大三下学期做的计算机网络的课程设计,文档格式文件,有用的朋友可以下载-junior next semester so the computer network of curriculum design, document format, useful friends can be downloaded
标签： curriculum computer semester network
上传时间： 2013-12-12
上传用户：lwwhust
This document provides guidelines and describes how to easily port S60 2nd Edition C++ applications to S60 3rd Edition. The document has been written based on experiences of porting regular S60 2nd Edition applications, such as the S60 Platform: POP/IMAP Example [4] that can be downloaded from Forum Nokia. Code snippets from the example are shown in Chapter 8, “Application build changes,” and in Appendix A, “Code example." In addition, Appendix B, "Commonly used functions that require capabilities," and Appendix C, "Commonly used interfaces that have been changed or removed," provide useful information on some frequently used functions and interfaces in third-party applications.
标签： application guidelines describes document
上传时间： 2017-01-29
上传用户：wang5829
Instead of finding the longest common subsequence, let us try to determine the length of the LCS. Then tracking back to find the LCS. Consider a1a2…am and b1b2…bn. Case 1: am=bn. The LCS must contain am, we have to find the LCS of a1a2…am-1 and b1b2…bn-1. Case 2: am≠bn. Wehave to find the LCS of a1a2…am-1 and b1b2…bn, and a1a2…am and b b b b1b2…bn-1 Let A = a1 a2 … am and B = b1 b2 … bn Let Li j denote the length of the longest i,g g common subsequence of a1 a2 … ai and b1 b2 … bj. Li,j = Li-1,j-1 + 1 if ai=bj max{ L L } a≠b i-1,j, i,j-1 if ai≠j L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
标签： the subsequence determine Instead
上传时间： 2013-12-17
上传用户：evil
Reads/writes text as a character stream, buffering characters so as to provide for the efficient reading/writing of characters, arrays, and lines. The buffer size may be specified, or the default size may be used. The default is large enough for most purposes. In general, each read request made of a Reader causes a corresponding read request to be made of the underlying character or byte stream. It is therefore advisable to wrap a BufferedReader around any Reader whose read() operations may be costly, such as FileReaders and InputStreamReaders.
标签： characters character buffering efficient
上传时间： 2017-02-20
上传用户：cjl42111
复接入，B/W双用户使用直接扩频序列 ％ >>>multiple access b/w 2 users using DS CDMA ％ >>>format is : cdmamodem(user1,user2,snr_in_dbs) ％ >>>user1 and user2 are vectors and they should be of equal length ％ >>>e.g. user1=[1 0 1 0 1 0 1] , user2=[1 1 0 0 0 1 1],snr_in_dbs=-50 ％ >>>or snr_in_dbs=50 just any number wud do ％ Waqas Mansoor ％ NUST , Pakistan
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上传时间： 2014-11-22
上传用户：zl5712176
先用C-均值聚类算法程序，并用下列数据进行聚类分析。在确认编程正确后，采用蔡云龙书的附录B中表1的Iris数据进行聚类。然后使用近邻法的快速算法找出待分样本X（设X样本的4个分量x1=x2=x3=x4=6；子集数l=3）的最近邻节点和3-近邻节点及X与它们之间的距离。
上传时间： 2014-01-23
上传用户：frank1234
This is the procedure for lab 1. This is a two-week lab. Prelab should be done BEFORE going to the lab session. In this lab the dielectric properties of materials and how these properties affect electric fields will be analized.
上传时间： 2013-12-23
上传用户：003030
为 了提高用户身份认证和授权管理的灵活性，从We b 应用系统的安全性角度出发，讨论了 一 种在． N E T F r a me w o r k下保证应用程序安全性的身份验证和授权模型，并给出了模型的具体实现方法。 该模型利用 F o r ms身份验证方法对用户的身份进行鉴别。在授权处理上，模型结合统一资源定位( u ． J f o r m R e s o u r c e L o c a t o r , U R L ) 授权模式和用户所具有的系统角色，分别从页面级和页面操作级对用户的访问进行 控制。该模型在企业局域网环境内能够提供比较灵活的身份认证和基于角色的授权服务。实际应用表明， 基于该模型的We b应用系统能够对用户的访问进行有效的控制，从而保证了系统的安全性
上传时间： 2013-12-31
上传用户：VRMMO
if the method is to be made void, then it cannot have a return statement, all it can do is set the value. Your SeatReserved variable is already a global variable, so what you have in the last post will work, it is known as a set or setter method. All it does is set a value, usually these are done accross classes.
标签： the statement cannot method
上传时间： 2013-12-22
上传用户：xaijhqx
课程设计： 1.求出在一个n×n的棋盘上，放置n个不能互相捕捉的国际象棋“皇后”的所有布局。 2.设计一个利用哈夫曼算法的编码和译码系统，重复地显示并处理以下项目，直到选择退出为止。 【基本要求】 1) 将权值数据存放在数据文件(文件名为data.txt，位于执行程序的当前目录中) 2) 分别采用动态和静态存储结构 3) 初始化：键盘输入字符集大小n、n个字符和n个权值，建立哈夫曼树； 4) 编码：利用建好的哈夫曼树生成哈夫曼编码； 5) 输出编码； 6) 设字符集及频度如下表： 字符 空格 A B C D E F G H I J K L M 频度 186 64 13 22 32 103 21 15 47 57 1 5 32 20 字符 N O P Q R S T U V W X Y Z 频度 57 63 15 1 48 51 80 23 8 18 1 16 1
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上传时间： 2017-04-24
上传用户：zhyiroy
