登录
首页 » Visual C++ » PROGRAM

PROGRAM

于 2012-03-21 发布 文件大小:32106KB
0 179
下载积分: 1 下载次数: 473

代码说明:

  本书是《程序员面试宝典》的第三版,在保留第二版的数据结构、面向对象、程序设计等主干的基础上,使用各大IT公司及相关企业最新面试题替换和补充原内容,以反映自第一版以来近几年时间所发生的变化。   欧立奇、刘洋、段韬编著的《程序员面试宝典》取材于各大公司面试真题(笔试、口试、电话面试、英语面试,以及逻辑测试和智商测试),详细分析了应聘程序员(含网络、测试等)职位的常见考点。本书不仅对传统的C系语言考点做了详尽解说,还根据外企出题最新特点,新增加了对友元、Static、图形/音频、树、栈、ERP等问题的深入讲解。最后本书着力讲述了如何进行英语面试和电话面试,并对求职中签约、毁约的注意事项及群体面试进行了解析。本书的面试题除了有详细解析和答案外,对相关知识点还有扩展说明。真正做到了由点成线,举一反三,对读者从求职就业到提升计算机专业知识都有显著帮助。   《程序员面试宝典》适合计算机相关专业应届毕业生阅读,也适合作为正在应聘软件行业的相关就业人员和计算机爱好者的参考书。(This book is the" canon" in the third edition, retained in the second version of the data structure, object-oriented program design, main basis, the use of each big IT companies and related enterprises to the latest interview questions substitution and supplement the original content, in order to reflect the since its first edition in recent years since time changes. Ou Liqi, Liu Yang, Duan Tao," canon" drawn from major companies interview questions ( written examination, interview, telephone interview, interview, and logical test and IQ test ), a detailed analysis of the application programmer (including network, testing ) positions of the common points. This book not only to the traditional C language test done a detailed explanation, according to the new characteristics of the new foreign title, added to the friend, Static, graphics/audio, tree, stack, ERP issues in depth explanation. The last book on described how the English interview and telephone interview, and the job of signin)

文件列表:

下载说明:请别用迅雷下载,失败请重下,重下不扣分!

发表评论

0 个回复

  • 1013_digital_root
    数字根算法,用最简单的代码计算数字根。如果把一个大数的各位数字相加得到一个和,再把这个和的各位数字相加又得一个和,再继续作数字和,直到最后的数字和是个位数为止,这最后的数称为最初那个数的“数字根”。(Digital root algorithm, with the most simple code to calculate the digital root. If you figure in a large numbers is the sum of a, then this and you figure the sum is again a, and then continue for digital and until the final figure and the single digits, the last number called initially that the number of "digital root".)
    2012-04-27 10:07:18下载
    积分:1
  • kaikuohaohebikuohaodepipei
    开括号和比括号的匹配,数据结构内容,很不错(More than brackets and brackets to open the match, the contents of data structure)
    2009-04-13 14:35:56下载
    积分:1
  • Tic-tac-toe-chess
    C语言实现井字棋博弈的过程,采用极大极小算法。(The C language game of tic-tac-toe chess, using the minimax algorithm.)
    2012-05-20 23:15:48下载
    积分:1
  • bracketmatching
    这个程序用来解决树结构与树之间的程序匹配的问题这个程序可以解决很多的问题(it is very good)
    2013-01-21 21:48:17下载
    积分:1
  • astar
    A*算法 1、将开始节点放入开放列表(开始节点的F和G值都视为0) 2、重复以下步骤: 在开放列表中查找具有最小F值的节点,并把查找到的节点作为当前节点 把当前节点从开放列表删除, 加入到封闭列表. (A* algorithm 1, will begin to node placed in the and opening up list of (the began to node of the F and G values ​ ​ are regarded as 0) 2, repeat the the following steps: to Find a the has a the the smallest F value of the node in the the and opening up list of, and put Find a to the node as the current node current node is removed from the open list, added to the closed list.)
    2013-04-04 11:06:42下载
    积分:1
  • Maze
    实现迷宫搜索程序。迷宫数据从文件输入。输出迷宫路径。 a) 实现从入口到出口的输出顺序; b) 能否打印行走方向(向左、向右、向前), 如: 1 先前 2 向右 6 向左 7; (Maze searching)
    2011-12-20 20:59:04下载
    积分:1
  • LCS
    the longest common serials
    2011-12-13 00:32:40下载
    积分:1
  • 9
    说明:  任意给定输入的一个小写英文字符串a1a2a3…an-1an (n≥5) 输出:字符串Ana1An-1a2An-2a3…A2an-1A1an,其中Ai为ai的大写形式 例如,输入aybdx,则递归程序输出XaDyBbYdAx (Any given input of a lowercase string a1a2a3 ... an-1an (n ≥ 5) Output: the string Ana1An-1a2An-2a3 A2an-1A1an, which Ai ai uppercase for example, input aybdx recursive program output XaDyBbYdAx)
    2012-05-30 10:59:39下载
    积分:1
  • maze
    通过学习奥赛培训课程,利用回溯加递归的编程技巧完成的迷宫问题(Training courses through learning Orsay, plus use of recursive programming skills back to complete the maze problem)
    2010-09-02 21:43:32下载
    积分:1
  • jgctgu
    流水线车间的零件加工排序问题,以达到最优的加工顺序,使加工时间最短或零件后期维护费用最少(Plant parts processing pipeline scheduling problem, in order to achieve optimal machining sequence, so that the processing time of the shortest or least parts later maintenance costs)
    2014-05-08 01:12:54下载
    积分:1
  • 696518资源总数
  • 106155会员总数
  • 8今日下载