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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
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xuanzexingjiegou
选择型结构c语言编程 (Choose the type of structure of C language programming)
- 2013-09-24 22:11:37下载
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huffman
数据结构(严蔚敏版)中哈夫曼树的操作程序源代码(Of data structure (Yan Wei Min version) operating procedures in the source code Huffman tree)
- 2010-05-24 23:19:39下载
- 积分:1
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american_heritage
USACO中的American heritage,用C++编写,运用了树的遍历知识(the American heritage of USACO, with C++ prepared, using a tree traversal knowledge)
- 2013-07-21 18:31:10下载
- 积分:1
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C1
说明: 由C程序实现的 数学上 滑雪问题 的代码实现,供大家学习用(Ski problem code for all to learn to use)
- 2011-05-28 14:51:40下载
- 积分:1
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chess
在N*N的棋盘上(1<=N<=10)填入1,2,...N*N共N*N个数,使得任意两个相邻的数之和为素数. (In the N* N chessboard (1 < = N < = 10) filled 1,2, ... N* N total number of N* N so that any two adjacent numbers and prime numbers.)
- 2009-06-01 11:21:39下载
- 积分:1
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PRIM
PRIM算法 对任意给定的网和起点,用PRIM算法的基本思想求解出所有的最小生成树。(PRIM algorithm for any given network and the starting point, PRIM algorithm used to solve the basic idea of all the minimum spanning tree.)
- 2009-01-11 19:05:21下载
- 积分:1
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sort.cpp
各种排序的模板实现,并分析了各种实现的时间复杂度及改进办法(Templates of various sort, and analyze the time complexity of the various implementations and ways to improve)
- 2012-04-14 03:08:18下载
- 积分:1
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FinalExam
C语言期末考试题,含答案。数组、链表、函数等的使用和C编程。(failed to translate)
- 2011-11-07 20:07:25下载
- 积分:1
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adms
阿当姆斯显式和隐式求解方法,用四阶龙格库塔作为起始,然后运用四阶阿当姆斯算法求解初值问题。主要程序包含在test2.cpp中,方法简单易懂。编译环境VC2010(Adam James explicit and implicit method for solving fourth-order Runge-Kutta as a start, then use the fourth-order A Williams algorithm for solving initial value problem. The main program contains in test2.cpp method is simple and easy to understand. Compile environment VC2010)
- 2012-06-09 11:27:18下载
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