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[Other resource] HD7279(c语言)驱动程序 DL : 0: HD7279(c语言)驱动程序作用：测试Sage Tech.开发板（MCP300）中hd7279键盘显示芯片是否能正常工作程序功能：按下1号键显示0、1、2、3，按下2号键显示c、d、e、f，按下3号键显示4、5 、6、7 按下4号键显示8、9、a、b，按下0号键点亮4个LED灯-HD7279 (c Language) driver role : Sage tests Tech. Development board (MCP300) hd7279 keyboard chip whether normal procedures functions : pressing keys on the 1st show 0,1,2,3? Press on the 2nd c, d, e , f, press the button on the 3rd show 4,5, 6,7 press the button shows on the 4th 8,9, a, b, press the button to light up the No. 0 4 LED lights
Date : 2008-10-13 Size : 1.73kb User : 刘浪
[Other resource] 附录B 对比C++和java DL : 0: C++和JAVA的对比,里面详细介绍的2种语言的不同点.-C and JAVA contrast, they described in detail two types of language differences.
Date : 2008-10-13 Size : 13.33kb User : 22
[Other resource] C语言开发 BTREE 数据文件索引程序库 DL : 0: 用c开发的B+树索引结构,能实现B+树基本的查找,删除,搜索功能-with the development of B-tree index structures to achieve the basic B-tree search, delete, search functions
Date : 2008-10-13 Size : 15.41kb User : 皇者之剑
[Other resource] 排序算法、字典和B-树的C++语言实现 DL : 0: 排序算法、字典和B-树的C++语言实现代码内容包括以下算法： qui.c sort: quicksort qsort.c sort: qsort ins.c sort: insert sort shl.c sort: shell sort has.c dictionary: hash tables bin.c dictionary: binary tree rbt.c dictionary: red-black trees skl.c dictionary: skip lists ext.c external sort btr.c btree 语言 C++ 编译平台 Visual C++ 作者 Thomas Niemann 备注来自Sorting and Searching Algorithms:A Cookbook-algorithm, dictionaries and B-Tree C language code include the following algorithm : qui.c sort : quicksort qsort.c sort : qsort ins.c sort : insert sort shl.c sort : shell sort has.c dictionary : hash tables bin.c dictionary : binary tree rbt.c dictionary : red-black trees skl.c dictionary : skip lists ext.c external sort btr.c BTREE Part C language Visual C platform translated author Thomas Niemann Remarks from the Sorting and S earching Algorithms : A Cookbook
Date : 2008-10-13 Size : 24.36kb User : 罗鹏魁
[Other resource] aaagchcv DL : 0: 源代码\\用动态规划算法计算序列关系个数用关系\"<\"和\"=\"将3个数a，b，c依次序排列时，有13种不同的序列关系： a=b=c,a=b<c,a<b=v,a<b<c,a<c<b a=c<b,b<a=c,b<a<c,b<c<a,b=c<a c<a=b,c<a<b,c<b<a 若要将n个数依序列，设计一个动态规划算法，计算出有多少种不同的序列关系，要求算法只占用O(n),只耗时O(n*n).-source \\ use dynamic programming algorithm sequence relationship with the number "
Date : 2008-10-13 Size : 8.24kb User : Sana
[Other resource] LCS_Dynamic_Programming DL : 0: LCS（最长公共子序列）问题可以简单地描述如下：一个给定序列的子序列是在该序列中删去若干元素后得到的序列。给定两个序列X和Y，当另一序列Z既是X的子序列又是Y的子序列时，称Z是序列X和Y的公共子序列。例如，若X={A，B，C，B，D，B，A}，Y={B，D，C，A，B，A}，则序列{B，C，A}是X和Y的一个公共子序列，但它不是X和Y的一个最长公共子序列。序列{B，C，B，A}也是X和Y的一个公共子序列，它的长度为4，而且它是X和Y的一个最长公共子序列，因为X和Y没有长度大于4的公共子序列。最长公共子序列问题就是给定两个序列X={x1,x2,...xm}和Y={y1,y2,...yn}，找出X和Y的一个最长公共子序列。对于这个问题比较容易想到的算法是穷举，对X的所有子序列，检查它是否也是Y的子序列，从而确定它是否为X和Y的公共子序列，并且在检查过程中记录最长的公共子序列。X的所有子序列都检查过后即可求出X和Y的最长公共子序列。X的每个子序列相应于下标集{1,2,...,m}的一个子集。因此，共有2^m个不同子序列，从而穷举搜索法需要指数时间。 -LCS (longest public sequences) can be simply described as follows : given a sequence of sequences in the sequence is deleted after some element of the sequence. Given two sequences X and Y, Z is another sequence of the X-Y sequence is the sequences, Z is said X and Y series of public sequences. For example, if X = (A, B, C, B, D, B, A), (Y = B, D, C, A, B, A), the sequence (B, C, A) X and Y is a public sequences, it is not X and Y in a public longest sequences. Sequence (B, C, B, A) X and Y is a public sequences, and its length is 4. but it is the X and Y for a public longest sequence, X and Y because no greater than the length of the four sequences in public. The longest public sequences problem is given two sequences X = (x1, x2, ...) xm and Y = (y1, y2, ... yn), to identify X and Y in a
Date : 2008-10-13 Size : 4.63kb User : 汤烈
[Other resource] dir2par DL : 0: ％直接型到并联型的转换％％[C,B,A]=dir2par(b,a) ％C为当b的长度大于a时的多项式部分％B为包含各bk的K乘2维实系数矩阵％A为包含各ak的K乘3维实系数矩阵％b为直接型分子多项式系数％a为直接型分母多项式系数％
Date : 2008-10-13 Size : 875byte User : 吴江华
