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[Other resourcecfd2

Description: 关于计算流体力学中方腔环流问题的求解粘性条件下-on computational fluid dynamics China cavity circulation problem solving viscous conditions
Platform: | Size: 1167 | Author: 和昆 | Hits:

[transportation applicationscfd2

Description: 关于计算流体力学中方腔环流问题的求解粘性条件下-on computational fluid dynamics China cavity circulation problem solving viscous conditions
Platform: | Size: 1024 | Author: 和昆 | Hits:

[Data structsRYC

Description: 问题描述:设计一个程序实现两个任意长的整数的求和运算。 基本要求:利用双向循环链表,设计一个实现任意长的整数进行加法运算的演示程序。要求输入和输出每四位一组,组间用逗号隔开。如:1,0000,0000,0000,0000。 -Problem Description: Design a program to achieve the two arbitrary long integer summation operator. Basic requirements: the use of two-way circulation list, the design of a realization of arbitrary long integer addition operations for the demo program. Input and output requirements for each of four groups of groups separated by commas. Such as: 1,0000,0000,0000,0000.
Platform: | Size: 603136 | Author: 迟浩东 | Hits:

[Windows Developyue

Description: 本程序针对约瑟夫问题的描述:编号为12,……n的n个人按顺时针方向围成一圈,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序时针方向自1开始顺序报数,报道m时停止报数,报m的人处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。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。 演示程序以用户和计算机的对话方式执行,即在计算机终端上显示“提示信息”下,用户输入循环人数n和每人的密码。-This procedure description specifically for Joseph problem: The serial number is 12, ... The n n individual surrounds the ready-made one circle according to the clockwise sense , each person holds (positive integer) a password, one positive integer beginning to assume a post of selections is upper limits ms numbering off, direction starts the hour hand beginning from the first people according to order from 1 numbering off in proper order , stop numbering off when reporting m, the person place column reporting m looks on his password as new m value , the individual begins to number off again from 1 , getting down like this from his time as soon as going ahead in clockwise sense, until proprietor all till leaving self s place in the ranks. Make use of one-way circulation chained list memory structure to simulate this process, according to that the order leaving self s place in the ranks prints out each person s serial number. Demonstrate the procedure interactive mode implementation wit
Platform: | Size: 1024 | Author: 张蔷 | Hits:

[Software Engineeringchengxushejishiyan4

Description: 一、实验题目: 实验4 程序的形式化推导 二、实验内容: (1) 根据给定的问题,建立前置谓词和后置谓词。 (2) 根据条件语句的推导策略,建立程序中的条件语句。 (3) 构造循环不变式,确立界函数。 (4) 根据循环语句的推导策略,建立程序中的循环语句。 (5) 面向目标,退到出所要的程序。 三、实验目的 根据条件语句的推导策略和循环语句的推导策略,基于最弱前置条件的概念,用比较形式化的方法,从设计要求达到的目标一步步推导出程序。 四、实验要求: 运用C++,编写一个程序实现N的阶乘。 运用循环不变式设计。 -First, the experimental subjects: Formal test procedures derived 4 Second, experimental elements: (1) According to a given problem, the establishment of pre-and post-predicate predicate. (2) Derivation of conditional statements in accordance with the strategy, the establishment of procedures for the conditional statements. (3) loop invariant structure, the establishment of community function. (4) According to the derived loop strategy, the establishment of procedures for the circulation statement. (5) goal-oriented, back to the process. Third, the experiment was Derivation of conditional statements in accordance with the strategy and the derived loop strategy, based on the concept of the weakest pre-conditions, a more formal way, from design requirements to achieve a step-by-step procedures derived. Fourth, the experimental requirements: The use of C++, the preparation of a program to achieve the factorial N. Design of the use of loop invariants.
Platform: | Size: 8192 | Author: 沈炯 | Hits:

[Program docCpracticereport

Description: C++实习报告:包括源代码、需求分析、解题说明、调试报告等,主要内容是:循环和分支,数组和函数,指针、引用和异常处理,结构体和类,派生、继承和多态,文件和模板及综合等7大类型。 -C++ internship report: including source code, needs analysis, problem-solving that debug report, the main contents are: circulation and branch, array and function, pointer, reference, and exception handling, structures and categories, derivative, inheritance and polymorphism, documents and templates, and integrated, such as type 7.
Platform: | Size: 281600 | Author: elsa | Hits:

