GraphicsExam-2010-A-solution
2010 AMC 10A 试题及答案解析
2010 AMC 10A problems and solutions.The test was held on February 8, 2010. The first link contains the full set of test problems. The rest contain each individual problem and its solution.Problem 1Mary’s top book shelf holds five books with the follow ing widths, incentimeters: , , , , and . What is the average book width, in centimeters?SolutionTo find the average, we add up the widths , , , , and , to get a total sum of . Since there are books, the average book width isThe answer is .Problem 2Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?SolutionLet the length of the small square be , intuitively, the length of the big square is . It can be seen that the width of the rectangle is .Thus, the length of the rectangle is times large as the width. The answer is .Problem 3Tyrone had marbles and Eric had marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?SolutionLet be the number of marbles Tyrone gave to Eric. Then,. Solving for yields and . The answer is .Problem 4A book that is to be recorded onto compact discs takes minutes to read aloud. Each disc can hold up to minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?SolutionAssuming that there were fractions of compact discs, it would take CDs to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs, in which case it would have minutes on each of the 8 discs. The answer is .Problem 5The area of a circle whose circumference is is . What is the value of ?SolutionIf the circumference of a circle is , the radius would be . Since the area of a circle is , the area is . The answer is . Problem 6For positive numbers and the operation is defined asWhat is ?Solution. Then, is The answer isProblem 7Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?SolutionCrystal first runs North for one mile. Changing directions, she runs Northeast for another mile. The angle difference between North and Northeast is 45 degrees. She then switches directions to Southeast, meaning a 90 degree angle change. The distance now from travelling North for one mile, and her current destination is miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to , which is equal to . The answer isTony works hours a day and is paid $per hour for each full year of his age. During a six month period Tony worked days and earned $. How old was Tony at the end of the six month period?SolutionTony worked hours a day and is paid dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is years old, he gets dollars a day. We also know that he worked days and earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. Because he earned dollars, we know that he was for some period of time, but not the whole time, because then the money earned would be greater than or equal to . This is why he was when he began, but turned sometime in the middle and earned dollars in total. So the answer is .The answer is . We could find out for how long he was and . . Then isand we know that he was for days, and for days. Thus, the answer is .Problem 9A palindrome, such as , is a number that remains the same when its digits are reversed. The numbers and are three-digit and four-digit palindromes, respectively. What is the sum of the digits of ?Solutionis at most , so is at most . The minimum value ofis . However, the only palindrome between and is , which means that must be .It follows that is , so the sum of the digits is .Marvin had a birthday on Tuesday, May 27 in the leap year . In what year will his birthday next fall on a Saturday?Solution(E) 2017There are 365 days in a non-leap year. There are 7 days in a week. Since 365 = 52 * 7 + 1 (or 365 is congruent to 1 mod 7), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year.For example: 5/27/08 Tue 5/27/09 WedHowever, a leap year has 366 days, and 366 = 52 * 7 + 2. So the same date (after February) moves "forward" two days in the subsequent year, if that year is a leap year.For example: 5/27/11 Fri 5/27/12 SunYou can keep count forward to find that the first time this date falls on a Saturday is in 2017:5/27/13 Mon 5/27/14 Tue 5/27/15 Wed 5/27/16 Fri 5/27/17 Sat Problem 11The length of the interval of solutions of the inequality is . What is ?SolutionSince we are given the range of the solutions, we must re-write the inequalities so that we have in terms of and .Subtract from all of the quantities:Divide all of the quantities by .Since we have the range of the solutions, we can make them equal to .Multiply both sides by 2.Re-write without using parentheses.Simplify.We need to find for the problem, so the answer isProblem 12Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?SolutionThe water tower holds times more water than Logan's miniature. Therefore, Logan should make his towertimes shorter than the actual tower. This ismeters high, or choice .Problem 13Angelina drove at an average rate of kph and then stopped minutes for gas. After the stop, she drove at an average rate of kph. Altogether she drove km in a total trip time of hours including the stop. Which equation could be used to solve for the time in hours that she drove before her stop?SolutionThe answer is （）because she drove at kmh for hours (the amount of time before the stop), and 100 kmh for because she wasn't driving for minutes, or hours. Multiplying by gives the total distance, which is kms. Therefore, the answer isProblem 14Triangle has . Let and be on and , respectively, such that . Let be the intersection of segments and , and suppose that is equilateral. What is ?SolutionLet .Since ,Problem 15In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.Brian: "Mike and I are different species."Chris: "LeRoy is a frog."LeRoy: "Chris is a frog."Mike: "Of the four of us, at least two are toads."How many of these amphibians are frogs?SolutionSolution 1We can begin by first looking at Chris and LeRoy.Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but clearly Chris is not a frog, and we have a contradiction. The same applies if Chris is a frog.Clearly, Chris and LeRoy are different species, and so we have at least frog out of the two of them.Now suppose Mike is a toad. Then what he says is true because we already have toads. However, if Brian is a frog, then he is lying, yet his statement is true, a contradiction. If Brian is a toad, then what he says is true, but once again it conflicts with his statement, resulting in contradiction.Therefore, Mike must be a frog. His statement must be false, which means that there is at most toad. Since either Chris or LeRoy is already a toad, Brain must be a frog. We can also verify that his statement is indeed false.Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have frogs total.Solution 2Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.As Mike is a frog, his statement is false, hence there is at most one toad.As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad. Hence we must have one toad and three frogs.Problem 16Nondegenerate has integer side lengths, is an angle bisector, , and . What is the smallest possible value of the perimeter?SolutionBy the Angle Bisector Theorem, we know that . If we use the lowest possible integer values for AB and BC (the measures of AD and DC, respectively), then , contradicting the Triangle Inequality. If we use the next lowest values (and ), the Triangle Inequality is satisfied. Therefore, our answer is , or choice .Problem 17A solid cube has side length inches. A -inch by -inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?SolutionSolution 1Imagine making the cuts one at a time. The first cut removes a box . The second cut removes two boxes, each of dimensions, and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is .Therefore the volume of the rest of the cube is.Solution 2We can use Principle of Inclusion-Exclusion to find the final volume of the cube.There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has cubic inches. However, we can not just sum their volumes, as the central cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.Hence the total volume of the cuts is.Therefore the volume of the rest of the cube is.Solution 3We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.Each edge can be seen as a box, and each corner can be seen as a box..Problem 18Bernardo randomly picks 3 distinct numbers from the setand arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?SolutionWe can solve this by breaking the problem down into cases and adding up the probabilities.Case : Bernardo picks . If Bernardo picks a then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a is .Case : Bernardo does not pick . Since the chance of Bernardo picking is , the probability of not picking is .If Bernardo does not pick 9, then he can pick any number from to . Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger.Ignoring the for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers.We get this probability to beProbability of Bernardo's number being greater isFactoring the fact that Bernardo could've picked a but didn't:Adding up the two cases we getProblem 19Equiangular hexagon has side lengthsand . The area of is of the area of the hexagon. What is the sum of all possible values of ?SolutionSolution 1It is clear that is an equilateral triangle. From the Law of Cosines, we get that . Therefore, the area of is .If we extend , and so that and meet at , and meet at , and and meet at , we find that hexagon is formed by taking equilateral triangle of side length and removing three equilateral triangles, , and , of side length . The area of is therefore.Based on the initial conditions,Simplifying this gives us . By Vieta's Formulas we know that the sum of the possible value of is .Solution 2As above, we find that the area of is .We also find by the sine triangle area formula that, and thusThis simplifies to.Problem 20A fly trapped inside a cubical box with side length meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?SolutionThe distance of an interior diagonal in this cube is and the distance of a diagonal on one of the square faces is . It would not make sense if the fly traveled an interior diagonal twice in a row, as it would return to the point it just came from, so at most the final sum can only have 4 as the coefficient of . The other 4 paths taken can be across a diagonal on one of the faces, so the maximum distance traveled is.Problem 21The polynomial has three positive integer zeros. What is the smallest possible value of ?SolutionBy Vieta's Formulas, we know that is the sum of the three roots of the polynomial . Also, 2010 factors into. But, since there are only three roots to the polynomial, two of the four prime factors must be multiplied so that we are left with three roots. To minimize , and should be multiplied, which means will be and the answer is .Problem 22Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point insidethe circle. How many triangles with all three vertices in the interior of the circle are created?SolutionTo choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. Therefore, the answer is which is equivalent to 28,Problem 23Each of 2010 boxes in a line contains a single red marble, and for , the box in the position also contains white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let be the probability that Isabella stops afterdrawing exactly marbles. What is the smallest value of for which ?SolutionThe probability of drawing a white marble from box is . Theprobability of drawing a red marble from box is .The probability of drawing a red marble at box is thereforeIt is then easy to see that the lowest integer value of that satisfies the inequality is .Problem 24The number obtained from the last two nonzero digits of is equal to . What is ?SolutionWe will use the fact that for any integer ,First, we find that the number of factors of in is equal to. Let . The we want is therefore the last two digits of , or . Since there is clearly an excess of factors of 2, we know that , so it remains to find .If we divide by by taking out all the factors of in , we canwrite as where where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form is replaced by , and every number in the form is replaced by .The number can be grouped as follows:Using the identity at the beginning of the solution, we can reducetoUsing the fact that (or simply the fact that if you have your powers of 2 memorized), we can deduce that . Therefore.Finally, combining with the fact that yields.Problem 25Jim starts with a positive integer and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with , then his sequence contains numbers:Let be the smallest number for which Jim’s sequence has numbers. What is the units digit of ?SolutionWe can find the answer by working backwards. We begin with on the bottom row, then the goes to the right of the equal's sign in the row above. We find the smallest value for whichand , which is .We repeat the same procedure except with for the next row and for the row after that. However, at the fourth row, wesee that solving yields , in which case it would be incorrect since is not the greatest perfect square less than or equal to . So we make it a and solve . We continue on using this same method where we increase the perfect square until can be made bigger than it. When we repeat this until we have rows, we get:Hence the solution is the last digit of , which is .。
怎样申请MCT证书成为微软认证
以下是根据⼴⼤学员咨询的怎么申请成为⼀个合格的微软认证获得MCT证书的要求和步骤，任何在微软相关技术和演讲⽅⾯有⼀定能⼒和兴趣的⼈员都有机会成为微软认证(MCT)。
如果您⼀直致⼒于研究或使⽤微软产品和解决⽅案、拥有相当的实际经验并具备良好的表达能⼒和整体形象，您可以按照以下5步完成微软认证(以下简称MCT)的申请。
第1步：获得微软认证专家(MCP)或微软商业解决⽅案(MBS)资格证书 根据微软公司全球统⼀标准，在申请成为 MCT 之前，您必须获得下列证书之⼀。
此项标准的⽬的是为了核实 MCT申请者具备经过证实的技术实⼒和教学⼯作能⼒。
⼀旦获得某项微软认证，即表明您已经在与您的证书相关的领域或⼯作岗位中展⽰了您运⽤微软产品和技术开展⼯作的熟练程度。
Microsoft Certified Desktop Support Techniciann (MCDST) Microsoft Certified Systems Administrator (MCSA) on Microsoftn Windows 2000 Microsoft Certified Systems Administrator (MCSA) on Microsoftn Windows Server 2003 Microsoft Certified Systems Engineer (MCSE) onn Microsoft Windows 2000 Microsoft Certified Systems Engineer (MCSE) onn Microsoft Windows Server 2003 Microsoft Certified Application Developern (MCAD) for Microsoft .NET Microsoft Certified Professional Developern (MCPD) Microsoft Certified Solution Developer (MCSD) on Visual Studion 6.0 Microsoft Certified Solution Developer (MCSD) for Microsoft .NETn n Microsoft Certified Database Administrator (MCDBA) on Microsoft SQL Server 2000 Microsoft Certified IT Professional (MCITP)n Microsoft Officen Specialist (MOS) – pass at least one of the following exams: • Word 2003 Expert • Excel 2003 Expert • PowerPoint 2003 • Access 2003 • Outlook 2003 Microsoft Certifiedn Application Specialist – pass at least one of the following exams: • Exam 70-601: Using Microsoft Office Word 2007 • Exam 70-602: Using Microsoft Office Excel 2007 • Exam 70-603: Using Microsoft Office PowerPoint 2007 • Exam 70-604: Using Microsoft Office Outlook 2007 • Exam 70-605: Using Microsoft Office Access 2007 • Exam 70-600: Using Microsoft Windows Vista Microsoft Certifiedn Professional: Exam 70-301, Managing, Organizing, and Delivering IT Projects by Using the Microsoft Solutions Framework Microsoft Certified Businessn Management Solutions Professional – Application for Microsoft Dynamics GP Microsoft Certified Business Management Solutions Professional – Installation & Configuration for Microsoft Dynamics GP Microsoft Certified Businessn Management Solutions Professional – Application for Microsoft Dynamics SL Microsoft Certified Business Management Solutions Professional – Installation & Configuration for Microsoft Dynamics SL Microsoft Certified Businessn Management Solutions Professional – Application for Microsoft Dynamics CRM Microsoft Certified Business Management Solutions Professional – Installation & Configuration for Microsoft Dynamics CRM Microsoft Certifiedn Business Management Solutions Professional – Developer for Microsoft Dynamics CRM Microsoft Certified Business Management Solutions Professional –n Application for Microsoft Dynamics NAV Microsoft Certified Businessn Management Solutions Professional – Installation & Configuration for Microsoft Dynamics NAV Microsoft Certified Business Management Solutionsn Professional – Developer for Microsoft Dynamics NAV Microsoft Certifiedn Business Management Solutions Professional – Application for Microsoft Dynamics AX Microsoft Certified Business Management Solutions Professional –n Installation & Configuration for Microsoft Dynamics AX Microsoftn Certified Business Management Solutions Professional—Developer for Microsoft Dynamics AX Microsoft Retail Management System MCT Designationn。
