猴子选大王，有N只猴子，从1～N进行编号。它们按照编号的顺时针方向，排成一个圆圈，然后从第一只猴子开始报数。第一只猴子报1，以后每只猴子报的数字都是它前面猴子所报数字加1。如果一只猴子报的数字是M，则该猴子出列，下一只猴子重新从1开始报数。剩下的猴子继续排成一个圆圈报数，直到全部的猴子都出列为止。最后一个出列的猴子胜出。按照上述规则，编写程序选出排成一圈的猴子中的大王。
Monkey King, N monkeys, numbered from 1 to N. They are in accordance with the number of clockwise, arranged in a circle, and then start off from the first monkey. The first monkey reported 1, and then each monkey reported the number of monkeys in front of it, plus 1. If a monkey's numbers is M, the monkey out of a monkey from 1 to start off again. The rest of the monkey to row into a circle off, until all the monkeys are out of date. The last one out of the monkey. According to the above rules, the program is written to select the king of monkeys in a circle. (2017-11-20, Visual C++, 1KB, 下载1次)
约瑟夫斯（Josephus）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，从他在顺时针方向下一个人开始重新从1报数，如此下去，直至所有的人全部出列为止。试设计一个程序，按出列顺序印出各人编号。
利用单向循环链表存储结构模拟此过程，按照出列的顺序印出各人的编号。
Josephus (Josephus) a description of the problem is: Numbers for 1, 2,..., n n individual clockwise around a circle, each holding a password (positive integer).At first choose a positive integer as the upper limit count off m, since the first man since 1 count off, clockwise when reporting for m stop count off.To dequeue m, his password as the new m values, in a clockwise direction him the next person started to count off 1, so go on, until all the people all dequeue.Try to design a program to print out according to the order number. (2016-03-24, Visual C++, 1KB, 下载1次)
约瑟夫斯（Josephus）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，从他在顺时针方向下一个人开始重新从1报数，如此下去，直至所有的人全部出列为止。试设计一个程序，按出列顺序印出各人编号。
Josephus (Josephus) a description of the problem is: Numbers for 1, 2,..., n n individual clockwise around a circle, each holding a password (positive integer).At first choose a positive integer as the upper limit count off m, since the first man since 1 count off, clockwise when reporting for m stop count off.To dequeue m, his password as the new m values, in a clockwise direction him the next person started to count off 1, so go on, until all the people all dequeue.Try to design a program to print out according to the order number. (2016-03-24, Visual C++, 1KB, 下载2次)
Dijkstra(迪杰斯特拉)算法是典型的单源最短路径算法，用于计算一个节点到其他所有节点的最短路径。主要特点是以起始点为中心向外层层扩展，直到扩展到终点为止。Dijkstra算法是很有代表性的最短路径算法，在很多专业课程中都作为基本内容有详细的介绍，如数据结构，图论，运筹学等等。注意该算法要求图中不存在负权边。
Dijkstra (Dijkstra) algorithm is a typical single source shortest path algorithm for computing a node to all other nodes of the shortest path. The main feature is the starting point for the expansion of the center outward, until the end of the extension to the end. Dijkstra algorithm is very representative of the shortest path algorithm, in many professional courses as the basic content of a detailed introduction, such as data structure, graph theory, operations research, etc.. Note that there is no negative edge in the graph. (2016-03-22, Visual C++, 6KB, 下载1次)
约瑟夫斯（Josephus）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，从他在顺时针方向下一个人开始重新从1报数，如此下去，直至所有的人全部出列为止。
A description of Josephus (Josephus) question is: numbered 1,2, ..., n n individuals clockwise direction around a circle, each holding a password (positive integer). Choose a positive integer beginning as reported upper limit m, began a clockwise direction from the first person to stop self-reported the number 1 to start off, report m. M out of newspaper columns, his password as the new value of m, from his start in the clockwise direction from the next person to re-report the number 1, and so on, until all the people all of the columns so far. (2014-05-14, Visual C++, 1KB, 下载3次)