[VHDL-FPGA-Verilogkey

Description: Verilog HDL编写的键盘扫描程序,考虑了判断按键弹起的问题。程序按一定的频率用低电平循环扫描行线,同时检测列线的状态,一旦判断有一列为低则表示有键被按下,停止扫描并保持当前行线的状态,再读取列线的状态从而得到当前按键的键码;等待按键弹起:检测到各列线都变成高点平后,重新开始扫描过程,等待下一次按键。-Written in Verilog HDL keyboard scanner, taking into account to determine key bounce problem. Program according to a certain frequency of scan lines with low-level circulation lines, while testing out the state line, once the judge has said there is a classified as low-key is pressed, stop the scan and to maintain the current line-line state, and then read out line state to get the current keys key codes to wait for key pop-up: To detect the lines at all out into a high level after the re-start the scanning process, waiting for the next key.
Platform: | Size: 2201600 | Author: 力文 | Hits:

[Data structs33

Description: 实验目的:熟练掌握字符串处理以及循环控 问题描述 实现移位密码:k表示26个字母向前循环位移的字符个数 加密: 输入:k,以及小写表示的明文字母,输出:大写字母表示的密文 解密:输入:k,以及大写字母表示的密文,输出:小写字母表示的明文 -Experimental Objective: To master the string processing and circulation control description of the problem to achieve the shift Password: k said that 26 letters forward in the number of encryption cycles displacement of characters: Input: k, as well as the lowercase letters that express the output: capital letters expressed ciphertext decryption: Input: k, and uppercase letters of the ciphertext, output: lowercase letters of the plaintext
Platform: | Size: 7168 | Author: 吴海 | Hits:

[Compress-Decompress algrithms2

Description: 2. 用循环链表解Josephus问题。设有n个人围坐在一个圆桌周围,现从第1个人开始报数,数到第m的人出列,然后从出列的下一个人重新开始报数,数到第m的人又出列,…,如此反复直到所有的人全部出列为止。Josephus问题是:对于任意给定的n和m,求出按出列次序得到的n个人员的序列-2. Josephus solution of the problem with the circulation list. With n individuals sitting around a round table is started by a report from the first number, the number of people out to the first m columns, and then from the next person out of the column began to count off again, several to the first m out of those people who have column, ..., and so forth until all of the columns were all out of date. Josephus The question is: For any given n and m, find out the columns according to the order received sequence of n-persons
Platform: | Size: 1024 | Author: sprite | Hits:

[Web ServerWEBOA

Description: web保存word文件,在公文流转中很有用处,这些都是测试代码。没有任何问题-web to save word documents, in the circulation of useful documents, these are test code. There is no problem
Platform: | Size: 744448 | Author: henanlzy | Hits:

[Other2-5-2

Description: 使用一般循环单链表(设立指向表位节点的指针)解决约瑟夫问题的算法-Single linked list using general circulation (the establishment of the node pointer points to the table spaces) to solve the problem of the algorithm Joseph
Platform: | Size: 2048 | Author: 安全卫士 | Hits:

[Data structsJosephus_CircList

Description: 用循环链表求解约瑟夫问题,采用C++语言,内含详细注释,个人认为用数据结构方式解决约瑟夫问题比采用数组更容易理解-Joseph solving the problem with the circulation list, using C++ language, containing detailed comments, personally think that the data structure used to resolve the problem Joseph easier to understand than using an array
Platform: | Size: 1010688 | Author: weiw | Hits:

[Data structsxz

Description: c++用循环链表解决约瑟夫问题课程设计代码-c++ problem with the circulation list address curriculum design code Joseph
Platform: | Size: 3072 | Author: 火炬 | Hits:

[Data structsMoney100Chicken100

Description: 历史上最简单的百钱百鸡问题求解程序。采用单层次循环,至多循环五次就能够搜索到百钱百鸡问题的4个可行答案。-The easiest way to solve the probelm of "100 money and 100 chickens". Single-level circulation, circulation up to five times the money to be able to find one hundred one hundred chickens four possible answers to the problem.
Platform: | Size: 1024 | Author: 张新红 | Hits:

[Data structsKNAPSACK

Description: 0-1背包问题,用动态规划做的,很不错,供大家传阅。算法设计的原型。我给具体实现了。-0-1 knapsack problem, using dynamic programming to do, very good for your circulation. Algorithm design prototype. I gave a concrete realization.
Platform: | Size: 586752 | Author: 尚启昌 | Hits:

[Algorithmdaxiezhuanhuan(VCPPP6.0)

Description: 一个将阿拉伯数字转大写汉字的命令行下的程序,可以将-99999999~99999999范围内的阿拉伯数字转大写汉字。自己测试过无数次,所以输出结果基本上是没有问题的。只是程序连续执行时如果输入的不是数字程序会进入死循环,一直没找到原因,希望有了解的高手能指点下。-Wrote a number of Chinese characters turn Arab capital under the command line program that will-99999999 ~ 99999999 within the scope of the Arabic Numbers turn capital characters. His numerous times test, so the output is basically no problem. Just for the execution of the program if the input is not digital program will go to dead circulation, didn t find a reason, hope has the understanding of the superior
Platform: | Size: 246784 | Author: 赵云 | Hits:

[Other Riddle gamesA-wise-man-live

Description: 问题的提出:17世纪法国数学家加斯帕的一本《数学的游戏问题》描述了许多有趣问题,约瑟问题就是其中之一: 15名基督教徒和15名异教徒同乘一船航行,途中风浪大作,危机万分,领航者告诉大家,只要将全船的一半人投入海中,其余人就能幸免。大家都同意这个办法,并协定者30人围成一圈;由第一个人起报数,每数至第9人便把他投入海中,下一个接着从1开始报数,第9人又被投入海中,依次循环,直至剩下15人为止。问题是如何排法使投入海中的人全为异教徒? 编程的思路:此问题有人把它的答案隐示于下列詩句中: From numbers’ aid and art, (依靠数字的和魔力,) Never will fame depart! (声誉决不会离你而去!) 这句诗中元音字母依次为:oueaiaaeeiaeea,分别用1、2、3、4、5代替a、e、i、o、u,便得一排数(划线得数字表示基督教徒,不划线的表示异教徒)。 4 5 2 1 3 1 1 2 2 3 1 2 2 1 故所求排法诗4个基督教徒,5个异教徒,再2个基督教徒,1个异教徒,…可以证明,这种排法所有的基督教徒都是安全的。 -The problem put forward: 17 th century French mathematician gaspar s book "the game of mathematics problems describes many interesting questions, Joseph is one of them: 15 Christian and pagan 15 with a ship sailing by, on the wind was blowing and crisis extremely, pilot told everyone, as long as the whole ship half of them into the sea, and the rest will survive. We all agree with this method, and the agreement is 30 a circle By the first man up to number off, each number to 9 people gave him into the sea, and the next then from 1 began to sound off, who was put into the sea and 9, in circulation, until 15 people left so far. The question is how to row in the sea to input method for all pagan? Programming ideas: this was the answer to the question it concealed in the following lines: From Numbers aid and art, (depending on the number and the magic,) Never will fame depart! (reputation will never leave you!) This poem of vowels in order as: oueaiaaeeiaeea, respectively for 1, 2,
Platform: | Size: 11264 | Author: jim | Hits:

[AlgorithmAlgorithm-analysis-and-design

Description: 计算机实验室 内容: 1、分治法,maxmin算法 2、动态规划,矩阵连乘 3、贪心法, 1)背包问题,2)装载问题 4、回溯法,N皇后问题的循环结构算法和递归结构算法。-Computer laboratory content: 1, partition method, maxmin algorithm 2, the dynamic planning, LianCheng matrix 3, greedy method, 1) knapsack problem, 2) the loading problem 4, back in the method, the circulation of the queen s problems N structure algorithm and recursive structure algorithm.
Platform: | Size: 4096 | Author: 王越瑾 | Hits:

[Database systemvfp-wuliu

Description: 随着计算机技术的广泛应用,在现代流通企业管理中引入计算机管理技术,成为一个值得深入研究的问题。 本文首先概要的论述了数据库的有关知识与现状。之后,对当代计算机数据库技术的现状,尤其是对关系型数据库作了系统的描述。在此基础上,论文重点对数据库的开发环境 — Microsoft Visual FoxPro 6.0 作了详细的介绍,然后,对开发一个小型的现代流通企业综合管理系统所用到的语句、函数、对象进行了详细的阐述。最后,结合自己编写的现代流通企业综合管理系统,给出了完整的、可实际使用的数据库管理系统。并给出了该系统的使用方法和设计方法。 -With the extensive application of computer technology, the modern circulation enterprise management, the introduction of computer technology become a problem worthy of further study. This paper first discusses the knowledge of the database with the status quo. After that, the status quo of contemporary computer database technology, especially relational database systematic description. On this basis, the paper focuses on database development environment- Microsoft Visual FoxPro 6.0 is introduced in detail, then, statements, functions, objects used in the development of a small modern circulation integrated enterprise management system detail. Finally, the combination of modern circulation integrated enterprise management system, write your own complete, the actual use of the database management system. And given the use of the system and the design method.
Platform: | Size: 968704 | Author: liuyan | Hits:

[ELanguageLinkTable

Description: 数据结构中线性链表,及循环链表的应用,这是对约瑟夫问题的求解。-Linear linked list data structure, and the application of the circulation list,For Joseph problem solving.
Platform: | Size: 5120 | Author: 李婉婷 | Hits:
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