信息处理技术员 上半年试卷新
全国计算机技术与软件专业技术资格全国计算机技术与软件专业技术资格（（水平水平））考试考试20201010年上半年半年 信息处理技术员信息处理技术员 上午试卷上午试卷（考试时间 9:00～11:30 共150分钟）请按下述要求正确填写答题卡1. 在答题卡的指定位置上正确写入你的姓名和准考证号，并用正规 2B 铅笔在你写入的准考证号下填涂准考证号。
2. 本试卷的试题中共有75个空格，需要全部解答，每个空格 1分，满分75分。
3. 每个空格对应一个序号，有A、B、C、D 四个选项，请选择一个最恰当的选项作为解答，在答题卡相应序号下填涂该选项。
4. 解答前务必阅读例题和答题卡上的例题填涂样式及填涂注意事项。
解答时用正规 2B 铅笔正确填涂选项，如需修改，请用橡皮擦干净，否则会导致不能正确评分。
例题● 2010年上半年全国计算机技术与软件专业技术资格（水平）考试日期是 （88） 月 （89） 日。
（88）A.4 B.5 C.6 D.7 （89）A.19 B.20 C.21 D.22因为考试日期是“5月22日”，故（88）选B，（89）选D，应在答题卡序号（88）下对 B 填涂，在序号（89）下对 D 填涂（参看答题卡）。
● 下列关于信息特性的叙述，不正确的是 （1） 。
（1）A. 信息必须依附于某种载体进行传输B. 信息是不能被识别的C. 信息能够以不同的形式进行传递，并且可以还原再现D. 信息具有时效性和时滞性● 下列叙述中，正确的是 （2） 。
（2）A. 数据是指记录下来的事实，是客观实体属性的值B. 信息是对事实、概念或指令的一种特殊表达形式C. 数据的驻留地称为信宿D. 数据是对各种事物的特征、运动变化的反映● 信息处理链中的第一个基本环节是 （3） 。
（3）A. 信息的采集 B. 信息的存储 C.信息的加工 D.信息的传输● 某班级有学生60名，语文平均成绩为85分，按成绩将学生划分为优秀和非优秀，优秀学生的平均成绩为90分，非优秀学生的平均成绩为75分，则优秀学生的人数是 （4） 。
2024年3月GESP编程能力认证Python等级考试三级真题(含答案)
2024年3月GESP编程能力认证Python等级考试三级真题(含答案)一、单选题（每题2分，共30分）第1题，小杨的父母最近刚刚给他买了一块华wei手表，他说手表上跑的是鸿meng，这个鸿meng是？（C）A. 小程序B. 计时器C. 操作系统D. 神话人物第2题，中国计算机学会（CCF）在2024年1月27日的颁奖典礼上颁布了王选奖，王选先生的重大贡献是（C）。
A. 制造自动驾驶汽车B. 创立培训学校C. 发明汉字激光照排系统D. 成立方正公司第3题，下列流程图的输出结果是？（B）A. 25B. 30C. 35D. 55第4题，在Python语言中，hex函数可以将十进制数转换成十六进制数。
执行语句hex(2024)后，显示的运行结果是？（C）A. 0x7e6B. 0x7e7C. 0x7e8D. 0x7e9第5题，下列说法错误的是？（D）A. 字典是通过key（键）来进行访问的，跟列表、元组不同。
B. a={}是一个空字典。
C. {(1,2):"123"}的数据类型是字典。
D. 字典一旦创建不可以被修改。
第6题，使用tuple函数创建元组错误的是？（C）A. a=tuple("1234")B. a=tuple([10,20,30,40])C. a=tuple(20,30)D. a=tuple({10,20,30,40})第7题，Python代码print(list(range(1,10))[1::2])，输出结果是（D）？A. 上述代码执行报错B. [1,3,5,7,9]C. [2,4,6,8,10]D. [2,4,6,8]第8题，下列Python赋值语句错误的是（A）？A. a={10,20,30}*3B. a=(10,20,30)*3C. a="123"*3D. a=[10,20,30]*3第9题，执行下列Python代码，输出结果是（C）？A. (1,3,3,6,8,9)B. (9,8,6,3,3,1)C. [9,8,6,3,3,1]D. [1,3,3,6,8,9]第10题，对于字典data={'apple':3.5,'banana':4.0,'orange':6.0}，删除'banana':4.0键值对的正确操作是（A）？A. del data['banana']B. del data['banana':4.0]C. del dataD. data.remove('banana')第11题，`在三位数的自然数中，找出至少有一位数字是5的所有整数，应在横线填入代码是（B）。
Visual Basic考试卷2010
教育技术专业Visual Basic 课程试卷（A）2009~20101.程序流程图中带有箭头的线段表示的是（C ）A.图元关系B.数据流C.控制流D.调用关系2.下列描述中正确的是（H ）E.程序就是软件F.软件开发不受计算机系统的限制G.软件既是逻辑实体，又是物理实体H.软件是程序、数据与相关文档的集合3.在窗体（名称为Form1）上画一个名称为Text1的文本框和一个名称为Command1的命令按钮，然后编写一个事件过程。
程序运行以后，如果在文本框中输入一个字符，则把命令按钮的标题设置为“计算机等级考试”。
以下能实现上述操作的事件过程是（I ）I.Private Sub Text1_Change()Command1.Caption=”计算机等级考试”End SubJ.Private Sub Command1_Click()Caption=”计算机等级考试”End SubK.Private Sub Form1_Click()Text1.Caption=”计算机等级考试’End SubL.Private Sub Command1_Click()Text1.Text=”计算机等级考试”End Sub4.设a=5,b=4,c=3,d=2,表达式(3>2*b Or a=c And b<>c Or c>d)的值是（B ）A.1B.TrueC.FalseD.25.设a=”Microsoft Visual Basic”，则以下使变量b的值为”Visual Basic”的语句是（B ）A.b=Left(a,10)B.b=Mid(a,10)C.b=Right(a,10)D.b=Mid(a,11,10)6.假定有如下的窗体事件过程：Private Sub Form_()a$=” Microsoft Visual Basic”b$=Right(a$,5)c=Mid(a$,1,9)Msgbox a$,34,b$,c$,5End Sub程序运行后单击窗体，则在弹出的信息框中的标题栏中显示的信息是（C ）A.Microsoft VisualB.MicrosoftC.BasicD. 57.为了使命令按钮（名称为Command1）右移200，应使用的语句是（C ）mand1.Move-200mand1.Move 200mand1.Left=Command1.Left+200mand1.Left=Command1.Left-2008.在窗体上画一个文本框，然后编写如下事件过程：Private Sub Form_Click()x=InputB ox(“请输入一个整数”)Print x+Text1.TextEnd Sub程序运行时，在文本框中输入456，然后单击窗体，在输入对话框中输入123，单击“确定”按钮后，在窗体上显示的内容是：( D )A.123B.456C.579D.1234569.在窗体上画一个文本框和一个计时器控件，名称分别为Text1和Timer1,在属性窗口中把计时器的Interval属性设置为1000，Enabled属性设置为False,程序运行后，如果单击命令按钮，则每隔一秒种在文本框中显示一次当时的时间。
az-104 exam simulation 题
一、az-104考试概述az-104是微软推出的一项专业考试，主要测试参与者在Azure管理方面的技能和知识水平。
参加该考试的人员需要具备一定的IT基础知识以及相关Azure服务的使用经验，通过该考试可以证明个人在Azure管理方面的能力和水平，对于个人职业发展和就业竞争具有重要意义。
二、考试内容概述az-104考试的内容涵盖了Azure基础架构、安全与监控、解决方案和管理等多个方面。
以下是该考试的主要内容：1. Azure基础架构：包括虚拟网络、存储、虚拟机、应用服务等方面的知识和技能要求。
2. 安全与监控：考察参与者在Azure安全管理、身份与访问管理、安全策略实施、监控与报警等方面的能力。
3. 解决方案和管理：主要涉及Azure资源管理、数据存储解决方案、Azure计算解决方案、应用架构解决方案等内容。
以上内容是az-104考试的主要知识点，考生需要系统学习并掌握这些内容，通过模拟考试进行练习，提高自己的应试能力。
三、az-104模拟考试的重要性1. 模拟考试可以帮助考生了解考试形式和题型，熟悉考试流程和时间分配，提前适应考试环境，降低考试焦虑感。
2. 通过模拟考试可以评估个人的学习效果和知识掌握程度，及时发现不足之处，有针对性地进行复习和提高。
3. 模拟考试还可以帮助考生总结经验，找出解题技巧和答题规律，提高应试技巧和答题速度，增加通过考试的把握。
四、az-104模拟考试的准备方法1. 充分了解考试大纲和考试指南，梳理考试知识点，制定合理的学习计划，有条不紊地进行系统学习。
2. 参考冠方推荐的教材和资料，结合上线课程和实际操作，全面系统地掌握考试所需的知识和技能。
3. 寻找优质的模拟考试题库，进行反复练习和模拟，提高解题速度和准确率，增强应试能力。
4. 参加模拟考试培训班或讲座，向专业的老师请教，了解考试经验和技巧，不断提升自己的备考水平。
五、总结az-104考试是一个全面考察参与者在Azure管理方面的综合能力的重要考试，通过模拟考试进行练习对于备考具有重要意义。
2024年9月GESPC++三级试卷附答案
2024年9月 GESP C ++三级试卷（共27题，满分100分，时间90分钟）第1 题下列代码中，输出结果是（）。
A. EqualB. Not equalC. 程序不能正确执行D. 没有输出第2 题关于计算机中的编码，下列说法中正确的是（）。
A. 机器数的形式值和真值是一致的B. 原码就是符号位加上真值C. 机器数是带符号的D. [-1]=[1000 0001]原=[1111 1111]反第3 题8进制数3703转换成16进制数是（）。
A. 7C3B. 7A3C. 7B3D. 7D3第4 题 0.8125变成二进制是（）。
A. 0.1110B. 0.1 1111 1011 1101C. 0.1 1111 1011 1100D. 0.1101第5 题下面说法正确的是（）。
A. （22&01）==（22&&01）成立B. (23|11)==30C. (23|10)==31D. (23|01)==31 第6题下列说法正确的是（）A. 2>>1和1>>1的值是一样的B. (2>>2)和(1>>1)的值是一样的C. (11^00)和(1^0)的值是一样的D. (~0的输出值是1)第 7 题下列代码实现的是（）A. a和b的异或B. a和b的同或C. a和b的值交换D. a和b的高低位互换第8 题a&~1运算实现的是（）。
A. 使a的最低位为1B. 使a的最高位为1C. 使a的最低位为0D. 使a的最高位为0第 9 题a=1010 1110,a<<2,下面关于这个说法正确的是（）。
A.a的值变为1011 1000B. a的值变为1011 1010C. a的值变为0101 1101D. a的值变为0101 1100第 10 题下列程序中，result和result2输出分别是 ( )A. 123 -1B. 123 -123C. 1 -1D. 1 -123第 11 题陈A歹种植一批农作物，第一天需要浇水一次，随后的两天（第2、第3天），每天需要浇水2次，再随后的3天（第4、第5、第6天），需要每天浇水3次，这样持续下去，随后的n天，每天需要浇水n次。
国家信息化计算机教育认证项目（CEAC）网络应用工程师试题及答案
测试科目:典型企业网络设计2201试题共计40题单选题：请从四个选项里找出最正确的答案。
1. (本题2分) DNS的什么改进功能使得在DNS复制期间不必传输整个区域文件。
A. 动态更新B. 增加的区域传输C. 更新通知D. 反向查找B2. (本题2分) 从一个主DNS服务器到一个次级DNS服务器移动记录的过程叫什么?A. 复制B. 区域传输C. 中间拷贝D. 通知 B3. (本题2分) 当你要进行活动目录创建时，应该运行什么命令：A. DcpromoupB. Winnt32C. DcpromoD. Setup C4. (本题2分) 当一个组策略被删除了：A. 策略不再存在了B. GPO link不在存在了C. 可以选择D. 无法删除组策略 C5. (本题2分) 公司有2个地点，一个在天津，一个在北京，北京是总部，在公司内部实现活动目录管理，域的控制器在北京，但是天津用户抱怨，每天在登录时要等很长一段时间，公司现在使用ISDN进行连接，怎样处理才能解决这个问题：A. 在天津的公司内部安装一台DHCP服务器，并且设置好自动更新。
B. 在天津的公司内部安装一台DC，创建2个站点，分别把2个地点的Dc放置不同的站点里，设置好子网，配置好站点连接C. 在天津的公司把用户的DNS地址都指向北京的地址D. 重新设计域的结构在天津重新创建一个DC C6. (本题2分) 你需要配置微软的Visio 2000到Engineering OU，同时又不想用户通过点击.VST文件来安装V isio 2000。
需要怎样做彩能够达到以上目的？A. 在文件扩展名激活选项中清除掉“自动安装该应用程序”一项，从而可禁止文件的安装。
B. 在文件扩展名激活选项中选中“自动安装该应用程序”一项，从而可禁止文件的安装。
C. 在文件扩展名激活选项中清除掉“不要自动安装该应用程序”一项，从而可禁止文件的安装。
D. 在文件扩展名激活选项中选中“不要自动安装该应用程序”一项，从而可禁止文件的安装。
全国计算机技术与软件专业技术资格(水平)考试
2010年下半年系统集成项目管理工程师上午试卷第1页（共15页）全国计算机技术与软件专业技术资格（水平）考试2010 年下半年系统集成项目管理工程师上午试卷（考试时间9:00～11:30 共150 分钟）请按下述要求正确填写答题卡1．在答题卡的指定位置上正确写入你的姓名和准考证号，并用正规2B 铅笔在你写入的准考证号下填涂准考证号。
2．本试卷的试题中共有75 个空格，需要全部解答，每个空格1 分，满分75 分。
3．每个空格对应一个序号，有A、B、C、D 四个选项，请选择一个最恰当的选项作为解答，在答题卡相应序号下填涂该选项。
4．解答前务必阅读例题和答题卡上的例题填涂样式及填涂注意事项。
解答时用正规2B 铅笔正确填涂选项，如需修改，请用橡皮擦干净，否则会导致不能正确评分。
例题●2010年下半年全国计算机技术与软件专业技术资格（水平）考试日期是（88）月（89）日。
（88）A．12 B．11 C．10 D．9（89）A．10 B．11 C．12 D．13因为考试日期是“11月13日”，故（88）选B，（89）选D，应在答题卡序号（88）下对B 填涂，在序号（89）下对D 填涂（参看答题卡）。
2010年下半年系统集成项目管理工程师上午试卷第2页（共15页）●以下（1）不属于系统集成项目。
（1）A．不包含网络设备供货的局域网综合布线项目B．某信息管理应用系统升级项目C．某软件测试实验室为客户提供的测试服务项目D．某省通信骨干网的优化设计项目●关于计算机信息系统集成企业资质，下列说法错误的是（2）。
（2）A．计算机信息系统集成的资质是指从事计算机信息系统集成的综合能力，包括技术水平、管理水平、服务水平、质量保证能力、技术装备、系统建设质量、人员构成与素质、经营业绩、资产状况等要素B．工业和信息化部负责计算机信息系统集成企业资质认证管理工作，包括指定和管理资质认证机构、发布管理办法和标准、审批和发布资质认证结果C．企业已获得的系统集成企业资质证书在有效期满后默认延续D．在国外注册的企业目前不能取得系统集成企业资质证书●某计算机系统集成二级企业注册资金2500 万元，从事软件开发与系统集成相关工作的人员共计100 人,其中项目经理15 名,高级项目经理10 名。
imc2010-day2-solutions
