Shannon编码步骤：
1、将信源符号按概率从大到小的顺序排列，
p(a1)≥ p(a2)≥…≥ p(an)
2、确定满足下列不等式的整数Ki，
－log2 p(ai) ≤ Ki < 1－log2 p(ai)
3、令P(a1)=0，用Pi表示第i个码字的累加概率，
Pi=sum_{k=1}^{i-1} p(ak)
4、将Pi用二进制表示,并取小数点后Ki位作为符号ai的编码。
Shannon coding steps:
1. The source symbols are arranged in order of probability from large to small
2. Determine the integer ki satisfying the following inequality
3. Let P(A1)= 0, and PI denotes the cumulative probability of the i-th codeword,
4. PI is represented in binary system,and Ki bit after decimal point is taken as the code of symbol AI. (2020-09-06, C/C++, 1KB, 下载0次)
客户账单管理是电信计费系统必备的重要功能模块，主要负责对电信各类客户每月账单的增加、修改、删除、查询、备份等管理工作。本课题以中国电信企业客户账单管理模块原型参照，要求基于单链表结构对文件存储的客户账单数据进行排序、查找、计算、显示等造作。通过此可以，熟练掌握单链表结构、文件读写、函数调用等知识，以及查找、排序典型算法的设计与应用。
Customer billing management is an important prerequisite of telecommunications billing system modules, is mainly responsible for a variety of customers monthly telecom bills add, modify, delete, query, backup management. The topics to corporate customers of China Telecom billing management module prototype reference, based on the requirements of the customer billing data files stored on a single list structure sort, search, calculate, display and other pretentious. By this can master a single linked list structure, document literacy, function calls, such as knowledge, as well as search, sort and application design typical of the algorithm. (2013-12-22, Visual C++, 587KB, 下载3次)
任务分配问题
问题描述：
N个任务分配给n个人，任务j分配给人i的成本是C[I,j]，希望完成所有任务的成本最低，算法如何设计？
编程任务：
给定任务表示如下表，编程计算所需完成任务最低的成本。
数据输入：
请自己定义。
结果输出:
程序运行结束时，将计算结果输出到文件output.txt中。
Task allocation problem Problem Description: the N tasks assigned to the n individuals, the cost of task j allocation gives i C [I, J] want to complete all the tasks of the lowest cost, the algorithm is how to design? Programming task: given task the following table, programming required to complete the task the lowest cost. Data input: your own definition. Output: the end of the run, the calculation results output to the file output.txt. (2013-04-21, Visual C++, 1KB, 下载42次)
通过对源代码中那些未作文档说明的关键数据结构和方法进行描述，《深入理解MySQL核心技术》为您提供了很好的机会，帮助您了解这个广经企业考验的数据库的内部运作。不论您是开发人员、数据库管理员、程序员、软件供应商还是学生，本书将指导您探索和改进大型数据库
The source code of those documents to illustrate the key data structure and method are described, " in-depth understanding of the MySQL core technology " to provide you with a good opportunity, help you understand this widely by the enterprise internal operation test database. Whether you are a developer, database administrators, programmers, software vendors or student, this book will guide you through the exploration and improvement of large database (2013-01-09, Java, 47827KB, 下载3次)
1)自选存储结构,输入含n个顶点（用字符表示顶点）和e 条边的图G；
(2)求每个顶点的度,输出结果；
(3)指定任意顶点x为初始顶点,对图G作DFS遍历,输出DFS 顶点序列(提示:使用一个栈实现DFS)；
(4)指定任意顶点x为初始顶点,对图G作BFS遍历,输出BFS 顶点序列(提示:使用一个队列实现BFS)；
(5)输入顶点x,查找图G:若存在含x的顶点,则删除该结点及 与之相关连的边,并作DFS遍历(执行操作3)；否则输出信 息“无x”；
(6)判断图G是否是连通图，输出信息“YES”/“NO”；
(7)如果选用的存储结构是邻接矩阵,则用邻接矩阵的信息生 成图G的邻接表,即复制图G,然再执行操作(2)；反之亦然。
i don t know. (2012-11-09, C/C++, 2KB, 下载5次)
本项目旨在通过一个简化的企业管理信息系统（Enterprise Management Information System，
EMIS）项目，使学生在完成对C/C++程序设计语言和基本数据结构与算法课程的学习后，综合运用
所学到的语法和算法知识，构建一个接近实际应用场景的软件系统，以达到复习和巩固前期课程内容
并为后续课程奠定基础的目的。
