的成绩单,内容包括各校所取得的每项成绩的项目号,名次,姓名和得分 产生团体总分报表,内容包括校号,男子团体总分,女子团体总分和团体总分.概要设计:1. 为实现上述程序功能,应以线性表表示集合.2. 本程序包含3个模块:(1) 各集合定义模块(2) 线性表实现模块(3) 主程序模块-topics : participation in the Games N-school No. 1 ~ N. contestants were divided into men item M and W-woman project, a number of M and M ~ ~ 1 M W. As projects vary greatly in the number of participants, some of the items from the former five, scoring sequence of 7,5,3,2,1 also just take some items before the three scores sequence of incidents. write a variety of statistical procedures and report card scoring statements. the basic requirements : have the school report card, which includes school admission the results of each item, ranking names and scores scores statements groups, including schools, the men"s team scores, the women"s team scores and scores groups. summary of design : 1. to achi

数据库VFP编程模块-Database Programming Module

访问FTP下载数据，通过数值分析，并把数据以及结果存放到Oracle数据库中。FTP访问；数据分析；Oracle连接、操作。-FTP access to download data, through numerical analysis, and the data and the results stored into the Oracle database. FTP access data analysis Oracle connection, operation.