IMC2010,Blagoevgrad,BulgariaDay2,July27,2010Problem1.(a)A sequence x1,x2,...of real numbers satisfiesx n+1=x n cos x n for all n≥1.Does it follow that this sequence converges for all initial values x1?(b)A sequence y1,y2,...of real numbers satisfiesy n+1=y n sin y n for all n≥1.Does it follow that this sequence converges for all initial values y1?Solution1.(a)NO.For example,for x1=πwe have x n=(−1)n−1π,and the sequence is divergent.(b)YES.Notice that|y n|is nonincreasing and hence converges to some number a≥0.If a=0,then lim y n=0and we are done.If a>0,then a=lim|y n+1|=lim|y n sin y n|=a·|sin a|, so sin a=±1and a=(k+12)π≤|y n|<(k+1)πfor all n>n0.Then all the numbers y n0+1,y n0+2,...lie in the union of the intervals(k+12)π .Depending on the parity of k,in one of the intervals (k+12)π the values of the sine function is positive;denote this interval by I+.In the other interval the sine function is negative;denote this interval by I−.If y n∈I−for some n>n0then y n and y n+1=y n sin y n have opposite signs,so y n+1∈I+.On the other hand,if If y n∈I+for some n>n0then y n and y n+1 have the same sign,so y n+1∈I+.In both cases,y n+1∈I+.We obtained that the numbers y n0+2,y n0+3,...lie in I+,so they have the same sign.Since|y n|isconvergent,this implies that the sequence(y n)is convergent as well.Solution2for part(b).Similarly to thefirst solution,|y n|→a for some real number a.Notice that t·sin t=(−t)sin(−t)=|t|sin|t|for all real t,hence y n+1=|y n|sin|y n|for all n≥2. Since the function t→t sin t is continuous,y n+1=|y n|sin|y n|→|a|sin|a|=a.Problem2.Let a0,a1,...,a n be positive real numbers such that a k+1−a k≥1for all k= 0,1,...,n−1.Prove that1+1a1−a0 ···1+1a0 1+1a n .Solution.Apply induction on n.Considering the empty product as1,we have equality for n=0.Now assume that the statement is true for some n and prove it for n+1.For n+1,the statement can be written as the sum of the inequalities1+1a1−a0 ···1+1a0 ··· 1+1a0 1+1a n−a0 ·1a0 ··· 1+1a n+1.(1)Hence,to complete the solution it is sufficient to prove(1).1To prove(1),apply a second induction.For n=0,we have to verify1a1−a0≤ 1+1a1.Multiplying by a0a1(a1−a0),this is equivalent witha1≤(a0+1)(a1−a0)a0≤a0a1−a201≤a1−a0.For the induction step it is sufficient that1+1a n+2−a0≤ 1+1a n+2.Multiplying by(a n+2−a0)a n+2,(a n+1−a0+1)a n+2≤(a n+1+1)(a n+2−a0)a0≤a0a n+2−a0a n+11≤a n+2−a n+1.Remark1.It is easy to check from the solution that equality holds if and only if a k+1−a k=1for all k.Remark2.The statement of the problem is a direct corollary of the identity1+ni=0 1x j−x i =n i=0 1+1It is well known that|Gx|=|G||Gx|=|G||G xi|(|G xi|−1),hence1−1|G xi| .(6)If for some i,j∈{1,...,k}|G xi |,|G xi|≥2thenki=1 1−12 + 1−1|G| which contradicts with(6),thus we can assume that|G x1|=...=|G xk−1|=1.Then from(6)we get|G xk |=|G|,hence G xk=G.Problem4.Let A be a symmetric m×m matrix over the two-elementfield all of whose diagonal entries are zero.Prove that for every positive integer n each column of the matrix A n has a zero entry.Solution.Denote by e k(1≤k≤m)the m-dimensional vector over F2,whose k-th entry is1and all the other elements are0.Furthermore,let u be the vector whose all entries are1.The k-th column of A n is A n e k.So the statement can be written as A n e k=u for all1≤k≤m and all n≥1.For every pair of vectors x=(x1,...,x m)and y=(y1,...,y m),define the bilinear form(x,y)= x T y=x1y1+...+x m y m.The product(x,y)has all basic properties of scalar products(except the property that(x,x)=0implies x=0).Moreover,we have(x,x)=(x,u)for every vector x∈F m2.It is also easy to check that(w,Aw)=w T Aw=0for all vectors w,since A is symmetric and its diagonal elements are0.Lemma.Suppose that v∈F m2a vector such that A n v=u for some n≥1.Then(v,v)=0. Proof.Apply induction on n.For odd values of n we prove the lemma directly.Let n=2k+1and w=A k v.Then(v,v)=(v,u)=(v,A n v)=v T A n v=v T A2k+1v=(A k v,A k+1v)=(w,Aw)=0.Now suppose that n is even,n=2k,and the lemma is true for all smaller values of n.Let w=A k v;then A k w=A n v=u and thus we have(w,w)=0by the induction hypothesis.Hence, (v,v)=(v,u)=v T A n v=v T A2k v=(A k v)T(A k v)=(A k v,A k v)=(w,w)=0.The lemma is proved.3Now suppose that A n e k=u for some1≤k≤m and positive integer n.By the Lemma,we should have(e k,e k)=0.But this is impossible because(e k,e k)=1=0.Problem5.Suppose that for a function f:R→R and real numbers a<b one has f(x)=0for all x∈(a,b).Prove that f(x)=0for all x∈R ifp−1k=0f y+kN =0 .Notice that0∈J N,any linear combinations of any elements in J N is in J N,and for every P(x)∈J N we have xP(x)∈J N.Hence,J N is an ideal of the ring R[x].By the problem’s conditions,for every prime divisors of N we havex N−1x N/p−1are those N throots of unity whose order does not divide N/p.The roots of the greatest common divisor is the intersection of such sets;it can be seen that the intersection consist of the primitive N th roots of unity.Therefore,gcd x N−1N <b−a.It is well-known that lim infN→∞ϕ(N) N =0for all x∈R.If x∈[b,b+1N .On the right-hand side,all numbers x−ϕ(N)−kN).It can be obtained similarly that f(x)=0for all x∈(a−1N,b+1。
Visual Studio 2010 测试管理解决方案
ӄMicrosoft Testing SolutionMicrosoft Testing Solution 微软测试解决方案Microsoft Visual Studio 2010 带来了全新的软件测试解决方案，并使其整合在整个应用生命周期管理体系(ALM)之中。
作为应用生命周期最重要的环节，软件质量保证历来是软件企业最迫切期望得到改善和提高的部分。
微软测试解决方案包括快速进行有关测试的实际与开发、测试用例管理，与Visual Studio Team Foundation Server 集成的测试计划，以确保所有变更和迭代都被正确的测试和覆盖。
这些功能贯穿于整个测试周期：测试计划、测试执行、进度跟踪和结果分析。
Microsoft Visual Studio 2010 测试工具集可以用于创建测试计划、管理测试用例、运行测试用例、管理和分析测试结果，以及与微软新一代开发测试云集成等特性。
微软测试解决方案真正做到了和开发团队的有效协作，是目前最流行的敏捷方法 Agile/Scrum 的最佳实践。
Microsoft Visual Studio 2010 测试框架Microsoft T est Manager 微软测试管理解决方案微软测试管理框架Microsoft Visual Studio 2010 测试管理包含架构设计、项目模板、测试管理、项目门户等；对于MS Project 、SharePoint 、Hyper-v 虚拟化平台及云计算平台进行了全面的集成。
可以大幅提升软件质量,集成的测试工具集提供了完整的”计划-测试-跟踪” 工作流。
使用丰富的诊断工具为开发人员归档大量bug 。
充分利用以人物为导向的用户界面和手动测试快进等功能, 可以获得所有团队角色之间的上下文协作, 大大增加了对整个项目的可视性, 同时提供了对用户需求的完整可跟踪性、进度报告和实施质量度量。
有助于您做出明智、及时的决策, 并降低与软件发布相关的风险。
办公软件应用OFFICE2010高级操作员ATA部分题解课件.doc
办公软件应用OFFICE2010高级操作员ATA题解编辑排版：陈锦老师第一单元操作系统应用（共10 分）第1 题：启动“资源管理器”（1 分）题解：右击‘任务栏左端的开始’—打开Windows 资源管理器——完成——下一题第2 题：在C盘下新建文件夹，文件夹名为“4000001”。
（1 分）题解：右击‘任务栏左端的开始’—打开Windows 资源管理器—打开C：，文件——新建——文件夹——输入4000001 ——回车第3 题：将C盘下的“KSML”2文件夹内KS1-1.DOCX、KS1-2.DOCX、KS1-3.DOCX、KS1-4.DOCX、KS1-5.DOCX、KS4-1.XLSX、KS4-2.XLSX、KS7-1.DOCX、一次性复制到“C4:000001 ”文件夹中，并分别重命名为A1.DOCX、A2.DOCX、A3.DOCX、A4.DOCX、A5.DOCX、A6.XLSX、A7.XLSX、A8.DOCX。
（2 分）题解：右击‘任务栏左端的开始’—打开Windows 资源管理器—打开C：；（1）打开KSML2，按住Ctrl+ 分别单击KS1-1.DOCX、KS1-2.DOCX、KS1-3.DOCX、KS1-4.DOCX、KS1-5.DOCX、KS4-1.XLSX、KS4-2.XLSX、KS7-1.DOCX；编辑——复制；（一次性复制、不能选错）（2）打开C:\4000001，编辑——粘贴；（3）选定要改名的文件KS1-1.DOCX，文件——重命名——输入A1——回车；（用同样方法将KS1-2.DOCX、KS1-3.DOCX、KS1-4.DOCX、KS1-5.DOCX、KS4-1.XLSX、KS4-2.XLSX、KS7-1.DOCX重命名为A2、A3、A4、A5、A6、A7、A8）；完成后单击下一题ATA DEMO模拟练习盘，第4、第5 题题解ATA模拟练习盘试卷（KS1-1）第4 题（3 分）：将程序“截图工具”先附到「开始」菜单上，然后再锁定到任务栏中。
Pearson Edexcel Level 3 GCE Candidate信息表说明书
*P62608A0132*Turn over Pearson Edexcel Level 3 GCECandidate surname Other namesTotal MarksCentre Number Candidate NumberPlease check the examination details below before entering your candidate informationP62608A©2020 Pearson Education Ltd.1/1/1Instructions • Use black ink or ball-point pen.• Fill in the boxes at the top of this page with your name,centre number and candidate number.• Answer one question in Section A and one question in Section B on yourchosen texts.• Answer the questions in the spaces provided– there may be more space than you need .Information • The total mark for this paper is 72.• The marks for each question are shown in brackets– use this as a guide as to how much time to spend on each question .Advice • Read each question carefully before you start to answer it.• Check your answers if you have time at the end.You must have:Prescribed texts (clean copies)English LiteratureAdvanced SubsidiaryPaper 1: Poetry and DramaPaper Reference 8ET0/01Morning (Time: 2 hours)Friday 15 May 2020*P62608A0232*2SECTION A: PoetryAnswer ONE question and write your answer in the space provided.EITHER1 Compare the ways in which poets explore power in The Gun by Vicki Feaver and oneother poem of your choice from Poems of the Decade: An Anthology of the ForwardBooks of Poetry2002–2011.In your answer you should consider the following:• the poets’ development of themes• the poets’ use of language and imagery• the use of other poetic techniques.(Total for Question 1 = 24 marks) OR2 Compare the ways in which poets use settings in The Lammas Hireling by Ian Duhigand one other poem of your choice from Poems of the Decade: An Anthology of theForward Books of Poetry2002–2011.In your answer you should consider the following:• the poets’ development of themes• the poets’ use of language and imagery• the use of other poetic techniques.(Total for Question 2 = 24 marks)*P62608A0332*Turn over3*P62608A0432*4*P62608A0532*Turn 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2010.Nov.CIMA.E3.Answers
