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  • I want to provide an example file system driver for Windows NT/2000/XP. For some time I have worked

    I want to provide an example file system driver for Windows NT/2000/XP. For some time I have worked on an implementation of RomFs. RomFs is a small filesystem originally implemented in Linux, because of its simple disk layout its a good choice for an example driver. The current status is a working read-only driver that supports caching of file data, the create functionallity still needs some work but I m releasing it due to the high public demand.

    标签: provide Windows example driver

    上传时间: 2013-12-19

    上传用户:zsjzc

  • 一个基于GTK+的单词数值计算器

    一个基于GTK+的单词数值计算器,1、 按照规则计算单词的值,如果 A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 26个字母(全部用大写)的值分别为 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26,如: WINJACK这个单词的值就为:W+I+N+J+A+C+K=23+9+14+1+3+11=71% HARDWORK=H+A+R+D+W+O+R+D=8+1+18+4+23+15+18+11=98% LOVE=L+O+V+E=12+15+22+5=54% LUCK=L+U+C+K=12+21+3+11=47% ATTITUDE= A+T+T+I+T+U+D+E=1+20+20+9+20+24+4+5=100% 2、对程序的界面布局参考如下图所示,在第一个单行文本框输入一个单词,点击“计算”按钮,按照以上算法计算出该单词的值。 3、如果在最下面的单行文本框输入一个文件路径,此文件每行记录一个单词,那么经过程序计算出各个单词的值,并把结果输出到当前目录下result.txt文件中。如果文件不存在,应该提示错误。

    标签: GTK 数值 计算器

    上传时间: 2014-01-11

    上传用户:康郎

  • Neural Networks at your Fingertips.rar =============== Network: Adaline Network ==============

    Neural Networks at your Fingertips.rar =============== Network: Adaline Network =============== Application: Pattern Recognition Classification of Digits 0-9 Author: Karsten Kutza Date: 15.4.96 Reference: B. Widrow, M.E. Hoff Adaptive Switching Circuits 1960 IRE WESCON Convention Record, IRE, New York, NY, pp. 96-104, 1960

    标签: Network Fingertips Networks Adaline

    上传时间: 2014-12-22

    上传用户:lizhizheng88

  • Instead of finding the longest common subsequence, let us try to determine the length of the LCS.

    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

  • The files included implement the M3 Enigma machine. Rotors.m implements the GUI and enigma.m impleme

    The files included implement the M3 Enigma machine. Rotors.m implements the GUI and enigma.m implements the cipher substitutions. The implementation can be used as an aid to teaching the enigma machine. Also, the source can be used to show how to implement an enigma machine. The files are thoroughly commented and the application can run in MATLAB by just typing rotors

    标签: implements implement the included

    上传时间: 2017-02-24

    上传用户:lixinxiang

  • sbit CS = P1^0 sbit SCK = P1^6 sbit SIN = P1^7 sbit SOUT = P1^5 sbit WP = P1^1

    sbit CS = P1^0 sbit SCK = P1^6 sbit SIN = P1^7 sbit SOUT = P1^5 sbit WP = P1^1 void XReady(void) void XSendByte(unsigned char b) unsigned char XGetByte(void) void XWriteEn(void) void XWriteDis(void) unsigned char XReadStatus(void) void XWipPoll(void) void XWriteStatus(unsigned char b) void XReadData(unsigned char *m,unsigned char x,unsigned char n) void XWriteData(unsigned char *m,unsigned char x,unsigned char n) #define XResetDog() {CS=0 CS=1 } void InitX5045(void)

    标签: sbit SOUT SCK SIN

    上传时间: 2014-01-17

    上传用户:lijinchuan

  • nearest neighbour clustering is an algoithm that puts data in categories. This algotithm is a data m

    nearest neighbour clustering is an algoithm that puts data in categories. This algotithm is a data mining subject. Similar data is put in the same cluster.

    标签: data clustering categories algotithm

    上传时间: 2017-03-27

    上传用户:181992417

  • 为 了提高用户身份认证和授权管理的灵活性

    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

    标签: 用户 授权 身份认证

    上传时间: 2013-12-31

    上传用户:VRMMO

  • 课程设计: 1.求出在一个n×n的棋盘上

    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

    标签:

    上传时间: 2017-04-24

    上传用户:zhyiroy

  • Maya Calendar During his last sabbatical, professor M. A. Ya made a surprising discovery about the

    Maya Calendar During his last sabbatical, professor M. A. Ya made a surprising discovery about the old Maya calendar. From an old knotted message, professor discovered that the Maya civilization used a 365 day long year, called Haab, which had 19 months. Each of the first 18 months was 20 days long, and the names of the months were pop, no, zip, zotz, tzec, xul, yoxkin, mol, chen, yax, zac, ceh, mac, kankin, muan, pax, koyab, cumhu. Instead of having names, the days of the months were denoted by numbers starting from 0 to 19. The last month of Haab was called uayet and had 5 days denoted by numbers 0, 1, 2, 3, 4. The Maya believed that this month was unlucky, the court of justice was not in session, the trade stopped, people did not even sweep the floor.

    标签: A. M. sabbatical surprising

    上传时间: 2014-01-05

    上传用户:libenshu01