-
With a worldwide community of users and more than a million dedicated programmer...
With a worldwide community of users and more than a million dedicated programmers, Perl has proven to be the most effective language for the latest trends in computing and business. Every programmer must keep up with the latest tools and techniques. This updated version of Advanced Perl Programming from O Reilly gives you the essential knowledge of the modern Perl programmer. Whatever your current level of Perl expertise, this book will help you push your skills to the next level and become a more accomplished programmer. -With a worldwide community of users and more than a million dedicated programmers, Perl has proven to be the most effective language for the latest trends in computing and business. Every programmer must keep up with the latest tools and techniques. This updated version of Advanced Perl Programming from O Reilly gives you the essential knowledge of the modern Perl programmer. Whatever your current level of Perl expertise, this book will help you push your skills to t
- 2022-04-17 03:59:15下载
- 积分：1
-
约瑟夫（Joeph）问题的一种描述是：编号为1,2,…,n的n个人按顺时针方向围坐一圈，每人持有一个密码（正整数）。一开始任选一个正整数作为报数上限值m，从第一...
约瑟夫（Joeph）问题的一种描述是：编号为1,2,…,n的n个人按顺时针方向围坐一圈，每人持有一个密码（正整数）。一开始任选一个正整数作为报数上限值m，从第一个人开始按顺时针方向自1开始顺序报数，报到m时停止报数。报m的人出列，将他的密码作为新的m值，从他在顺时针方向上的下一个人开始重新从1报数，如此下去，直至所有人全部出列为止。试设计一个程序求出出列顺序。-Joseph (Joeph) a description of the problem is: numbered 1,2, ..., n, n individuals sitting clockwise circle, each holding a password (positive integer). An optional start a positive integer as the number of reported upper limit of m, from the first person to embark on the clockwise direction starting from a reported sequence number, to report the number of m stop reported. Those who reported m out of the column, his password as the new m value, from his home in a clockwise direction on the next person started to re-reported from a few, and so on, until all of the column until all the up. Try a program designed to find out the column order.
- 2022-08-14 04:29:17下载
- 积分：1
-
Understanding the design, recommend books, upcoming sentiment design, experience...
感悟设计，好书推荐，即将上市的感悟设计，由实战经验丰富的一线工程师编写。推荐。-Understanding the design, recommend books, upcoming sentiment design, experienced by the actual preparation of the first-line engineers. Recommend.
- 2023-08-12 02:25:03下载
- 积分：1
-
00ic开发板(s3c44b0)的iic测试源代码
00ic开发板(s3c44b0)的iic测试源代码-err
- 2022-03-19 04:48:53下载
- 积分：1
-
即插急用的的设备的配置
即插急用的的设备的配置-194,900 of the portable equipment configuration
- 2022-08-13 07:47:24下载
- 积分：1
-
ADS_错误代码解释
ADS_错误代码解释-ADS_ error code explained
- 2023-01-26 12:30:04下载
- 积分：1
-
DevelopersGuide Delphi Kylix
英文版linux的delphi
DevelopersGuide Delphi Kylix
英文版linux的delphi-DevelopersGuide Delphi Kylix English version of linux delphi
- 2022-04-21 16:22:14下载
- 积分：1
-
这是一个用c语言实现了实现区间数相关操作的程序，区间树具有动态创建和动态调整的特点，具有很多应用...
这是一个用c语言实现了实现区间数相关操作的程序，区间树具有动态创建和动态调整的特点，具有很多应用-This is a language used to achieve the realization of c interval number associated operating procedures, with a dynamic range of the tree to create and adjust the dynamic characteristics, with many applications
- 2022-03-01 13:24:36下载
- 积分：1
-
《AFS用户指南》
Andrew文件系统(AFS)是一个分布式网络文件系统，其能够使来自任何通过这个国家的AFS机器能够在文件一经在本地存储就能访问。AFS...
《AFS用户指南》
Andrew文件系统(AFS)是一个分布式网络文件系统，其能够使来自任何通过这个国家的AFS机器能够在文件一经在本地存储就能访问。AFS是一个稳定的单元，其每个单元代表文件空间中一个独立的管理部分。单元连接在一起组成了一个在根/afs目录下的一个庞大的UNIX文件系统。
本文是供AFS用户使用的指南性文件。-"AFS user guide" Andrew File System (AFS) is a distributed network file system , it can come from any country through the AFS machines to a paper by the local store will be able to visit. AFS is a stable unit, each unit space on behalf of an independent management component. Modules linked to a form of the root/afs directory of a large UNIX file system. This article is for AFS users guide documents.
- 2022-04-13 01:54:17下载
- 积分：1
-
Money to the c++ Programming source code and answers. Attached to books, can be...
钱能的c++程序设计的源代码和答案.
附带图书，可以进行很好的学习-Money to the c++ Programming source code and answers. Attached to books, can be good learning
- 2022-02-28 23:10:25下载
- 积分：1