The Examiner's Answers – November 2010E3 - Enterprise StrategySECTION AAnswer to Question OneRequirement (a)Porter's model of competitive advantage is predicated on the following concepts: •Overall cost leadership•Differentiation•FocusThese concepts are combined to give the following strategic options:1. Overall cost leadership2. Differentiation3. Cost focus4. Differentiation focusPorter maintains that the model is generic, in that it provides a systematic means of classification which can be applied to all firms and all strategies. If the model is applied successfully in the development of strategy it can lead to long-term or sustainable competitive advantage where the firm outperforms others in its industry.Applying these ideas of Porter to the situation of DEF Airport gives the following:(i)Overall cost leadershipThis option would require DEF to be the lowest cost airport operator with respect to its competitors. TUV airport which is approximately100 kilometres away is regarded by DEF as being its major competitor and there are three other airports within 80 kilometres of DEF which are lesser competitors. It would be useful for the Board to commission a competitor analysis for DEF if this has not been done recently. The Board would also have to consider the nature of the competition which it faces because its competition will not be monolithic, that is, there is not one competitor competing on one basis against DEF. This is because DEF is competing in a number of different markets as shown by the analysis of its revenue which is made up of the following components:•Aviation income•Retail concessions•Car parking•Other income (including property income, cargo handling and other fees and charges)It is likely that competitor analysis will show that DEF has a range of competitive performance when compared against the different aspects of the revenue stream, for example, it may be very successful in respect of its retail concessions because it has total control over the granting of these concessions. However, as regards car parking, this may not be as profitable because of the competition from parking operations outside, but close to, the airport boundaries.In order to apply successfully a strategy of overall cost leadership, DEF needs to analyse the cost structures of its various activities, identify those which are most competitive and where this is likely to be sustainable. It would also benefit from analysing its competitors’ cost structures where this is possible. Where the market in which it competes is mature and the competition is intense, it is unlikely that DEF will be successful in pursuing a policy of overall cost leadership because it will not possess any unique cost advantage.(ii)DifferentiationThis strategy depends upon creating something that is regarded as being unique or special and, because of this, customers are willing to pay a premium price. There are many ways in which differentiation can be achieved but before pursuing it DEF needs to consider the factors explained in (i), namely that it is unlikely to be able to apply this strategy to every aspect of its business. However, DEF does possess some unique aspects: the first is its geographical position and secondly, within the airport’s boundaries, it has a degree of monopoly power in that every airline, retail outlet and service agent is only present at the airport by agreement with DEF. No one has a right to trade at DEFwithout DEF’s permission.One area of possible sustainable competitive advantage for DEF is offered by the findings of the recent survey of European air passengers. The survey revealed that business and first class passengers valued ‘distinctiveness’. Business and first class passengers are analogous to ‘customers willing to pay a premium price’. If DEF could provide a service to these passengers which is ‘distinctive’ or as one respondent expressed it, ‘Offering something which nobody else offers’, DEF could achieve long-term competitive advantage. The survey identified the characteristics of ‘distinctiveness': having the leading airlines operating from the airport not the low-priced airlines, luxury passenger lounges, high class restaurants and extensive duty free shops with the best exclusive brands. If this strategy is to be chosen it is likely to involve DEF in significant capital expenditure as it would need to up-grade its facilities.DEF could use Johnson and Scholes ‘three tests’ to evaluate the suitability, acceptability and feasibility of following a strategy of differentiation based on ‘distinctiveness’ designed to deliver sustainable competitive advantage. However, a differentiation strategy within Porter’s model has a broad scope: it should be directed to the entire industry’s business and first class passengers: this may not be feasible. (See Differentiation focus below).(iii)Cost focusThis strategy would involve DEF offering services at a lower cost to a particular segment of a market. As it serves a number of markets, as evidenced by the components of its revenue stream, this gives the possibility of successfully pursuing a cost focus strategy in one or more segments. A cost focus, or niche strategy, implies that DEF can provide a service better to a particular market segment than rivals who are attempting the whole of the market. Aviation income is the single most important component of DEF’s revenue and DEF already has one low-cost carrier as a client. It may be possible to offer a cost focus strategy to one or more low-priced airlines but there will be a limit to the number to which this can apply. The lower costs extended by DEF to its clients need to be outweighed by benefits, such as higher turnover and complementary business if its overall profitability is not to be adversely affected. There is also the danger that the pursuit of a cost focus strategy could change into overall cost leadership as clients who have not been offered the low cost deal ask for it to be extended to themselves also.A further problem could arise if DEF decides to pursue a differentiation strategy based on the concept of ‘distinctiveness’. One aspect of distinctiveness identified by the recent passenger survey was that it meant ‘having the leading airlines operating from the airport not the low-priced airlines’ and this is in direct contradiction to a cost focus strategy embracing low cost airlines.This dilemma was anticipated by Porter who warned of the dangers of being ‘stuck in the middle’: that is, a business should decide whether it will compete by low costs or by differentiation. To attempt to follow both strategies is likely to end in failure.Differentiation focusLike the cost focus strategy, differentiation focus relies on DEF being able to address itself better to a market segment than its broader scope rivals can. DEF should identify market segments where it can provide services that its rivals do not. This could be within its region, say within a 20 mile radius of the airport, or to a particular buyer group, for example, dedicated cargo handling service for exporters. A differentiation strategy based on‘distinctiveness’ was discussed above. It may not be feasible for DEF to offer this strategy to the entire industry, for example, it may not be attractive to all its customers if DEF is not located near the country’s capital city. However, the strategy may be feasible as a focus strategy when it addresses the market segment that live within commuting distance of the airport.Requirement (b)(i)The Board has produced a development plan which ‘... reviews the development of DEF