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于 2020-10-14 发布 文件大小:11604KB
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  中科大数据结构全部上机实验的实验程序以及报告,内容包括多项式,图,栈,SQL语言……(All experimental procedures USTC data structures as well as reports on experiments, including polynomial, maps, stack, SQL language ......)

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  • SHU
    为了点分 求分下载好麻烦 c++二叉排序树(In order to find points dotted download good trouble c++ binary sort tree)
    2013-07-25 01:35:56下载
    积分:1
  • Kruskal
    克鲁斯卡尔算法思想.cpp int seekedge(EDGE * (&a), int n) void bubble(EDGE * (&a), int size) void searchandjoin(EDGE * (&a1), int k) // a1是图边集头指针,a2是最小生成树边集的头指针,k为所要删的边 void storepicture(EDGE * (&a), int n, VERTEX *b, int m) (Thinking of Kruskal algorithm. Cpp int seekedge (EDGE* (& a), int n) void bubble (EDGE* (& a), int size) void searchandjoin (EDGE* (& a1), int k) // a1 edge set is the first pointer map, a2 is the minimum spanning tree of the first indicators margination, k for the edges to be deleted void storepicture (EDGE* (& a), int n, VERTEX* b, int m) )
    2009-06-27 01:43:39下载
    积分:1
  • adms
    阿当姆斯显式和隐式求解方法,用四阶龙格库塔作为起始,然后运用四阶阿当姆斯算法求解初值问题。主要程序包含在test2.cpp中,方法简单易懂。编译环境VC2010(Adam James explicit and implicit method for solving fourth-order Runge-Kutta as a start, then use the fourth-order A Williams algorithm for solving initial value problem. The main program contains in test2.cpp method is simple and easy to understand. Compile environment VC2010)
    2012-06-09 11:27:18下载
    积分:1
  • example-program
    里面有50例经典算法,其中包括水仙花素,兔子的繁殖等古典问题的求解(There are 50 cases of classic algorithms, including daffodils prime, rabbit breeding classical problem solving)
    2013-02-03 11:14:12下载
    积分:1
  • 3.4.6
    单链表中的数据元素含有三种字符(字母、数字、其他字符),算法实现构造三个循环链表,是每个循环链表中只含有同一类字符,且利用原表中的结点空间作为这三个表的结点空间。(Singly linked lists of data element contains three characters (letters, Numbers and other characters), the algorithm constructs three circular linked list, each cycle list contains only the same type of character, and make use of the primary node space in a table as a node of the three table space.)
    2014-04-15 13:11:37下载
    积分:1
  • zhebanchazhao
    实现折半查找,非递归形式,从而实现数据结构所要的结果。(To achieve binary search, the non-recursive form)
    2012-08-29 12:35:25下载
    积分:1
  • ANT
    A*算法通过启发式探索下一结点,逐步逼近目标结点,且使得路径最短。(A* algorithm heuristic to explore the next node, and gradually approaching the target node, and that the shortest path.)
    2012-05-17 09:14:21下载
    积分:1
  • cycQueue
    单循环链表,用C++实现 提供入队,出队,判断空等方法(Single cycle linked list, with C++ into the team, the team judge waited in vain for methods)
    2012-10-08 20:02:15下载
    积分:1
  • PROGRAM
    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)
    2012-03-21 11:28:19下载
    积分:1
  • yuesefu
    数据结构经典算法 约瑟夫问题 c++ 已经测试通过(josephus c++)
    2012-12-02 16:23:02下载
    积分:1
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