用一个线性表模拟电子词典的使用，线性表中每个数据元素记录一个单词及其使用次数（频率）。为了提高查找速度，要求将经常使用的单词存放在线性表的前部。第一次使用的单词(还不在电子词典中)插在线性表的表尾，使用频率为1；以后每使用(查找)1次，其使用频率加1，并根据其频率将该单词前移到线性表的相应位置，使线性表中的单词一直以频率为关键字递减排列，频率相同的根据查找的先后顺序排列
Table with a linear analog electronic dictionary use, linear table for each data element to record a word and its frequency of use (frequency). In order to improve search speed required to store frequently used words in the front line of the table. The first use of the word (not in electronic dictionary) footer linear interpolation table, using a frequency of 1 after every use (to find) 1, its frequency by 1, and the word forward according to their frequency linear form to the appropriate location, so that the linear form of words has been descending frequency of keywords, to find the same frequency arranged according to the order (2014-04-23, Visual C++, 1KB, 下载4次)
1)编写一个程序，接受用户输入的一个字符串，统计其中出现过的所有字符，按其出现频率从高到低的顺序排列输出。(注：请使用这样的方法编写程序，使用一个链表存放出现过的字符，链表的每个结点的数据域中有两个变量，一个存放该字符，一个存放字符出现的次数。每检索到原字符串的一个字符时，先在链表中查找，若找到此字符，则将其出现次数加l，否则创建一个新的结点链接在链表的末尾。)
2)对已经建立好的链表进行如下处理：删除出现次数小于3的字符节点，并输出删除之后的链表信息。
1) Write a program that accepts user input a string, statistics which appeared in all the characters, their frequency of occurrence in descending order of output. (Note: the use of such method of preparation, the use of a stored list of characters appeared, the linked list data field of each node has two variables, a storage of the character, a number of characters stored in each retrieved original string of one character, the first in the list to find, if it finds this character, it is the number of occurrences plus l, or create a new node in the list at the end of the link.) 2) have already been carried out to establish a good list handled as follows: Delete occurrences of characters is less than 3 nodes, and outputs the list after deletion of information. (2013-07-30, Visual C++, 1KB, 下载1次)
Dijkstra(迪杰斯特拉)算法是典型的单源最短路径算法，用于计算一个节点到其他所有节点的最短路径。本算法时间复杂度完全符合要求。
Dijkstra (Dijkstra) algorithm is a typical single-source shortest path algorithm, is used to calculate a node to all other nodes in the shortest path. The main features is the starting point as the center outward expansion layers until the extension to the end point. Dijkstra algorithm is very representative of the shortest path algorithm, the basic content as described in detail in many professional courses, such as data structures, graph theory, operations research, and so on. Dijkstra s general statements are generally two ways, one permanent and temporary label, one is using the OPEN, CLOSE table, there are permanent and temporary label. Note that the algorithm requires that the figure does not exist in the negative right side. (2013-05-12, Visual C++, 1KB, 下载2次)
针对带附加头结点的单链表，试编写下列函数。
（1）定位函数Locate:在单链表中寻找第i个结点。若找到，则函数返回第i个结点的地址；若找不到，则函数返回NULL。
（2）求最大值函数max：通过一趟遍历在单链表中确定值最大的结点。
（3）统计函数number：统计单链表中具有给定值x的所有元素。
（4）建立函数create：根据一维数组a[n]建立一个单链表，使单链表中各元素的次序与a[n]中各元素的次序相同，要求该程序的时间复杂性为O(n)。
Singly linked list with additional head node, try to write the following function.
(1) orientation function Locate is: in a single linked list to find the i-th node. If found, the function returns the i-node address if no, then the function returns NULL.
(2) the maximum function max: traversing through a trip to determine the value of the largest node in a singly linked list.
(3) The number of statistical functions: statistical singly linked list with all the elements of a given value of x.