Airport and its forecast growth in passengers for the next two decades’. Clearly, the Board is taking a long-term view of the airport and its strategy.It is also stated that ‘future development of the airport will be phased and gradual in order not to create unexpected consequences for the local communities and local industry. The development plan has also introduced the concept of ‘Sustainable Aviation’ which is designed to reduce the effects of flying on the environment’ and which restricts night-time flying operations. The mission statement also makes reference to ‘conform(ing) to the highest ethical standards’. These policies indicate the Board’s acceptance of its responsibilities to its various stakeholders rather than solely to its shareholders.Another important environmental aspect influencing the development plan is that the existing four shareholders are local state governments (LSGs) and the Board’s Non-Executive Chairman is always drawn from one of the four LSGs. It is likely, therefore, that the Chairman is a local politician and that the LSGs will be interested in the Airport as a means of promoting their social and economic objectives. The Airport is not listed on a stock exchange and is not, consequently, directly responsible to financial markets.IVBThe culture and objectives of IVB are very different to those of The Board. The manner of working of the Board and its objectives and strategy are likely to be challenged by IVB.IVB, the investment bank, describes itself as ‘the business turnaround experts’. It looks for businesses which have underperformed in terms of profits and changes these businesses by severe cost-cutting which boosts their profitability. Its severe cost-cutting has led to large scale job losses in the businesses it has turned around. At the conclusion of the turnaround process IVB ‘rapidly’ sells its businesses to third parties. IVB is a short-term investor as evidenced by its track record since 2007 whereas the Board, as evidenced by its published mission statement and strategic development plan are long-term investors. The Board has also expressed its concern for 'Sustainability' and its desire to be a 'good corporate citizen'. Neither of these concerns has been expressed by IVB whose sole concern, judging by its track record, is to make good short-term profits for its shareholders; it seemingly has no long-term interest in developing any of the businesses which it buys into.Requirement (b)(ii)The Board of DEF Airport has established six strategic objectives in its development plan and IVB's likely attitude to each of these as follows:1. Create a planning framework which enables DEF Airport to meet the demands of the forecast passenger numbers; This objective does not appear to be too contentious. Such a framework will be a requirement to help the Board deal with the expansion in passenger numbers. It may be the case that IVB will not be too interested in this objective as it is not normally a long-term investor. However, IVB would be interested if the creation of the planning framework impacts upon the Airport's short-term profits due to any increase in costs.2. Reduce to a minimum the visual and audible impacts of the operation of the airport on the local environment; This objective may be open to challenge by IVB as it implies that the Board will have to invest to achieve the objective. IVB’s emphasis has always been on cost-reduction to enable it to boost profitability with a view to a rapid realisation of its investment. Any expenditure by the Board with respect to this objective will depress profit in the short-term and so IVB is likely to oppose this. However, if the objective could be realised successfully in the short-term it might be a factor in IVB achieving a higher price when it resells its investment and so it would not oppose it.3. Ensure that the airport is financially secure; No shareholder would oppose this objective. However, it is very general and the way it is interpreted operationally may lead IVB to challenge it.4. Improvement in land based access to the airport; IVB's attitude to this objective will be similar to its attitude to objective 1. IVB will only be interested in this objective's effect on profitability in the short-term.5. Minimise the pollution effects of the operation of the airport; One feature of the development plan is the requirement for 'Sustainable Aviation' which is reflected in objective5. As a short-term investor IVB's attitude to this will reflect its attitude to objectives 2, 3 and 4.6. Maintain/increase employment opportunities for people living close to the airport; This objective is most likely to be strongly challenged by IVB. IVB's 'raison d'être' is cost-cutting by large-scale job losses in order to increase profitability. This is directly contrary to objective 6 which it will, therefore, oppose.Requirement (b)(iii)If IVB buys the shareholding of one or more of the LSGs, it will become a significant shareholder in DEF Airport and a member of the Board. There are many potential consequences for the Airport which include the loss of a long-standing shareholder (whichever LSG sells out to IVB). The complexion of the Board will change and, as discussed in (b)(ii) above, IVB is likely to challenge some of the strategic objectives established in the development plan. IVB is also likely to propose its own objectives, for example cost-cutting in the form of job losses and sale of the Airport to a third party. However, it should be noted when IVB becomes a shareholder, it will be a minority shareholder, albeit a significant one. Therefore, it will not have control of the Board and cannot insist on anything but must rely on persuasion.An important factor which could have significant consequences will be the price which IVB pays for its shareholding. If the remaining shareholders consider the price paid to be an attractive one they may also consider selling their shareholdings to IVB and so it is possible that IVB could secure control of DEF Airport. In that case, the Board and the Airport would be subject to radical change. IVB would try to implement its ‘turnaround’ philosophy which would involve severe cost-cutting and job losses and a rapid resale.If IVB does not have control it is unlikely that it would accept or strongly support the development plan in its entirety. This could have some positive benefits to DEF Airport as IVB could bring a modernising influence to bear on the Board and its strategy for the Airport. IVB may also be able to introduce the Board to sources of finance which had not been available toit previously.Requirement (c)The expected value of the increased profits resulting from the introduction of cargo handling services is 2% above the criterion of D$200,000 suggested by the Finance Director. However, in the context of a forecast result this difference is marginal. It is not clear in the case how binding or formal the D$200,000 criterion is: the context suggests that this is an informal criterion and so may not be adhered to rigidly.The expected value is dependent upon numerous forecasts, namely:•The theoretical amount of cargo handling income in2010 and its subsequent growth rate •Fixed costs•The three levels of contribution margin•The probability estimatesThe expected value does not give clear accept/reject guidance. It is likely that the decision will be made on other grounds, for example, the Director of Facilities Management’s credibility, alternative projects for profit improvement and the project’s fit with the airport development plan. However, as the project does just satisfy the profitability criterion it is recommended for acceptance.SECTION BAnswer to Question TwoRequirement (a)DLC's control system's usefulnessIn order for DLC to operate a traditional budgetary control system it must engage in forecasting and, following on from this, planning. This then results in a system in which budgeted expenditure is compared to actual expenditure. The system imposes discipline within DLC