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Description: palm编成,这种书很少,有兴趣看看
Title: Palm Programming: The Developer s Guide
URL: http://safari.oreilly.com/JVXSL.asp?x=1&mode=section&sortKey=rank&sortOrder=desc&view=book&xmlid=1-56592-525-4&open=false&srchText=palm+programming&code=&h=&m=&l=1&catid=&s=1&b=1&f=1&t=1&c=1&u=1&page=0
ISBN: 1-56592-525-4
Author: Julie McKeehan/ Neil Rhodes
Publisher: O Reilly
Page: 478
Edition: 1st edition (December 1998)
Catalog: PDA programming / Palm
Format: pdf
Size: 2.06M
Supplier:
Summary: Emerging as the bestselling hand-held computers of all time, PalmPilots have spawned intense developer activity and a fanatical following. Used by Palm in their developer training, this tutorial-style book shows intermediate to experienced C programmers how to build a Palm application from the ground up. Includes a CD-ROM with source code and third-party developer tools-palm establishment, and this little book, is interested Title : Palm Programming : The Developer's Guide URL : http://safari.oreilly.com/JVXSL.asp x = 1
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Size: 1563571 |
Author: xiaojun |
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Description: American Gladiator,You are consulting for a game show in which n contestants are pitted against n gladiators in order to see which contestants are the best. The game show aims to rank the contestants in order of strength this is done via a series of 1-on-1 matches between contestants and gladiators. If the contestant is stronger than the gladiator, then the contestant wins the match otherwise, the gladiator wins the match. If the contestant and gladiator have equal strength, then they are “perfect equals” and a tie is declared. We assume that each contestant is the perfect equal of exactly one gladiator, and each gladiator is the perfect equal of exactly one contestant. However, as the gladiators sometimes change from one show to another, we do not know the ordering of strength among the gladiators. -American Gladiator. You are consulting for a game show in which n cont estants are pitted against n gladiators in orde r to see which contestants are the best. The game show aims to rank the contestants in order of str KV this is done via a series of 1-on-1 matches between contestants and gladiators. If the con testant is stronger than the gladiator. then the contestant wins the match otherwise, the gladiator wins the match. If the contestant and gladiator have equal strength. then they are "perfect equals" and a tie is decla red. We assume that each contestant is the perfe ct equal of exactly one gladiator. and each gladiator is the perfect exact equal of ly one contestant. However, as the gladiators sometimes change from one sho w to another, we do not know the ordering of stre
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Size: 1858 |
Author: zhoushusen |
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Description: 通过该程序可创建一个8元素数组，可对数组进行排序-adoption of the procedures will help to create an 8 element array, the array of rank
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Size: 2947 |
Author: 白学飞 |
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Description: 《学生成绩管理系统》的设计与实现
每一条记录包括一个学生的学号、姓名、3门成绩、平均成绩。
2、输入功能：可以一次完成无数条记录的输入。
3、显示功能：完成全部学生记录的显示。
4、查找功能：完成按姓名查找学生记录，并显示。
5、排序功能：按学生平均成绩进行排序。
6、插入功能：按平均成绩高低插入一条学生记录。
7、保存功能：将学生记录保存在任何自定义的文件中，如保存在：c:\\score。
8、读取功能：将保存在文件中的学生记录读取出来。
9、有一个清晰美观界面来调用各个功能-"student performance management system," Design and Implementation of every one, including a record of the number of students, name, three results, average scores. 2, the input feature : can a complete record of numerous input. 3, the display function : all students complete records show. 4, the search function : the completion of student records identifying the names and show. 5, sorting functions : average scores by students rank. 6, insert functions : according to the average level of achievement students to insert a record. 7, the depositary functions : student record keeping, since the definition of any document, such as the preservation of : c : \\ score. 8, read function : the preservation of the documents, the records of students read out. 9, there is a clear aesthetic int
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Size: 60243 |
Author: 小宇 |
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Description: 《学生成绩管理系统》的设计与实现
1、每一条记录包括一个学生的学号、姓名、3门成绩、平均成绩。
2、输入功能：可以一次完成无数条记录的输入。
3、显示功能：完成全部学生记录的显示。
4、查找功能：完成按姓名查找学生记录，并显示。
5、排序功能：按学生平均成绩进行排序。
6、插入功能：按平均成绩高低插入一条学生记录。
7、保存功能：将学生记录保存在任何自定义的文件中，如保存在：c:\\score。
8、读取功能：将保存在文件中的学生记录读取出来。
9、有一个清晰美观界面来调用各个功能-"student performance management system," Design and Implementation of an every one, including a record of the number of students, name, three results, average scores. 2, the input feature : can a complete record of numerous input. 3, the display function : all students complete records show. 4, the search function : the completion of student records identifying the names and show. 5, sorting functions : average scores by students rank. 6, insert functions : according to the average level of achievement students to insert a record. 7, the depositary functions : student record keeping, since the definition of any document, such as the preservation of : c : \\ score. 8, read function : the preservation of the documents, the records of students read out. 9, there is a clear aesthetic
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Size: 59310 |
Author: 小宇 |
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Description: 关联规则3
Rank-correlated set mining (Download)
This implementation is based on the rank-correlated set mining technique for numerical attributes as described in the paper \"Mining rank-correlated sets of numerical attributes\" (Calders, Goethals, Jaroszewicz, 2006).