[Other resource] PCBmake DL : 0: PCB可制造性设计探讨 Discuss on Manufacturable P C B Design PCB可制造性设计探讨-PCB design for manufacturing explore Discuss on Manufacturable P C B. Design for Manufacturability PCB Design
Date : 2008-10-13 Size : 260.36kb User : rf
[Other resource] CMP000-C-Tutorial.tar DL : 0: This section contains a brief introduction to the C language. It is intended as a tutorial on the language, and aims at getting a reader new to C started as quickly as possible. It is certainly not intended as a substitute for any of the numerous textbooks on C. 2. write a recursive function FIB (n) to find out the nth element in theFibanocci sequence number which is 1,1,2,3,5,8,13,21,34,55,…3. write the prefix and postfix form of the following infix expressiona + b – c / d + e * f – g * h / i ^ j4. write a function to count the number of nodes in a binary tr-This section contains a brief introductio n to the C language. It is intended as a tutorial o n the language, and aims at getting a reader new to C started as qu ickly as possible. It is certainly not intended as a substitute for any of the numerous textbook 's on C. 2. write a recursive function FIB (n) to fi nd out the nth element in sequence theFibanocci number which is 1,1,2,3,5,8,13,21,34,55, ... 3. write the prefix and postfix form of the follo wing infix expressiona b-c / d e f * - g * h / i ^ j4. wr ite a function to count the number of nodes in a bi nary tr
Date : 2008-10-13 Size : 21.79kb User : jim
[Other resource] C8051_I2C_E2PROM DL : 0: 摘要：本文件是C8051单片机I2C总线读写测试程序；将跳线器JP6短接，使用外部22.1184MHz晶振. 功能：定义 0 ~ 9 键为数字键， A ~ F 为功能键。按 A 键后，可按0 ~ 9 数字键，从零地址开始存储该键值，并送LED数码管上显示该键值。按 C 键停止后，按 B 键，从零地址开始读取数据值，并送LED数码管上显示。读取速度每秒一次。在按 C 键停止后，可按 A 键重新输入。 -Abstract : This document is C8051 MCU I2C Bus literacy test procedures; will be short jumper for JP6 Next, using external 22.1184 MHz oscillator. function : 0 ~ 9 definition of the key figures Key, A-F function keys. A key by, according to figures from 0 to 9 keys, started from zero addresses stored in the pair, and sent digital LED tube revealed that the pair. Key stop by the C, B, according to bond, starting from zero address read data values, and sent LED digital tubes show. Read one per second speed. By the C bond, after the cessation of A bond may re-enter.
Date : 2008-10-13 Size : 76.55kb User : Lance Yin
[Other resource] Programming.Embedded.Systems.With.C.and.Gnu.Develo DL : 0: Programming Embedded Systems By Michael Barr, Anthony Massa ............................................... Publisher: O Reilly Pub Date: October 01, 2006 ISBN: 0-596-00983-6 Pages: 304 -Embedded Systems Programming By Michael B. arr. Anthony Massa ...................... ................... Publisher : O Reilly Pub Date : October 01, 2006 ISBN : 0-596-00983-6 Pages : 304
Date : 2008-10-13 Size : 540.63kb User : 刘健华
[Other resource] C++classicgood1 DL : 0: 《C++ 程式語言經典本》，The C++ Programming Language, 3rd edition 中譯本的序、第一章、第二章、第三章、附錄Ｂ的電子檔此處採用的是由華康科技所開發的 DynaDoc 格式。內附DynaDoc 閱覽器-"C programming language for the classic," The C Programming Language, 3rd edition of the injunction order, the first chapter, Chapter II, Chapter III, Appendix B of the electronic file, adopted here by DynaComware DynaDoc developed by the format. Keeping containing DynaDoc
Date : 2008-10-13 Size : 3.03mb User : 李艷
[Other resource] 1800Decorations DL : 0: 该源码是一个问题的解决方法。问题是给你个长为L的串，串中可以出现n种字符，还给出m个子串，求有多少个长为n的只由这些字串组成的串。输入例子：4 5 6 ABB BCA BCD CAB CDD DDA 结果为2.而5 4 5 E D C B A的结果为625-source of the problem is a solution. The problem is that you have the head of the L series, the series can occur n characters, but also gives m tall Series, for the number of ministers to the n only by those strings of string. Input examples : 4 5 6 ABB BCA BCD CAB CDD DDA results of two. and 5 4 5 E D C B A to the results of the six 25
Date : 2008-10-13 Size : 7.91kb User : 姚秋林