This project aims through a simplified corporate management information systems (Enterprise Management Information System, EMIS) project, to enable students to learn the the C/C++ programming language and basic data structures and algorithms courses learning, integrated use of knowledge of grammar and algorithms, to build a software system close to the actual application scenarios to review and consolidate the the preliminary course content and the purpose of laying the groundwork for subsequent courses. (2012-10-16, Unix_Linux, 600KB, 下载14次)
为了使用Visual C语言编程直接读取dbf文件。需要了解dbf文件的二进制文件格式，下面给出简要的说明。 表文件由头记录及数据记录组成。头记录定义该表的结构并包含与表相关的其他信信息。头记录由文件位置 0 开始。数据记录1紧接在头记录之后（连续的字节），包含字段中实际的文本。 记录的长度（以字节为单位）等于所有字段定义的长度之和。表文件中存储整数时低位字节在前。
In order to use the Visual C language programming directly read dbf file. The need to understand the dbf file binary file format, a brief description is given below. The table file pretext of records and data records. The header record defines the structure of the table and contains other letter tables. The header record file position 0. Data records immediately after the header record (consecutive bytes), contains the actual text in the field. Recording the length (in bytes) is equal to the sum of the lengths of all the fields defined. Table stored in the file integer little endian. (2012-08-25, Visual C++, 19KB, 下载6次)
列主消元法练习 一个城镇有三个主要企业：煤矿、电厂和铁路作为它的经济系统。
生产价值1元的煤(产品1)，需消耗0.3元的电费(中间产品2)和0.2元的运输费(中间产品3)；
生产价值1元的电(产品2)，需消耗0.4元的煤费(中间产品1)、 0.1元的电费和0.1元的运输费；
提供价值1元的铁路运输服务(产品3)，需消耗0.3元的煤费和0.2元的电费和0.2元运输费。
在某个星期内，除了三个企业间的彼此需求，煤矿得到6万元的订单(即最终产品1)，电厂得到3万元的电量供应要求(即最终产品2)，而地方铁路得到价值5万元的运输需求(即最终产品3)。
Column elimination law practice in a town has three main businesses: coal mines, power plants and railways as its economic system. production of $ 1 worth of coal (1), 0.3 yuan in electricity consumption (intermediate product) and 0.2 yuan of transportation costs (intermediate products) the production value of $ 1 electricity (2), need to consume coal fee of 0.4 yuan (intermediate products), the tariff of 0.1 yuan and 0.1 yuan of transportation costs $ 1 worth of rail transport services (products), the cost of the coal charge of 0.3 yuan and 0.2 yuan in electricity transportation fees and 0.2 yuan. In a week, in addition to the three enterprises each other needs, coal 6 million orders (ie, the final product 1), power plant to get $ 30,000 in electricity supply requirements (ie, the final product 2), while the local railway 5 million worth of transportation needs (ie, the final product 3). (2012-05-30, Visual C++, 51KB, 下载5次)
本书是《程序员面试宝典》的第三版，在保留第二版的数据结构、面向对象、程序设计等主干的基础上，使用各大IT公司及相关企业最新面试题替换和补充原内容，以反映自第一版以来近几年时间所发生的变化。 欧立奇、刘洋、段韬编著的《程序员面试宝典》取材于各大公司面试真题(笔试、口试、电话面试、英语面试，以及逻辑测试和智商测试)，详细分析了应聘程序员(含网络、测试等)职位的常见考点。本书不仅对传统的C系语言考点做了详尽解说，还根据外企出题最新特点，新增加了对友元、Static、图形/音频、树、栈、ERP等问题的深入讲解。最后本书着力讲述了如何进行英语面试和电话面试，并对求职中签约、毁约的注意事项及群体面试进行了解析。本书的面试题除了有详细解析和答案外，对相关知识点还有扩展说明。真正做到了由点成线，举一反三，对读者从求职就业到提升计算机专业知识都有显著帮助。 《程序员面试宝典》适合计算机相关专业应届毕业生阅读，也适合作为正在应聘软件行业的相关就业人员和计算机爱好者的参考书。
This book is the" canon" in the third edition, retained in the second version of the data structure, object-oriented program design, main basis, the use of each big IT companies and related enterprises to the latest interview questions substitution and supplement the original content, in order to reflect the since its first edition in recent years since time changes. Ou Liqi, Liu Yang, Duan Tao," canon" drawn from major companies interview questions ( written examination, interview, telephone interview, interview, and logical test and IQ test ), a detailed analysis of the application programmer (including network, testing ) positions of the common points. This book not only to the traditional C language test done a detailed explanation, according to the new characteristics of the new foreign title, added to the friend, Static, graphics/audio, tree, stack, ERP issues in depth explanation. The last book on described how the English interview and telephone interview, and the job of signin (2012-03-21, Visual C++, 32106KB, 下载484次)