as regards its future plans and also in the collection of data regarding actual expenditure and the monthly reporting thereon. In order to produce DLC’s budgets there will be internal consultation and discussion which should result in a coherent and co-ordinated operating plan. The targets inherent within this system will also provide a degree of motivation for DLC’s budget managers. DLC also computes its Return On Capital Employed (ROCE) monthly. ROCE attempts to capture the totality of the business’s performance and provides a view of the effectiveness of DLC’s investments.DLC’s control system’s limitationsTraditional control systems have been widely criticised in recent years. Budgeting is frequently done on a ‘Historic Cost plus inflation’ basis which means that it is backward- looking. It is not clear how frequently, if at all, DLC revises its budget estimates, for example, by use of quarterly plans. Neither does DLC appear to flex its budget for actual levels of activity nor take account of seasonal variations in expenditure patterns. Although DLC reports on its sales achievement against budget in aggregate, given the size of DLC’s revenue, $24 million, this is unlikely to yield sufficient information for detailed management control.X relies heavily on the use of ROCE to reflect DLC’s performance. However, a number of flaws are manifest within ROCE. It is arguable that ROCE is not suitable for an organisation like DLC which is not yet mature. DLC will have had start-up costs which would depress ROCE but which will not recur. Therefore, historic comparisons of ROCE achievement are liable to suffer distortion. Another problem arises because of DLC’s substantial level of capital investment. New capital investment will also depress ROCE which might restrict managers’ willingness to invest. However, this would be a mistake as DLC requires substantial and continued capital investment to be successful.DLC has no system for reporting upon the two critical areas of Technological Innovation and Customer Service. This is a serious omission as these two areas are the basis of DLC’s success. They, therefore, should be reported upon explicitly, regularly and contemporaneously. Although ROCE reflects DCL’s success in the areas of Technological Innovation and Customer Service it too is aggregated and probably too late to act as a control measure.The final omission concerns DLC’s customers. DLC is now too large for X to know everything about every customer and contract and she has recognised that this is a weakness. It would be expected that a company like DLC would have systems to report contract and customer profitability. DLC’s lack of these is a weakness.Requirement (b)Although X has not articulated a formal strategy she has an implicit one based upon the following objectives: expansion within the same market/business segment, personal enrichment and the provision of secure well paid jobs for DLC’s staff. It is also implicit that DLC will endeavour to achieve these objectives by excellence in Technological Innovation and Customer Service. As DLC’s control system is exclusively a financial one it is unable to capture the richness of DLC’s business experience and enable it to maintain and improve its performance in the two areas critical for its success.X would, therefore, be assisted in the achievement of her strategic objectives by formalising excellence in Technological Innovation and Customer Service as Critical Success Factors(CSFs) and then developing Key Performance Indicators (KPIs) which would indicate the extent to which the CSFs are being achieved. KPI’s can be both financial and non-financial: a range of non-financial KPI’s could provide a creative measure of feedback about DLC’s performance which is not currently available from the financial control system.The KPI’s should be reported upon within a time-scale which is appropriate to the parameter which is being measured. Thus, for example, it may be appropriate to report against a KPI for customer complaints (which would be supporting the CSF excellent Customer Service) daily. However, a KPI measuring innovation in the installation time for a telephone exchange (which would be supporting the CSF excellent Technological Innovation) would be more appropriately reported upon against a longer timescale. The KPI’s should not be too numerous otherwise X is likely to be overwhelmed by them: she should also continually keep their usefulness under review. If DLC is continually meeting or beating a KPI it may be appropriate to discard this one and introduce a new one so that X may make the best use of her time.X has acknowledged that the lack of control information which she has about sales is a weakness. Non-financial performance measures could address this weakness and contribute towards X achieving her strategic objectives. KPI’s could also be constructed for sales performance as they are for CSF’s.Requirement (c)Currently X does not know the profit which DLC makes from each customer. As the annual revenue in 2010 is $24 million, this is too large for one person to know every detail about it as she did when the business started.One of X’s aims is for her business to continue expanding within the same market/business segment: it is implicit that she wants this expansion to be profitable. In the present circumstances X is not aware which of her current customers are profitable and which ones unprofitable. When DLC expands there is a risk that some new customers will also prove to be unprofitable ones and, possibly, that she might turn away profitable business.Customer Profitability Analysis (CPA) is defined by CIMA as the ‘Analysis of the revenue streams and service costs associated with specific customers or customer groups’. If X introduces CPA to her company it will reveal which customers are profitable and, therefore, should be retained and nurtured. Conversely, CPA will suggest which customers DLC should consider ceasing to do business with. As X wants to expand her business within the same market segment CPA could give insights as to the attractiveness of this segment and guidance as to which new customers and contracts she should accept. CPA could also help X to evaluate whether other market segments may provide more attractive expansion prospects in the future.Answer to Question ThreeRequirement (a)(i)Organic growthJKL has chosen in its strategy to grow organically. It has been influenced in this choice because of its recent experience with an acquisition which resulted in failure. It could be argued that organic growth is less risky than growth by acquisition. This is substantiated by the empirical evidence which demonstrates in the majority of cases when a company is acquired, it is the acquired company that benefits financially to the detriment of the acquiring company. Organic growth is usually achieved by reinvestment of profit which is then applied to the development of the company’s strengths. Therefore, organic growth will happen at a pace commensurate with the organisation’s ability to absorb and benefit from it. However, if organic growth is achieved by reinvestment then the speed of growth will be constrained to an extent by the amount of profits available for reinvestment. Organic growth will be suitable for a company where the culture is one of gradualism rather than radicalism: Evolution rather than Revolution.Requirement (a)(ii)AcquisitionIn many ways, an evaluation of growth by acquisition is the opposite of organic growth. Growth by acquisition can be fast, radical and transformational. It can offer opportunities to a company which otherwise would not be available. Thus, an acquisition target may have unique competences and capabilities, for example, it may own patents, licences and commercial and brand franchises, which are otherwise unavailable. Acquisition gives JKL the possibility of eliminating a competitor. However, the biggest downside to any acquisition is the empirical evidence which demonstrates that most