[Other resource] s3c44b0ledxianshi DL : 0: /*** *** *** *** *** *** **/ //**此映射表用来映射LED模块不译码时，显示的字符和必须输入的数据的关系 //**每段和对应比特位的关系见示意图 // g // --- --- // b | a |f | | <---显示0时点亮的段为gfedcb // --- // c | |e | | 那么写入数据为0x7e // --- --- // d // bit: 7 6 5 4 3 2 1 0 // 段位： g f e d c b a -/*** *** *** *** *** ***** *** *** *** *** *** *** ***/ / / ** mapping table to map LED module does not decoding, show characters and must import the data / / ** each and the corresponding bits see the relationship between the matrix / / g / / --- --- / / b | a | f
Date : 2008-10-13 Size : 194.01kb User : 简其平
[Other resource] C++Builder-Mideva DL : 0: Ma tla b 是当今使用最为广泛的数学软件，它具有相当强大的数值计算、数据处理、系统分析、图形显示、甚至符号运算功能，是一个完整的数学平台，但是它不能实现端口操作和实时控制Borland C + +Builder是一种新颖的可视化编程语言，可方便实现交互界面、数据采集和端口操作等，但是它在数值处理分析和算法工具等方面，效率远远低于Matlab语言-Ma tla b is the most widely used mathematical software, it is quite powerful numerical calculation, data processing, analysis, graphics, or even symbolic operation, is a complete mathematical platform However, it can not be achieved port operation and real-time control Borland C Builder is an innovative visualization programming language, will facilitate the realization of interactive interface, data acquisition and the port operation, However, in numerical analysis tools and algorithms, the efficiency is far below the Matlab
Date : 2008-10-13 Size : 62.35kb User : lijing
[Other resource] 24hoursC++b DL : 0: 《24学时精通c++》的随书源码的下半部分。欢迎下载学习。-"24 school proficiency c" as the source book to the second half. Welcome to download the study.
Date : 2008-10-13 Size : 75.55kb User : Legend
[Other resource] OReilly.Programming.Embedded.Systems.With.C.and.Gn DL : 0: Programming Embedded Systems By Michael Barr, Anthony Massa Publisher: O Reilly Pub Date: October 01, 2006 ISBN: 0-596-00983-6 Pages: 304 -Embedded Systems Programming By Michael B. arr, Anthony Massa Publisher : O Reilly Pub Date : October 01, 2006 ISBN : 0-596-00983-6 Pages : 304
Date : 2008-10-13 Size : 1.61mb User : mp
[Other resource] C++Primer4 DL : 0: 本书是久负盛名的C++经典教程，其内容是C++大师Stanley B. Lippman丰富的实践经验和C++标准委员会原负责人Josée Lajoie对C++标准深入理解的完美结合，已经帮助全球无数程序员学会了C++。本版对前一版进行了彻底的修订，内容经过了重新组织，更加入了C++先驱Barbara E.Moo在C++教学方面的真知灼见。既显著改善了可读性，又充分体现了C++语言的最新进展和当前的业界最佳实践。书中不但新增大量教学辅助内容，用于强调重要的知识点，提醒常见的错误，推荐优秀的编程实践，给出使用提示，还包含大量来自实战的示例和习题。　　　　对C++基本概念和技术全面而且权威的阐述，对现代C++编程风格的强调，使本书成为C++初学者的最佳指南；对于中高级程序员，本书也是不可或缺的参考书。
Date : 2008-10-13 Size : 1.58mb User : iceman
[Other resource] C++view1-7 DL : 0: 网络上最牛B的关于C编程的杂志，由于种种原因该杂志已经停刊了，所以这是绝版。-network most of the cattle B C Programming magazine, due to various reasons the magazine had folded. So it is out of print.
Date : 2008-10-13 Size : 3.35mb User : 王龙
[Other resource] C-PROGRAM DL : 0: 一、课程设计题目、内容、要求题目：《学生成绩管理系统》(第二套难度：2) 内容：按要求完成的基础上，增加了输入学号查询学生记录的功能，和错误输入报错功能要求： 1）输入将本班学生的信息（每个学生至少包括学号、姓名、科目成绩（三门）、学分、平均分）。 A．要求编写函数sort对个学生按某个条件（比如某科成绩，学号，学分）升序或降序（可选择）排序，并将结果（包括名次、学号、姓名、成绩、学分、平均分）显示出来。 B．编写函数find，用折半法查找某科某个成绩，并将结果打印出来。 C．要求编写函数fun找出所有及格的同学，并将他们的信息存显示出来。 -a curriculum design topic, and requested topic : "Student Performance Management System" (second difficulty : 2) : completed as required, on the basis of increase in the importation of Science inquiry student record function, input errors and mistakes in the functional requirements : 1) importation of Bumbuna student information (including each school students at least, names, subjects results (3), credits, average). A. Prepared function sort of a student by a certain condition (such as certain subjects, learning, credits) ascending or descending (optional) sequencing, and the results (including the rank of school, names, achievements, credits, average) is displayed. B. Prepared function find, with a half Finding out a success, and the results printed. C. Prepared function fun to
Date : 2008-10-13 Size : 3.78kb User : Kyle

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