-association rules three Rank-correlated set mining (Downlo ad) This implementation is based on the rank-co rrelated set numerical technique for mining at tributes as described in the paper "Mining rank - correlated sets of numerical attributes "(Ca lders, Goethals, Jaroszewicz, 2006).
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Size: 23050 |
Author: lion |
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Description: Performs Principal Component Analysis cross validation to determine the true rank of an X-block.-Performs Principal Component Analysis cr oss validation to determine the true rank of an X - block.
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Size: 1734 |
Author: 李楠 |
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Description: 排序及应用
题目一：统计成绩
1． 实验目的：掌握常用的排序方法，深刻理解排序的定义和各种排序方法的特点，并能灵活应用。
2． 实验内容：
（1） 输入n个学生的考试成绩，每个学生信息由姓名与分数组成；
（2）选择一种排序算法，按分数高低次序，打印出每个学生的名次，分数相同的为同一名次。
（3）按名次给出每个学生的姓名与分数。
3． 实验要求：任选一种高级程序语言编写源程序，并调试通过，测试正确。
题目二：按字典顺序排序
1． 实验目的：掌握常用的排序方法，深刻理解排序的定义和各种排序方法的特点，并能灵活应用到字符串排序中。
2． 实验内容
（1）输入若干个国家名称（所有的名称均用大写或小写表示）
（2）选择一种排序算法，将输入的国家进行排序；
（3）按字母顺序输出国家名称。
3． 实验要求：任选一种高级程序语言编写源程序，并调试通过，测试正确。
-ranking and a title application : a statistical results. Experimental Objective : To understand the sort commonly used methods, a profound understanding of the definition and sequencing various sort of characteristics, and can be applied flexibly. 2. Experiment : (1) importation n their examination grades, each student names and information from scores of composition; (2) choose a sorting algorithm, according to the scores they order, print out each student's scores, the same fraction of the same rank. (3) the scores of each student is given the names and scores. 3. Experimental requirements : a high-optional procedure language source files, and debugging through the test correctly. Two topics : according to a ranking order of the dictionary. Experimental Objective : To understand the
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Size: 10755 |
Author: jason |
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Description: 全国计算机等级考试一级WINDOWS 的模拟考试软件源代码 通过这个源码可以模拟等级考试-National Computer Rank Examination of a mock examinations Windows software source code through this source can be simulated grading examinations
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Size: 66207 |
Author: 孟丹 |
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Description: 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
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Size: 3873 |
Author: Kyle |
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Description: 1、任务：为宿舍管理人员编写一个宿舍管理查询软件, 程序设计要求：
（1）采用交互工作方式
（2）可以增加、删除、修改信息
（3）建立数据文件 ，数据文件按关键字（姓名、学号、房号）进行排序(用选择排序)
2、查询菜单: (用二分查找实现以下操作)
（1）按姓名查询
（2）按学号查询
（3）按房号查询
3、打印任一查询结果（可以连续操作）
-a task : to prepare dormitory management staff inquiries a dormitory management software, program design requirements : (1) interactive work (2) can add, delete, change information (3) Establish data files, data files by keyword (name, school, room number) rank (with choice of ranking) 2, inquiries menu : (2 hours with you to achieve the following operational) (1) the names of inquiries (2) the school inquiry (3) of the inquiry three, Print the results of an inquiry (capable of continuous operation)
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Size: 2095 |
Author: ahui |
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Description: Gabor-function convolution masks are increasingly used in image processing and computer vision. This function simply computes the cosine and sine masks for a given width, period and orientation. The masks returned are properly normalised.
