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[Windows Develop] 234156789097 DL : 1: 商业单位需要容易记忆的电话号码，有一些方法可以让电话号码变得更容易记忆。譬如，可以把电话号码写成单词或短语，如 TUT-GLOP 可以代表滑铁卢大学的电话。有时仅仅是把号码的一部分写成单词，如打 310-GINO 便可向 GINO 比萨饼店定购比萨。另一种让电话号码容易记忆的方法是将数字用一种容易记的方式组合起来，譬如 3-10-10-10 也可以代表 GINO 比萨饼店。电话号码的标准形式是七位十进制数字，在它的第三位和第四位之间用连字符连接(例如：888-1200)。电话的键盘提供了字符于数字之间的映射关系，如下所示： 2 A、B和C 3 D、E和F 4 G、H和I 5 J、K和L 6 M、N和O 7 P、R和S 8 T、U和V 9 W、X和Y Q 和 Z 没有映射到键盘，而连字符不需要被拨打并且可以根据需要添加和删除。TUT-GLOP 的标准形式是 888-4567，310-GINO 的标准形式是310-4466，3-10-10-10的标准形式也是 310-1010。如果两个电话号码有相同的标准形式，那么这两个电话号码是相同的。-commercial units need easy to remember telephone numbers, there are ways to allow more telephone numbers easy to remember. For example, telephone numbers can be written words or phrases, such as pseudodynamic - GLOP can represent the University of Waterloo calls. Sometimes numbers is only part of the written word. if the fight 310-GINO will be able to GINO pizza shop ordering pizza. Another phone number so easy to remember the method is to use a number of easy to remember the combination. For example 3-10-10-10 can also represent GINO pizza shop. The telephone number is the standard form seven decimal digits. in its third and fourth between connecting with characters (for example 888-320-4234-1200). Telephone keyboard characters in the digital mapping between, as follows : 2 A, B and C, D,
Date : 2008-10-13 Size : 925byte User : 张日天
[Windows Develop] luatest DL : 0: Lua游戏脚本，现在很多的公司都用着一种，这里仅是简单的说明lua与c++的使用-Lua game scripts, many companies have used a, where only a simple description of lua and c++ Use
Date : 2025-12-27 Size : 55kb User : 王方石
[Windows Develop] DOMCount DL : 0: 自己开发的XML DOM计数程序DOMCount，使用DOM API解析XML文件，构造DOM树，然后遍历DOM树并计算元素个数（仅使用一个API调用）。 DOMCount解析XML文件，计算其元素数并打印。用法： DOMCount [options]<XML file|List file> Options: -l,指示输入文件是一个列表文件，其中含有多个xml文件。默认情况下，该选项不存在，这时表明输入的是一个XML文件。 -v=xxx，Schema验证，xxx可以是：always，never或auto。默认情况下xxx=auto，如果DOCTYPE声明或schema声明出现在XML文档中，将进行验证：xxx=always，强制进行验证：xxx=never，不使用验证。 -n，允许处理名字空间。默认情况下不允许。 -s，允许处理大纲。默认情况下不允许。 -f，允许对整个大纲约束检查。默认情况下不允许。 -locale=ll_CC，指定本地编码，默认是en_US。 -?，显示帮助。这里是DOMCount的输出： DOMCount -v=always personal.xml personal.xml: 20 ms (37 elems) 注意，程序运行时间将会视你机器处理器的不同而不同。-err
Date : 2025-12-27 Size : 1.38mb User :
[Windows Develop] tinyxml_2_5_3 DL : 0: c++ xml parser lib(opensource)-c++ xml parser lib (opensource)
Date : 2025-12-27 Size : 237kb User : jdqxlin
[Windows Develop] log2 DL : 0: boost-log，一个牛人为boost库写的C++的日志库。boost作为一个庞大的C++基础库，一直缺少日志模块，改代码填补了这个空白，目前还在审核阶段-boost-log, a cow artificially boost library of C++ Writing the log database. a huge boost as the C++ foundation libraries, has been the lack of log module, change the code to fill the gaps in the current audit period is also
Date : 2025-12-27 Size : 980kb User : weihy
[Windows Develop] GamePhysicsEngineDevelopmentCode DL : 0: (C++) Game Physics Engine Development
Date : 2025-12-27 Size : 2.19mb User : llc
[Windows Develop] mushroom DL : 0: /* The sample demonstrates how to build a decision tree for classifying mushrooms. It uses the sample base agaricus-lepiota.data from UCI Repository, here is the link: Newman, D.J. & Hettich, S. & Blake, C.L. & Merz, C.J. (1998). UCI Repository of machine learning databases [http://www.ics.uci.edu/~mlearn/MLRepository.html]. Irvine, CA: University of California, Department of Information and Computer Science. */ // loads the mushroom database, which is a text file, containing // one training sample per row, all the input variables and the output variable are categorical, // the values are encoded by characters.
Date : 2025-12-27 Size : 4kb User : tofighi
[Windows Develop] C DL : 0: 实现一个简单加解密系统，要求如下： (1)输入要加密的原文，长度不小于30个字母； (2)输入密钥长度L及用来产生密钥的英文字母(个数>=L)；注：一个有效的密钥由L(3<=L<=15)个小写英文字母组成，至少有一个元音字母( a , e , i , o 或者 u )，至少有两个辅音字母(除去元音以外的字母)。 (3)选择加密功能，可对原文进行加密； (4)选择解密功能，可对密文进行解密。 -Implement a simple encryption and decryption system requirements are as follows: (1) Enter the encrypted text, the length is not less than 30 letters (2) Enter the key length L and used to generate the key in English letters (number> = L ) Note: a valid key from the L (3 < = L < = 15) composed of a lower-case letters, at least one vowel letter ( ' a' , ' e' , ' i' , ' o' or ' u' ), at least two consonant letters (other than the removal of vowel letters). (3) Select encryption can encrypt the original text (4) Select decryption functions, can decrypt the ciphertext.
