飞思卡尔智能车 车 模 介 绍
车模调校的主要参数
舵机控制方法
Freescale (2008-05-22, Visual C++, 1494KB, 下载195次)
校园导游咨询：编制一个为来访客人进行最短路径导游的程序从厦门大学漳州校区的平面图中选取10个有代表性的景点，抽象成一个无向带权图。以图中顶点表示景点，边上的权值表示两地之间的距离。
2.本程序的目的是为用户提供路径咨询。根据用户指定的始点和终点输出相应路径，或者根据用户指定的景点输出景点的信息。
Campus tour guides Advisory: the preparation of a visitors guide to the shortest path procedure from Zhangzhou Campus of Xiamen University (2008-03-30, Visual C++, 218KB, 下载39次)
本压缩包包含：
数据结构实验报告-校运动会-飞机票定票系统-猴子选大王-huffuman编码等源代码，欢迎参考！
the compression packages include : Experimental data structure report- school sports events- counting system set airfares- election monkey king-source coding huffuman code reference welcome! (2007-04-11, Visual C++, 20KB, 下载20次)
电梯模拟，模拟某校九层教学楼的电梯系统。该楼有一个自动电梯，能在每层停留。九个楼层由下至上依次称为地下层、第一层、第二层、……第八层，其中第一层是大楼的进出层，即是电梯的“本垒层”，电梯“空闲”时，将来到该层候命。
乘客可随机地进出于任何层。对每个人来说，他有一个能容忍的最长等待时间，一旦等候电梯时间过长，他将放弃。
模拟时钟从0开始，时间单位为0.1秒。人和电梯的各种动作均要消耗一定的时间单位（简记为t），比如：有人进出时，电梯每隔40t测试一次，若无人进出，则关门；关门和开门各需要20t；每个人进出电梯均需要25t；如果电梯在某层静止时间超过300t，则驶回1层侯命。
elevator simulation,Simulation of a nine-school teaching building elevator systems. This building has an automatic elevator in each stay. Followed by the bottom nine floors as the basement, first floor, second floor, ... ... the eighth floor, where the first layer is the layer out of the building, that is the elevator "home base layer," Elevator "free" when came to the floor stand.
Passengers can randomly access any layer. For everyone, he can tolerate a maximum waiting time, waiting for the elevator when the time is too long, he will give up.
Analog clock starts from 0, the time unit is 0.1 seconds. And lift a variety of actions have to consume a certain unit of time (abbreviated as t), for example: it was out of the elevator 40t test every time, without the movement of people, then closed close and open the door of the need to 20t each personal and out of elevators require 25t if the elevator stationary at a level more than 300t, then drove back a layer ready to go. (2011-08-12, Visual C++, 68KB, 下载13次)
把数据结构里面的字符串类实现了,可以方便大家用来学习。
袗斜歇 搂 袨薪袧褋袡袡袪褟校褑袝肖蟹卸袟褯袛袔袪褉歇袝携卸小蝎, 袩些胁写袟袧袘褍袛褤袦胁谐校袪袛斜袊携袗衼袃 (2007-09-17, Visual C++, 545KB, 下载12次)
ACM 2012多校暑假集训（7），标程+解题报告。
2012 Multi-University Training Contest- Contest 7 Tutorial. (2012-11-16, C/C++, 252KB, 下载12次)
模拟某校九层教学楼的电梯系统。该楼有一个自动电梯，能在每层停留。九个楼层由下至上依次称为地下层、第一层、第二层、……第八层，其中第一层是大楼的进出层，即是电梯的“本垒层”，电梯“空闲”时，将来到该层候命。
乘客可随机地进出于任何层。对每个人来说，他有一个能容忍的最长等待时间，一旦等候电梯时间过长，他将放弃。
模拟时钟从0开始，时间单位为0.1秒。人和电梯的各种动作均要消耗一定的时间单位（简记为t），比如：有人进出时，电梯每隔40t测试一次，若无人进出，则关门；关门和开门各需要20t；每个人进出电梯均需要25t；如果电梯在某层静止时间超过300t，则驶回1层侯命。
要求：
按时序显示系统状态的变化过程，即发生的全部人和电梯的动作序列。
Simulation of a nine-school teaching building elevator systems. This building has an automatic elevator in each stay. Followed by the bottom nine floors as the basement, first floor, second floor, ... ... the eighth floor, where the first layer is the layer out of the building, that is the elevator "home base layer," Elevator "free" when came to the floor stand.
