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Hash-table
哈希表
数据结构课程设计
1.1问题描述
针对自己的班集体中的“人名”设计一个哈希表，使得平均查找长度不超过R，完成相应的建表和查表程序。
1.2基本要求
假设人名为中国姓名的汉语拼音形式。待填入哈希表的人名共有30个，取平均查找长度的上限为2。构造哈希函数，用链表法处理冲突。
1.3测试数据
读取熟悉的30个人的姓名作测试。(Hash table
Data Structure Course Design
1.1 Description of the problem
For their own class group of the "names" to design a hash table, making the average search length does not exceed R, the completion of construction of tables and look-up table corresponding procedures.
1.2 Basic requirements
Assuming the name of the person called Chinese pinyin form. To be filled into a hash table names a total of 30, taking the average search length of the upper limit of 2. Construct a hash function, treatment with a list of conflict.
1.3 Test Data
Read the familiar names of 30 individuals tested.)
- 2011-12-19 15:39:00下载
- 积分：1
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Huffman
说明： 数据结构：Huffman编码。哈夫曼编码是哈夫曼树的一个应用。哈夫曼编码应用广泛，如JPEG中就应用了哈夫曼编码。 哈夫曼树又称最优二叉树，是一种带权路径长度最短的二叉树。(Data structure: Huffman coding. Huffman coding is an application of the Huffman tree. Huffman coding is widely used, such as JPEG in the application of the Huffman coding. Huffman tree, also known as the best binary tree is a weighted shortest path length binary tree.)
- 2011-03-29 22:44:59下载
- 积分：1
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Data-structure-diagram
数据结构中图的存储结构的建立与搜索
建立图的二种存储结构
在不同的结构上实现遍历
(Data structure diagram of the storage structure of the establishment of the search to establish two kinds of graph storage structure in the different structures to achieve through)
- 2011-06-27 09:29:40下载
- 积分：1
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CatchData
说明： 读取股票数据文件,C++源程序，非常好用。是学生做毕业设计的源料(Read the Stock Data File)
- 2010-03-22 20:19:35下载
- 积分：1
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binarytree
二叉树的操作，涉及到三种顺序的遍历，树的凹入表示法，转化为一维数组储存，节点的计数，以及中序线索化。(Binary operation, involving three kinds of the order of traversal, tree concave representation into a one-dimensional array of storage, the node count, as well as in the threaded sequence.)
- 2011-05-24 20:31:13下载
- 积分：1
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sstring
利用串来比较数据大小，代码中有两个串，通过串来比较大小(
10/5000
Use the string to compare the data size)
- 2016-12-20 17:05:29下载
- 积分：1
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Cpp1
校园导游，可以选择景点序号来了解景点，通过选择两点可计算最短距离和最近路线。(Campus tour guide, you can choose the attractions serial number to understand the attractions of by selecting the shortest distance can be calculated and the recently routes.)
- 2012-06-20 21:57:32下载
- 积分：1
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Strassen-code
思塞恩算法
常用算法之一
算法导论提及(strassen algorithm which can run correctly)
- 2014-11-28 14:33:16下载
- 积分：1
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chessboard-distance
一道poj上的问题的解答，棋盘上的距离问题。即实现棋盘上从一个点移动到王，后，兵之间最短步数的解答。(A poj answers to your questions on the chessboard distance problem. That is, move from one point to the king on the board, after the shortest number of steps between the soldiers answer.)
- 2012-06-20 10:10:45下载
- 积分：1
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Task1-trial--10211281--zly
描述：足球俱乐部包括球员、普通行政人员。球员有姓名、年龄、能力、年薪、转会费、进球总数、服役年限等重要参数；普通行政人员有姓名、年龄、能力、年薪等参数。你需要给相关管理机构开发一个管理程序，实现对众多足球俱乐部管理的基本功能。具体功能包括：
a) 俱乐部的增删改查，俱乐部的属性包括名称、现金、其下人员；
b) 可以对所有球员的各项属性进行简单搜索，支持输入多个条件，条件之间用&&连接，例如：
i. “年龄不大于25且服役年限大于5年且进球数大于100的球员”，查询表达式：!(@age>25) && @experience>5 && @kick>100
c) 支持球员在不同俱乐部间的买卖，买卖需符合以下条件：作为买方的俱乐部的现金>=球员的转会费，交易完成后，卖方将得到转会费。
要求：
d) 必须采用面向对象的方式，使用类组织数据结构，并注意类的继承关系；
e) 在题目要求基础上可以自行扩展功能，提供扩实用扩展功能者适当加分（加分不超过5分）。
(Description: Football club including players, general administrative staff. Player name, age, ability, salary, transfer fee, total number of goals, service life and other important parameters ordinary administrative staff name, age, ability, salary and other parameters. You need to give the relevant management agencies to develop a management program to achieve the number of football club management s basic functions. Specific features include:
a) the club CRUD, the name of the club s property, including cash, under which personnel
b) all players can perform a simple search of the property, supports input multiple conditions with && connection between conditions, for example:
i. "Age is not greater than 25 and greater than 5 years service life and goals more than 100 players," query expression:! (@ age> 25) && @ experience> 5 && @ kick> 100
c) Support the players traded between different clubs, the sale subject to the following conditions: As a buyer s club cash> = pl)
- 2013-08-20 10:09:29下载
- 积分：1