数据结构课程设计的，选存储结构,输入含n个顶点（用字符表示顶点）和e 条边的图G； (2)求每个顶点的度,输出结果； (3)指定任意顶点x为初始顶点,对图G作DFS遍历,输出DFS 顶点序列(提示:使用一个栈实现DFS)； (4)指定任意顶点x为初始顶点,对图G作BFS遍历,输出BFS 顶点序列(提示:使用一个队列实现BFS)； (5)输入顶点x,查找图G:若存在含x的顶点,则删除该结点及 与之相关连的边,并作DFS遍历(执行操作3)；否则输出信 息“无x”； (6)判断图G是否是连通图，输出信息“YES”/“NO”；
Data structures course, choose the storage structure, input containing n vertices vertex (character) and e edges in the graph G (2) Find the degree of each vertex, the output results (3) specify an arbitrary vertex xinitial vertex DFS traversal on the graph G, the output DFS sequence of vertices (hint: use a stack to achieve DFS) (4) specify an arbitrary vertex x is the initial vertex for BFS traversal of a graph G, the output the BFS vertex sequence (Hint: achieve the BFS uses a queue) (5) input vertices x, find the graph G: If the vertices containing x, then delete the node connected with the side, and for the DFS traversal (perform operations) otherwiseoutput "x" (6) to judge a graph G is a connected graph, output "yES"/"nO" (2012-02-13, Visual C++, 223KB, 下载3次)
企业发放的奖金根据利润提成。利润(I)低于或等于10万元时，奖金可提10 ；利润高 于10万元，低于20万元时，低于10万元的部分按10 提成，高于10万元的部分，可可提 成7.5 ；20万到40万之间时，高于20万元的部分，可提成5 ；40万到60万之间时高于 40万元的部分，可提成3 ；60万到100万之间时，高于60万元的部分，可提成1.5 ，高于 100万元时，超过100万元的部分按1 提成，从键盘输入当月利润I，求应发放奖金总数 掉不满足条件的排列。
Corporate bonuses based on the profit commission. Profit (I) less than or equal to 10 million, money can be raised 10 profit of more than 10 million, less than 20 million, less than 10 million part of the commission of 10 , higher than 100,000 yuan part, cocoa 7.5 commission 200000-400000 between, more than 20 million parts can commission of 5 400000-600000 among more than 40 million part time, can be 3 commission 600000-1000000 between, more than 60 million parts can commission of 1.5 , higher than 100 million, more than 100 million Part of the 1 commission, monthly profits from the keyboard input I, seeking to be the total number of bonuses out of the arrangement does not meet the conditions. (2011-09-03, C++, 3KB, 下载5次)
Flatopia是个很平的地方，政府想要修高速公路，以解决交通问题。高速公路系统能使得驾驶能够不出离此系统就能够在任意两城之间通行。城镇编号1~N，每条高速路连接两个城镇，所有高速路都是直线的，双向使用的。但是一个驾驶员只能在公路的尽头的城镇改变线路。当地政府想要尽量减少所需修建最长公路的长度，然而，必须保证每一个城镇都是可以达到的。
输入：
处理情况。
由一个矩阵代表了每两个镇之间的距离，每个情况末尾是空行。
二、 算法分析：
题目抽象后可得，给定一个图，要求找到一个可以让所有结点连接的方式，使得其中权值最大者最小。
Flatopia is a very flat place, the government wants to repair the highway in order to solve the traffic problem.Highway system allows the driver to leave the system is not able to travel between any two cities.Town No. 1 ~ N, each highway connecting the two towns, all highways are straight, bi-directional use.But a driver can only be the end of the road to change the town line.The local government wants to minimize the build up to the required length of roads, however, to ensure every town is achievable. (2011-08-08, Visual C++, 2KB, 下载6次)
试设计一个用回溯法搜索排列空间树的函数。该函数的参数包括结点可行性判定函数和上界函数等必要的函数，并将此函数用于解圆排列问题。
圆排列问题描述如下：给定n 个大小不等的圆c1 , c2 ,..., cn ，现要将这n 个圆排进一个矩形框中，且要求各圆与矩形框的底边相切。圆排列问题要求从n 个圆的所有排列中找出有最小长度的圆排列。例如，当n=3，且所给的3 个圆的半径分别为1，1，2 时，这3 个圆的最小长度的圆排列是1,2,1，其最小长度为2 + 4*sqr(2)。
编程任务：
对于给定的n(n≤10)个圆，编程计算最小长度排列。
Time Limit:50000MS Memory Limit:65536K