Passengers can randomly access any layer. For everyone, he can tolerate a maximum waiting time, waiting for the elevator when the time is too long, he will give up.
Analog clock starts from 0, the time unit is 0.1 seconds. And lift a variety of actions have to consume a certain unit of time (abbreviated as t), for example: it was out of the elevator 40t test every time, without the movement of people, then closed close and open the door of the need to 20t each personal and out of elevators require 25t if the elevator stationary at a level more than 300t, then drove back a layer ready to go.
Requirements:
Timing display s (2011-06-20, Visual C++, 223KB, 下载10次)
学校校历生成，按月分割，自动识别闰年，计算星期。
School school calendar generation, monthly partitions, automatic identification of a leap year, calculate week. (2008-05-10, Visual C++, 231KB, 下载9次)
数据结构与算法课程”讲义及复习重点
名校不错的考研资料
Data Structures and Algorithms Course (2007-08-12, Visual C++, 10780KB, 下载7次)
校园导航系统包含佛山科技学院本部校区的校园景点平面图。通过系统的各功能不仅可以查询任意景点的编号，名称，简介等信息，还可以查询任意两景点间的最短路径及长度和所有路径及长度。另外，还可以通过一系列操作增加，更改或删除任意的景点的各信息。
Contains the campus attractions plan of the campus Foshan Institute of Technology campus navigation system. Through the function of the system not only can query any spots number, name, profile and other information, can also query the shortest path and arbitrary length between two spots and all paths and the length of. In addition, can also increase through a series of operations, the information change or delete any of the attractions. (2015-05-09, Visual C++, 3KB, 下载7次)
：文章提出了一种应用于高等院校的自动排课算法
该算法针对高等院校排课要求的高易用性、高收敛性等特点将
启发式算法、禁忌搜索算法、回溯算法进行有机结合．充分发挥
启发式算法在利用应用领域经验和规则的优势，提高了自动排课
的资源搜索能力。通过实际的应用表明，该算法在解决复杂的高
校排课问题时有较好的效果。
关键词：自动排课 启发式算法 禁忌搜索 回溯算法
Automatic Timetable A applied to the institutions of higher learning: Posts algorithm high ease of use of the algorithm for universities Timetable requirements, high convergence characteristics of the organic combination of backtracking algorithm heuristic algorithm, tabu search algorithm . Give full play to the heuristic algorithm to make use of the advantage of the experience and rules of application areas, to improve Automatic Timetable resource search capabilities. Through the practical application show that the algorithm is better results in solving complex university course timetabling problem. Keywords: Automatic Timetable heuristic algorithm tabu search backtracking algorithm (2013-03-11, Visual C++, 242KB, 下载6次)
华为机试整理 2014华为上机题华为校招上机C&C++试题
huawei jishi (2015-07-08, Visual C++, 581KB, 下载6次)
ACM 2012多校暑假集训（8），标程+解题报告。
ACM 2012 Multi-University Training Contest- Contest 8 Tutorial. (2012-11-16, C/C++, 477KB, 下载5次)
ACM 2012多校暑假集训（3），标程+解题报告。
ACM 2012 Multi-University Training Contest- Contest 3 Tutorial. (2012-11-16, C/C++, 539KB, 下载5次)
本程序是用VC实现的模拟某校九层教学楼的电梯系统，电梯能在每层停留，电梯“空闲”时，将来到第一层候命，乘客可随机地进出于任何层，每个人有一个能容忍的最长等待时间，一旦等候电梯时间过长，他将放弃。程序将会按时序显示系统状态的变化过程，即发生的全部人和电梯的动作序列。