(4) the establishment of the create function: one-dimensional array a [n] to establish a single linked list, singly linked list in the order of the elements a [n] is the order of the elements of the same requirements of the program s time complexity is O (n). (2012-04-06, Visual C++, 1KB, 下载12次)
1.编写一个类，声明一个数据成员和一个静态数据成员，其构造函数初始化数据成员，并把静态数据成员加1，其析构函数把静态数据成员减1。
(1) 编写一个应用程序，创建该类的3个对象，然后显示其数据成员和静态数据成员，再析构每个对象，并显示它们对静态数据成员的影响。
(2) 修改该类，增加静态成员函数并访问静态数据成员，并声明静态数据成员为保护成员。体会静态成员函数的使用，静态成员之间与非静态成员之间互访问题。
1 write a class, declare a data member, and a static data member, its constructor initializes data members, and to add a static data member, its destructor decrements the static data member.
(1) preparation of an application, create a class of three objects, then displays its data members and static data members, then each object destructor, and shows their impact on the static data member.
(2) to modify the class, add a static member functions and access static data members and static data members declared as protected members. Understand the use of static member functions, static members and non-static members of the exchange of visits between the problem (2011-12-14, Visual C++, 3KB, 下载3次)
25.学生成绩管理系统
现有学生成绩信息文件1.txt，内容如下
姓名 学号 语文 数学 英语
张明明 01 67 78 82
李成友 02 78 91 88
张辉灿 03 68 82 56
王露 04 56 45 77
陈东明 05 67 38 47
学生成绩信息文件2.txt,内容如下:
姓名 学号 语文 数学 英语
陈果 31 57 68 82
李华明 32 88 90 68
张明东 33 48 42 56
李明国 34 50 45 87
陈道亮 35 47 58 77
试编写一管理系统,要求如下:
1)实现对两个文件数据进行合并,生成新文件3.txt
2)抽取出三科成绩中有补考的学生并保存在一个新文件4.txt
3)合并后的文件3.txt中的数据按总分降序排序(至少采用两种排序方法实现)
4)输入一个学生姓名后,能查找到此学生的信息并输出结果(至少采用两种查找方法实现)
5)要求使用结构体,链或数组等实现上述要求.
6)采用多种方法且算法正确者,可适当加分.
xue sheng guan li xi tong (2011-11-30, Visual C++, 2KB, 下载20次)
约瑟夫生者死者游戏的大意是：30个旅客同乘一条船，因为严重超载，加上风高浪大，危险万分；因此船长告诉乘客，只有将全船一半的旅客投入海中，其余人才能幸免遇难。无奈，大家只得同意这种办法，并议定30个人围成一圈，由第一个人开始，依次报数，数到第9人，便把他投入大海中，然后从他的下一个人数起，数到第9人，再将他投入大海，如此循环，直到剩下15个乘客为止。问哪些位置是将被扔下大海的位置。
Joseph living dead game to the effect that: 30 passengers with by boat, because of severe overload, coupled with high winds and waves, extremely dangerous so the captain told the passengers, only the whole ship half of the passengers into the sea, the rest can be survived. Helpless, we had to agree with this approach, and 30 people agreed to form a circle, starting from the first person, in turn reported the number, the number to nine, put him into the sea, and from a number of his next play, the number of to 9, and then he put into the sea, and so on, until the remaining 15 passengers so far. Q. What position is the position of the sea will be dropped. (2011-11-29, Visual C++, 3KB, 下载3次)
约瑟夫斯（Josephus）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，从他在顺时针方向下一个人 开始重新从1报数，如此下去，直至所有的人全部出列为止。试设计一个程序，按出列顺序印出各人编号。