Total Submit:704 Accepted:481 (2011-06-09, Visual C++, 1KB, 下载4次)
New Year s Day is one of important days for many people in the world during the year. Most people spend the New Year s Day in hotels. January 1st is considered as the New Year s Day. Most companies, shops, school, and government offices are closed during that time. People prepare for New Year s Day from late December. First, people spend a few days to clean their houses completely. Some families then put up some new painting from November to be sent in January. The New Year s meal is also prepared from the end of December. During the New Year s Day, people usually do not cook and relax at home. On New Year s Eve, it is common to have a bag dinner with family members or friends at home or in hotels and hear bells which informs us of the coming New Year. On New Year s Day, people greet each other. Some people wear new coats and visit temples to pray for happiness and health through out the New Year. Children are busy with getting the gifts from their parents and relatives.
New Year s Day is one of important days for many people in the world during the year. Most people spend the New Year s Day in hotels. January 1st is considered as the New Year s Day. Most companies, shops, school, and government offices are closed during that time. People prepare for New Year s Day from late December. First, people spend a few days to clean their houses completely. Some families then put up some new painting from November to be sent in January. The New Year s meal is also prepared from the end of December. During the New Year s Day, people usually do not cook and relax at home. On New Year s Eve, it is common to have a bag dinner with family members or friends at home or in hotels and hear bells which informs us of the coming New Year. On New Year s Day, people greet each other. Some people wear new coats and visit temples to pray for happiness and health through out the New Year. Children are busy with getting the gifts from their parents and relatives.
(2011-01-11, Visual C++, 2538KB, 下载1次)
客户帐单管理是电信计费系统必备的重要功能模块，主要负责对电信各类客户每月帐单的增加、修改、删除、查询、备份等管理工作。本课题以中国电信企业客户帐单管理模块原型参照，要求基于单链表结构对文件存储的客户帐单数据进行排序、查找、计算、显示等造作。通过此可以，熟练掌握单链表结构、文件读写、函数调用等知识，以及查找、排序典型算法的设计与应用。
Customer Billing Telecom Billing System management is an important function modules necessary, is responsible for telecommunications to increase the monthly bills of all types of customers, modify, delete, query, backup management. The topics to corporate customers of China Telecom Billing Management module prototype reference requirements are based on single-chain structure of the customer billing data file storage to sort, find, calculate and display pretentious. With this you can, master a single linked list, file read and write, function calls and other knowledge, as well as find, sort of typical design and application of the algorithm. (2010-12-14, Visual C++, 60KB, 下载24次)
客户帐单管理是电信计费系统必备的重要功能模块，主要负责对电信各类客户每月帐单的增加、修改、删除、查询、备份等管理工作。本课题以中国电信企业客户帐单管理模块原型参照，要求基于单链表结构对文件存储的客户帐单数据进行排序、查找、计算、显示等造作。通过此可以，熟练掌握单链表结构、文件读写、函数调用等知识，以及查找、排序典型算法的设计与应用。
Customer billing management is an important telecommunications billing systems necessary function modules, is responsible for all telecommunications customers to increase the monthly bills, modify, delete, query, backup management. The topics to corporate customers of China Telecom Billing Management module prototype reference, required a single list structure based on customer billing data file storage to sort, search, calculate, display and pretentious. With this you can familiarize themselves with a single linked list structure, document literacy, the knowledge of the function call, and find, sort of typical design and application of the algorithm. (2010-05-17, Visual C++, 3KB, 下载8次)