Date : 2025-12-27 Size : 1kb User : gqw
[Windows Develop] FFTtest DL : 0: 1. BCB建立lib，在命令提示字元下執行: implib -a fftw3-3.lib libfftw3-3.dll implib -a fftw3f-3.lib libfftw3f-3.dll implib -a fftw3l-3.lib libfftw3l-3.dll VC則 2.將 *.dll 複製到 c:\windows\system32 3.將fftw*.lib 加入專案 4.專案#include "fftw3.h"-FFT
Date : 2025-12-27 Size : 1.44mb User : chin
[Windows Develop] 123456 DL : 0: 模拟实现单机目录FAT的文件系统基本思路：用二进制文件空间模拟磁盘空间，用文件块操作模拟磁盘块操作。基本设计要求：1、实现如下文件系统功能（过程或函数）： a、打开文件系统 FILE *OPENSYS(char *filename) b、关闭文件系统 int CLOSESYS(FILE *stream) c、显示目录 void LISTDIR(void) d、建立文件 int FCREATE(char *filename) e、删除文件 int FDELETE(char *filename) f、打开文件 int FOPEN(char *filename) g、关闭文件 int FCLOSE(int fileid) h、文件块读 int FREAD(void *ptr, int n, int fileid) i、文件块写 int FWRITE(void *ptr, int n, int fileid) j、判断文件结束 int FEOF(int fileid) k、获取文件指针 long FGETPOS(int fileid) l、设置文件指针 int FSETPOS(int fileid, long offset) m、取得文件长度 long FGETLEN(char *filename) 2、提供文件系统创建程序 3、有功能检测模块 4、为简化程序设计，假定目录区域大小固定。 -Simulation to achieve stand-alone catalog FAT file system
Date : 2025-12-27 Size : 15kb User : 黎民
[Windows Develop] ls-l DL : 0: It is a simple program in C, that doing "ls -l" command.
Date : 2025-12-27 Size : 1kb User : Marek
[Windows Develop] example DL : 0: 基于flex&bison(yacc)的四则运算器的例子，附带mylex.l与myyacc.y，文件输入在exprTest.txt中进行-Based on the flex&bison (yacc) four examples of logic, and with mylex. L and myyacc. Y, file input in exprTest. TXT
Date : 2025-12-27 Size : 210kb User : Alex
[Windows Develop] N(i-k)-NURBS(I-K-J-L) DL : 1: 这个是c++求解B样条基函数的过程及代码，还有用B样条基函数求非均匀有理B样条函数上的点的过程的例子-This is the c++ B-spline basis functions for solving the process and code, as well as with the B-spline basis function evaluation of non-uniform rational B spline points on the example of the process
Date : 2025-12-27 Size : 142kb User : 徐志洋
[Windows Develop] l DL : 0: 介绍c++中的类与对象，为PPT格式，希望对你有帮助-c++ class
Date : 2025-12-27 Size : 1.07mb User : 杨
[Windows Develop] tcl_tk_programming DL : 0: t c l / t k的真正功能在于，利用t c l脚本语言几乎完全可以编写复杂的图形应用程序，因而避开了利用C语言编写界面时所遇到的界面编程的许多复杂性。-tcl/tk is the real function, use tcl scripting language can be written almost entirely complex graphics application, and thus avoid the use of C language interface, the interface encountered by many of the complexity of programming.
Date : 2025-12-27 Size : 901kb User : gaohuinew
[Windows Develop] GINA_DL-L DL : 0: C++自动登录资料,我没有用过，我也不知道好用不，反正就是试一下。-C++ automatic login information
Date : 2025-12-27 Size : 82kb User : 54t45
[Windows Develop] PATA1053 DL : 0: 给定一个带有根R的非空树，并赋给每个树节点TI的权重WI。从R到L的路径的重量被定义为从R到任何叶节点L的路径上所有节点的权重之和。(Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L. Now given any weighted tree, you are supposed to find all the paths with their weights equal to a given number. For example, let's consider the tree showed in Figure 1: for each node, the upper number is the node ID which is a two-digit number, and the lower number is the weight of that node. Suppose that the given number is 24, then there exists 4 different paths which have the same given weight: {10 5 2 7}, {10 4 10}, {10 3 3 6 2} and {10 3 3 6 2}, which correspond to the red edges in Figure 1.)
Date : 2025-12-27 Size : 3.03mb User : lld
[Windows Develop] Fwd%3a_RE%3a_INDP3_Groupe_AST_2017_-_2018 DL : 0: platforme iot pour gerer l 'absence
Date : 2025-12-27 Size : 765kb User : TMIM
[Windows Develop] EdnaConsoleApplication DL : 0: c#l连接eDNA数据库并且获取eDNA数据库中实时数据和历史数据(C# gets eDNA real time data)
Date : 2025-12-27 Size : 55kb User : 大道打
[Windows Develop] 相机拍照读取视频 DL : 0: 利用Halcon联合C#在电脑里拍照，可拍视频和照片(Use Halcon and C Chen to take photos in the computer)
Date : 2025-12-27 Size : 726kb User : 大家伙一起吃饭

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