Josephus (Josephus) a description of the problem is: numbered 1,2, ..., n, n individuals sitting around a circle clockwise, each holding a password (positive integer). Choose the beginning of a positive integer number as reported on the limit m, starting from the first person in a clockwise direction starting from a report number, report m, the number of stops reported. Reported m the person out of the column, his password as the new value of m, in a clockwise direction from the next person he began to re-reported from a number, it goes on, until all the people all of the columns so far. Design a program, according to the column order prints each number. (2011-11-04, Visual C++, 1KB, 下载2次)
(1)已知一棵完全二叉树存放于一个一维数组T[n]中，T[n]中存放的是各结点的值。试设计一个算法，从T[0]开始顺序读出各结点的值，建立该二叉树的二叉链表表示。
(2)二叉树的双序遍历(Double-order traversal)是指：对于二叉树的每一个结点来说，先访问这个结点，再按双序遍历它的左子树，然后再一次访问这个结点，接下来按双序遍历它的右子树。试写出执行这种双序遍历的算法。
(1) known to a complete binary tree stored in a one-dimensional array T [n] in, T [n] is stored in the value of each node. Try to design an algorithm, from T [0] started in order to read out the value of each node to build the binary tree binary list said. (2) two-order binary tree traversal (Double-order traversal) is: For each binary tree node, the first visit to this node, and then double-traversing its left subtree, and then visit this node, followed by two-order traversal of its right subtree. Write down the implementation of this two-order traversal algorithm. (2010-11-27, Visual C++, 1KB, 下载9次)
圣诞节就要到了，为了答谢新老客户的厚爱，位于普罗旺斯的Snow & Leaf咖啡屋给北航的师生邮寄来一个巨大圣诞树，上面挂满了各种各样的礼物。不同于别的圣诞树的是，上面的礼物都是一对一对的，每一对相同礼物都是用一根缘分之绳连起来的，而且对与对之间不重复。更令人兴奋的是，它长得还是特别像二叉树。
礼物人人都想拿，所以报名的人太多了。由于大家都是程序设计的高手，Snow & Leaf的老板娘就说，我当初准备缘分之绳花了很久的时间，如果你们能算出总共有多长，就可以在上面挑选一个礼物，并和选择了对应礼物的人，一起免费普罗旺斯一周游。
二叉树用广义表给出，礼物节点用’A’-‘Z’的字母表示，保证节点有偶数个，缘分之绳都沿着二叉树的边，每一个父子之间边长为1。
输入第一行为整数n，表示数据的组数。接下来的n行，每行一个字符串a，是用广义表表示的二叉树。
输出有n行，每行一个整数length，表示缘分之绳的总长度。
Christmas is coming, to thank the love of old and new customers located in the Snow & Leaf Cafe Provence to Northern students and teachers
Mail to a huge Christmas tree, hung above a variety of gifts. Unlike other Christmas tree is a gift above all
Is a one to one right, each pair of the same gifts are connected with a fate of the rope together, but not between right and on repeat. Even more
Exciting, it looks like a binary tree or as a special.
Everyone went to get a gift, so too many people signed up. Because we are a master of programming, Snow & Leaf s boss
Say, the fate of the rope I had prepared for a long time to spend, and if you can figure out how long in total, to be selected in the above
A gift, and gifts, and select the corresponding person, one week with free tour of Provence.
Generalized table shows the binary tree, gifts node with A - Z the letter said, to ensure that there is an even number of nodes, along all the strings of fate
The binary tree edges, (2010-11-14, Visual C++, 2KB, 下载6次)
Mark Allen Weiss
1987年在普林斯顿大学获得计算机科学博士学位,师从著名算法大师Robert Sedgewick,现任美国佛罗里达国际大学计算与信息科学学院教授.他曾经担任全美AP(Advanced Placement)考试计算机学科委员会的主席(2000-2004).他的主要研究方向是数据结构,算法,和教育学.