It is not highly optimised - e.g. the symmetries are not used to reduce the computation.
It is recommended that convolutions with these masks are done using CONVOLVE2 (available from Matlab File Exchange) to take advantage of their low rank.
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Size: 1637 |
Author: li |
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Description: 老外的超高效率压缩，High efficient Data Compression Library for use with Delphi.
Support Rank, ZIP, BZIP and PPM compression algorithms.
Compression ratio more than 1.5-2 times better than ZIP/RAR archiver.
Compression speed up to 8 Mb/sec (on PIII-600).
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Size: 317487 |
Author: emirsoft |
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Description: 目前所能找到的VCLSkin控件的源码最新版, 控件功能参见官方网站: http://www.link-rank.com/
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Size: 235589 |
Author: liushinwell |
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Description: 一个输入学生成绩并排名次的程序，教材上的例子程序
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Size: 1332 |
Author: Tom |
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Description: IEEE trans M 的文章，是关于体育视频的，它实现了，对精彩shot的rank，很厉害啊
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Size: 1483606 |
Author: 河南 |
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Description: 从键盘接收输入的一个班的学生成绩，并存放于50字grade数组中，其中grade+i保存学号为i+1的学生成绩。然后根据grade中的学生成绩，把学生名次填入50字rank字数组中，其中 rank+i的内容是学号为i＋1学生的名次。再按学号顺序把名次从终端上显示出来。(输入学生个数<=50，可以不定)
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Size: 4131 |
Author: 婷婷 |
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Description: 逆Bbroyden秩1法解非线性方程组实例-Bbroyden Rank an inverse method for solving systems of nonlinear equations example
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Size: 1151 |
Author: 李 |
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Description: Blind extraction of a cyclostationary signal using reduced-rank cyclic regression—A unifying approach
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Size: 1363472 |
Author: xiaozhang |
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Description: 游戏说明： 牌的大小顺序： 大王，小王,A,K,Q,j,9,8,7,6,5,4,3,2 出牌规则： 每次只能出一张牌，每局开始时随机由其中一人先出牌，以后由上一次赢的玩家出牌。首家可以出任意一张牌。首家出牌后，下家所出的牌的花色要与首家的相同，大小不限。如确实没有花色相同的牌，则可以出其它的牌。 胜负判定： 当每人都出了一张牌后，如果花色相同，则比较两张牌的大小，牌大的赢.如果花色不同，则先出牌的赢。最后经过统计赢的次数最多的 玩家最终赢得比赛。 The rules is following: 1) The Cards RANK (big to small) is: A K Q J 10 9 8 7 6 5 4 3 2 2) Once only can play one Card,if the suit is same and the Rank is big,they win or lost if you no the suit ,you must play any card but your s always small, 3) Now i think you know the rules ! Goog Luck ! printf(\"Please choice your cards Color: 0 means spade 1 means hearts 2 means diamonds 3 means clubs :\\n\") -games : licensing the order : King, Wang, A, K, Q, j, 9,8,7,6,5,4,3,2 out licensing rules : each time out only one card, at the beginning of each Board random by one of the first to be out later last won by the player card. The first to be sent to an arbitrary license. The first card, the home under the license by the first matched with the same open-size. If they really do not matched the same card, can produce other card. The results found : When everyone out of a license, if the same colors, compared to the size of two licenses, the licenses won. If the colors were different, first-out licensing of winning. After winning the final statistics of the number of players up to the final whistle. The rules is following : 1) The Cards RANK (big to small) is : A K Q J 10 9 8 7 6 5 4 3 2 2) Once
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Size: 4912 |
Author: 冰雨 |
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