This program is a simulation of a nine-story classroom building s elevator system, elevator to stay at each floor.When the elevator is "free",it will come to the first layer to wait for command.Passengers can come in and go out of every floor, each passengers have a maximum wait time, once the waiting elevator takes too long, he will give up. The program will display the diversification of the system according the sequence in time. (2012-03-05, Visual C++, 5KB, 下载5次)
[问题描述]
现要开发一个开放式的高校本科招生最低录取分数线的查询系统，供师生和学生家长等查询，高校自愿放入该校的信息，可能随时有高校加入。要求实现的查询功能有：
（1）查询等于用户给定分数的高校；
（2）查询大于（或小于）用户给定分数的高校；
（3）查询最低录取分数线在用户给定的分数段中的高校。
[基本要求]
（1）要求有简单的菜单界面；
（2）数据规模最少不得低于50。
[测试数据]
自行设定
[Problem Description] is to develop an open admission college undergraduate admissions lowest score of the query system for teachers and parents of students and other inquiries, the university voluntarily placed in the school' s information, colleges and universities may join at any time. Required to achieve the query functions are: (1) query the user for a given fraction is equal to the college (2) query is larger than (or less) given by the user scores of colleges and universities (3) check the minimum entry score in the score given by the user segment of the university. [Basic requirements] (1) requires a simple menu interface (2) Data shall not be less than the minimum size of 50. [Test data] set their own (2011-11-10, C/C++, 67KB, 下载4次)
/*
* Author: m_h_yang@163.com, on March 29th, 2012
* 问题一：删除子串；
* 问题二：对两个任意大正整数进行相加，得到其和。
* 注：此俩问题为某公司校招时的上机编程考试题。
*/
int delete_sub_str(const char*str, const char* sub_str, char* result_str)
void add(const char*num1, const char*num2, char*result) (2012-03-29, C/C++, 2KB, 下载4次)
ACM 2012多校暑假集训（10），标程+解题报告。
ACM 2012 Multi-University Training Contest- Contest 10 Tutorial. (2012-11-16, C/C++, 3940KB, 下载4次)
包含各大名校的数据结构习题和答案，真正体会数据结构算法方法精髓
Data structure that contains the major schools of exercises and answers, the real essence of understanding the data structure algorithm method (2011-05-24, C/C++, 905KB, 下载4次)
参加运动会的 个学校编号为 。比赛分成 个男子项目和 个女子项目，项目编号分别为 和 。由于各项目参加人数差别较大，有些项目取前五名，得分顺序为7，5，3，2，1；还有些项目只取前三名，得分顺序为5，3，2。写一个统计程序产生各种成绩单和得分报表。
2、要求产生各学校的成绩单，内容包括各校所取得的每项成绩的项目号、名次（成绩）、姓名和得分；产生团体总分报表，内容包括校号、男子团体总分、女子团体总分和团体总分。
3、测试数据：对于 ， ， ，编号为奇数的项目取前五名，编号为偶数的项目取前三名，设计一组实例数据。
The school serial number participating in Games is. Competition divides into men s event and women s event , the project serial number parts for the sum. The difference is bigger since every project participates in a number , some projects choose the first five , score order is 7 , 5 , 3 , 2 , 1 Still have some of projects taking the first three places , score only being 5 , 3 , 2 in proper order. Write a form for report counting procedure producing the various school report card and score. 2, demands the school report card producing every school , content includes every achievement project number , position in a name list (achievement) , full name and score per got by school Produce the group total points form for report , content including the school number , male person group total points , woman group total points and group total points. 3, testing data: Be that the odd number project chooses the first five to the serial number, the serial number is that the even number project c (2009-04-07, Visual C++, 135KB, 下载4次)