In this second edition of his successful book, experienced teacher and author Mark Allen Weiss continues to refine and enhance his innovative approach to algorithms and data structures. Written for the advanced data structures course, this text highlights theoretical topics like abstract data types and the efficiency of algorithms, as well as performance and running time. This... (展开全部) In this second edition of his successful book, experienced teacher and author Mark Allen Weiss continues to refine and enhance his innovative approach to algorithms and data structures. Written for the advanced data structures course, this text highlights theoretical topics like abstract data types and the efficiency of algorithms, as well as performance and running time. This edition also includes a new chapter on advanced data structures and material on the Standard Template Library that conforms to the new standard. In addition, all code has been updated and tested on multiple platforms and confo (2010-07-21, Visual C++, 65KB, 下载2次)
稀疏矩阵的运算程序。该设计程序的功能是由键盘输入稀疏矩阵中非零值的行标，列标和元素值构成的三元组顺序表，这样就确定了整个矩阵，再使两个矩阵以三元组表形式求和、差、乘积并将运行的结果输出，本程序主要用到的知识是以顺序表形式存储稀疏矩阵，再用链表对其进行操作，来通过三元组判断位置、非零值运算实现稀疏矩阵的加、减、乘等运算。
Sparse matrix computation program. The design process is the function of keyboard input sparse matrix by a non-zero value for the Industry Standard, the value of the column constitutes goals and elements of the order of the table triples, thus determining the entire matrix, and then to the two matrices to form the table triples sum, difference, product and running the resulting output, the main use of this procedure is based on the order form of knowledge stored sparse matrix, and then manipulate the list to determine location through the triple, non-zero value of computing to achieve sparse matrix addition, subtraction, multiplication and other operations. (2010-05-10, Visual C++, 114KB, 下载4次)
八皇后问题是一个古老而著名的问题，是回溯算法的典型例题。该问题是十九世纪著名的数学家高斯1850年提出：在8X8格的国际象棋上摆放八个皇后，使其不能互相攻击，即任意两个皇后都不能处于同一行、同一列或同一斜线上，问有多少种摆法。可以利用递归方法分别一一测试每一种摆法，直到得出正确的答案。
要求：
1、在处理每个题目时，要求从分析题目的需求入手，按设计抽象数据类型、构思算法、通过设计实现抽象数据类型、编制上机程序和上机调试等若干步骤完成题目，最终写出完整的分析报告。前期准备工作完备与否直接影响到后序上机调试工作的效率。在程序设计阶段应尽量利用已有的标准函数，加大代码的重用率。
2、.设计的题目要求达到一定工作量（300行以上代码），并具有一定的深度和难度。
3、程序设计语言推荐使用C/C++，程序书写规范，源程序需加必要的注释
4、每位同学需提交可独立运行的程序；
5 、每位同学需独立提交设计报告书（每人一份），要求编排格式统一、规范、内容充实，不少于10页（代码不算）；
6、课程设计实践作为培养学生动手能力的一种手段，单独考核。
err (2007-12-19, Visual C++, 16KB, 下载17次)
课程安排，用拓扑排序实现 4、实现课程的拓扑排序。(选)（加）问题描述：软件专业的学生要学习一系列课程，其中有些课程必须在其先修课程完成后才能学习，具体关系见下表：课程编号 课程名称 先决条件 C1 程序设计基础 无 C2 离散数学 C1 C3 数据结构 C1，C2 C4 汇编语言 C1 C5 操作系统 C3 假设每门课程的学习时间为一学期，试为该专业的学生设计教学计划，使
curriculum, and topological sorting achieve four to achieve courses topological sorting. (Excerpts) (Canada) Problem Description : software professional students to study a series of courses Some of the programs must, in its prevocational courses can be completed after the study, the specific relationship between the table below : Course No. 295 C1 prerequisite for the design process without foundation Discrete Mathematics C1 C2 C3 data structure C1, C2 C1 C4 C5 assembly language operating system C3 assumes that each course of study for a semester, test students for the professional design of the teaching plan, (2007-02-23, Visual C++, 188KB, 下载164次)
课程安排，用拓扑排序实现
4、实现课程的拓扑排序。(选)（加）
问题描述：软件专业的学生要学习一系列课程，其中有些课程必须在其先修课程完成后才能学习，具体关系见下表：
课程编号 课程名称 先决条件
C1 程序设计基础 无
C2 离散数学 C1
C3 数据结构 C1，C2
C4 汇编语言 C1
C5 操作系统 C3
假设每门课程的学习时间为一学期，试为该专业的学生设计教学计划，使他们能在最短的时间内修完这些课程。
curriculum, using topological sorting achieve four to achieve the topological sorting courses. (EAC) (Canada) Problem description : software professional students to study a series of courses Some of the programs must, in its prevocational courses can be completed after the study, the specific relationship between the table below : Curriculum prerequisite for the design procedures based on C1-C2 C3 Discrete Mathematics C1 C1 data structure, C2 C1 C4 C5 assembly language operating system C3 assumptions each course of study for a semester, Examination for Professional Teaching students design plans so that they can in the shortest time got these courses. (2006-06-02, Visual C++, 249KB, 下载49次)