1)自选存储结构,输入含n个顶点（用字符表示顶点）和e
条边的图G；
(2)求每个顶点的度,输出结果；
(3)指定任意顶点x为初始顶点,对图G作DFS遍历,输出DFS
顶点序列(提示:使用一个栈实现DFS)；
(4)指定任意顶点x为初始顶点,对图G作BFS遍历,输出BFS
顶点序列(提示:使用一个队列实现BFS)；
(5)输入顶点x,查找图G:若存在含x的顶点,则删除该结点及
与之相关连的边,并作DFS遍历(执行操作3)；否则输出信
息“无x”；
(6)判断图G是否是连通图，输出信息“YES”/“NO”；
(7)如果选用的存储结构是邻接矩阵,则用邻接矩阵的信息生
成图G的邻接表,即复制图G,然再执行操作(2)；反之亦然。
err (2009-01-04, Visual C++, 237KB, 下载12次)
在一个按照东西和南北方向划分成规整街区的城市里，n个居民点散乱地分布在不同的街区中。用x坐标表示东西向，用y坐标表示南北向。各居民点的位置可以由坐标(x,y)表示。街区中任意2点(x1,y1)和(x2,y2)之间的距离可以用数值|x1-x2|+|y1-y2|度量。
居民们希望在城市中选择建立邮局的最佳位置，使n个居民点到邮局的距离总和最小。
In a direction in accordance with the East-West and North-South divided into regular city blocks, n and the distribution of settlements scattered in different neighborhoods in. Things used to express the coordinates x, y coordinates used to express the North and the South. The location of the settlements by the coordinates (x, y) express. Arbitrary blocks 2:00 (x1, y1) and (x2, y2) can use the distance between the values | x1-x2 |+ | Y1-y2 | metrics. Residents want the city to select the best location to establish post offices, so that n settlements to the post office the sum of the smallest distance. (2008-05-22, Visual C++, 7KB, 下载2次)
在现代的企业管理中，火车管理系统有着重要作用。由于现代的竞争力的加强，讲究效率，不管是时间、还是地点，两地的生意常常因为距离而无法得到理想的解决。火车作为一个传统的交通工具，需要进行改革。由于订票、退票常常给乘客带来一定的麻烦，因此作为一个火车管理系统解决了这个问题。它方便乘客最快地订到自己想要的票也同时解决了退票的问题。
in modern enterprise management, train management system plays an important role. Contemporary the strengthening of competitiveness, focusing on efficiency, regardless of time or location, two of business because of distance and often can not be the ideal solution. Train as a traditional means of transport, the need for reforms. Since booking, a refund to passengers often cause some trouble, as a train management system to solve this problem. It is the convenience of passengers to the fastest set of the votes they want to also address the issue of refunds. (2006-04-11, Visual C++, 3KB, 下载6次)
一元稀疏多项式计算器[加法和乘法] 问题描述： 设计一元系数多项式计数器实现两个多项式间的加法、减法。 基本要求： （1） 输入并建立多项式 （2） 输出多项式，输出形式为整数序列：n，c1，e1，c2，e2……cn，en，其中n是多项式的项数，ci，ei分别为第i项的系数和指数。序列按指数降序排列。 （3） 多项式a和b相加，建立多项式a+b，输出相加的多项式。 （4） 多项式a和b相减，建立多项式a-b，输出相减的多项式。 用带表头结点的单链表存储多项式。 测试数据： (1) (2x+5x8-3.1x11)+(7-5x8+11x9) (2) (6x-3-x+4.4x2-1.2x9)-(-6x-3+5.4x2+7.8x15) (3) (x+x2+x3)+0 (4) (x+x3)-(-x-x-3)
one yuan sparse polynomial calculator [Addition and multiplication] Problem description : Design one yuan polynomial coefficient counter achieve the two polynomials addition, subtraction. Basic requirements : (a) input and the establishment of polynomial (2) output polynomial, the output form of integer sequence : n, c1, e1, c2, e2 ... cn, en, where n is the polynomial Number, ci, ei for the first item i the coefficients and indexes. By sequencing index in descending order. (3) a and b polynomial addition, the establishment of a polynomial b, the combined output polynomial. (4) a and b polynomial subtraction, the establishment of a polynomial-b, the output phase by polynomials. With the first table with a single node Chain store polynomial. Test data : (1) (2x 5x8- 3.1x11) (7-5x8 swath) ( (2005-07-07, C++, 37KB, 下载85次)