acquisitions do not benefit the acquirer. Requirement (b)1. JKL and LMN had very different accounting and control systems and these had not been satisfactorily combinedAlthough at the start of the acquisition process the accounting and control systems of the two companies will be different they must be made to converge as soon as possible. This is necessary to preserve the integrity of the published financial statements, for example, it would not be acceptable for the two companies to value their stocks on different bases. If managers are subjected to different control systems in the two companies, for example, JKL pays a bonus if a manager meets her budget targets but XYZ does not, there would be motivational problems if this continued. JKL would need to review the two systems, identify the differences and plan for their convergence. The change agent could facilitate this with the help of the Management Accountant.2. JKL and LMN had very different corporate cultures and this had posed many difficulties which were not resolvedIn any acquisition JKL will encounter a different culture to its own in the target company. JKL needs to survey and explicitly recognise the cultural differences and categorise those which are vital for the success of the acquisition. It should then outline an operational strategy to reconcile or accommodate these differences. Achievement against the plan should be regularly reviewed. This process of reconciliation and accommodation could be managed by the change agent.3. JKL had used an autocratic management style to manage the acquisition and this had been resented by the employees of both companiesAn autocratic management style implies that there is no delegation of responsibility: everything is driven from the centre. JKL’s use of this style had led to resentment and this, in turn, would probably lead to resistance. There are a number of alternative management styles that JKL could use, for example, participation, negotiation and facilitation. JKL should use a contingency approach and not assume that one management style will be suitable for all situations and all times.Requirement (c)A change agent is a person, or a group of people, who help an organisation to achieve strategic change. In the matter of JKL acquiring a French company, XYZ the change agent could carry out the following useful tasks:1. Define any possible difficulties presented by the proposed acquisitionOne obvious difficulty presented by the proposed acquisition consists of the potential problems in communication between the two companies as none of JKL’s staff speak fluent French or are able to correspond in French. XYZ is in a similar position as only a small number of its staff speak English fluently: none are able to correspond in English. The change agent would be able to highlight this difficulty for the attention of JKL’s management and to point to the areas of communication which are likely to cause the most problems: for example, phone conversations may be most problematic because of their ephemeral nature, e-mails a lesser problem because of their more permanent nature.2. Examine what causes the problem and diagnose how this can be overcomeIn this particular example, the change agent would have no difficulty in diagnosing the cause of the problem: it is the lack of complementary language skills in both companies. The solution would be to increase the language skills in both English and French in both companies.The more creative role for the change agent in this situation is to propose ways in which this problem could be overcome. Depending on the background of the change agent he/she may have to seek expert advice from a linguist experienced in both the English and French languages. There are a variety of ways to achieve the desired solution that the change agent could propose. One possible way would be for JKL to employ a language tutor to work with both companies.3. Arrive at alternative solutionsIn order to give the management of JKL some alternative ways of overcoming the lack of relevant language skills in the two companies the change agent could suggest alternative solutions, such as:•JKL could specify that all new appointments at senior management level are bi-lingual in French and English.•JKL could use the English speaking members of staff at XYZ to coach their counterparts in JKL in the French language.•JKL could send its senior managers at both companies to external language training.4. Implement solutionsOnce the management of JKL, after consultation with the change agent, have arrived at their preferred solution it will need to be implemented. The change agent could usefully take the lead in this process as he/she will be very well-informed about the preferred solution and the reasons why it has been chosen. If JKL’s chosen solution is to employ a language tutor to work with both companies the change agent could draft the job specification and assist with the recruitment and induction of the tutor.5. Transmit the learning process that allows the organisation to deal with change on an ongoing basis by itself in the futureThe change agent should document the learning process and discussions which JKL has undergone whilst resolving the lack of foreign language skills in both companies. This experience should be disseminated throughout both companies and the change agent should take the lead in this by organising workshops for managers and information sessions for employees.。
2024年3月GESP编程能力认证Python等级考试一级真题(含答案)
2024年3月GESP编程能力认证Python等级考试一级真题(含答案)一、单选题（每题2分，共30分）第1题，小杨的父母最近刚刚给他买了一块华wei手表，他说手表上跑的是鸿meng，这个鸿meng是？（C）A. 小程序B. 计时器C. 操作系统D. 神话人物第2题，中国计算机学会（CCF）在2024年1月27日的颁奖典礼上颁布了王选奖，王选先生的重大贡献是（C）。
A. 制造自动驾驶汽车B. 创立培训学校C. 发明汉字激光照排系统D. 成立方正公司第3题，Python表达式(3-2)*3+5的值是（B）。
A. -13B. 8C. 2D. 0第4题，Python语句print("5%2=",5%2)执行后的输出是（D）。
A. 2 2B. 1 1C. 5%2=2D. 5%2= 1第5题，执行Python语句a=input()时如果输入5+2，下述说法正确的是（B）。
A. 变量a将被赋值为整数7。
B. 变量a将被赋值为字符串，字符串内容为5+2。
C. 语句执行将报错，不能输入表达式。
D. 语句执行将报错，因为input()函数的括号内没有提示字符串作为参数。
第6题，下面Python代码执行后的输出是（B）。
A. a+1= 2B. a+1=2C. 2=2D. 2= 2第7题，下面Python代码执行时输入21后，有关描述正确的是（A）。
A. 仅有代码4行被执行。
B. 第4和第7行代码都被执行。
C. 仅有代码第7行被执行。
D. 第8行代码将被执行，因为input()输入为字符串。
第8题，下面Python代码第2行，总共被执行次数是（D）。
A. 0B. 10C. 19D. 20第9题，下面Python代码执行后的输出是（D）。
A. 0B. 7C. 18D. 20第10题，下面Python代码执行后的输出是（B）。
A. 9#6#3#B. 9#6#3#0#C. 8#7#5#4#2#1#D. 10#8#7#5#4#2#1#第11题，下面Python代码用于判断键盘输入的整数是否为质数。
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Algorithm Design and Analysis 华南理工大学期末考试
《计算机图形学与虚拟现实》试卷A
注意事项：1. 考前请将密封线内填写清楚；
2. 所有答案请直接答在试卷上(或答题纸上)； 3．考试形式：闭卷；
4. 本试卷共 5大题，满分100分;考试时间95分钟(2:00~3:35pm)。
题 号 1 2 3
4
5
6
7
8 总分 得 分 评卷人
1. (24分)名词解释.
(a) 线段光栅化(Line Segmentation Rasterization)
答:在光栅显示设备上输出数学表示的直线段时，需要先转换成点阵表示，这就是所谓的光栅转换。
线段光栅化算法有DDA 和Bresenham 等算法。
(b) Phong 光照模型(Phong Lighting Model)
答:Phong 光照模型定义了曲面上点的颜色计算方式。
它包含环境光、漫反射和高光反射三个部分。
(c) 纹理映射(Texture Mapping)
答:在真实感绘制中，为了描述模型的细节特征而引入的一种技术。
通过把图象中的颜色解释为模型的几何属性来达到细节绘制的效果。
纹理映射主要包括凹凸映射、法向映射和位移映射。
(d) 灭点(Vanishing Point)
答:一组与投影面不平行的平行线被投影到投影面上时会交于一点，这个交点称为灭点.与坐标轴平行的平行线的灭点称为主灭点。
2. (16分)3D 模型从建模到显示设备输出需要经过多次几何变换. 请描述一个典型的三维变换流程. 答：右图所示为一个
典型的三维变换 流程
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姓名 学号
学院 专业 座位号
( 密 封 线 内 不 答 题 ) ………………………密………………………………………………封………………………………………线……………………………………
Algorithm Design and Analysis
3. (15分)在图1所示的网格中, i v 表示网格的顶点, i f 表示三角形面, ij e 表示半边. 请完成下表.
半边
起点
相邻 半边
所在面
下条半边
e 3,1 e 3,2 e 4,1 e 4,2
解:
半边
起点
相邻 半边
所在面 下条半边
e 3,1
3v
e 3,2
f 1 e 1,1 e 3,2 2v e 3,1 f 2 e 4,1 e 4,1 3v e 4,2 f 2 e 5,1 e 4,2 5v
e 4,1
f 3
e 6,1
4. (15分) Bresenham 画线算法中, 一旦当前象素点(x , y )已经确定, 对线段斜率在0,1之间的情形, 下一个象素点只能是(x +1, y )和(x +1, y +1)中的一个. 算法利用这两个象素点到直线段的垂直方向上距离之差来决定取其中的一个. 利用Bresenham 画线算法给出直线段(3,4)—(8,7)的光栅扫描转换。
请给出计算过程，并在图2中画出相应的象素（注: 图2中水平和竖直线段的交点为象素中心）。
解：Bresenham 画线算法要考虑前一点的选取情况，再确定下一个点，因此更为简洁. 按照按算法可得：
(1)x ∆=8-3=5，y ∆=7-4=3。
(2)直线斜率：m=y ∆/x ∆=0.6<1。
(3)2y ∆=6,x y ∆∆2-2=6-10=-4。
(4)初始点)4,3(),(00=y x ,由x y p ∆-∆=20=1>0 可知
下一个点是初始点左上角的点=),(11y x )5,4()1,1(00=++y x 。
图2
图1
Algorithm Design and Analysis (5)根据递推式:
（i ）0>i p 时,取点)1,1(),(11++=++i i i i y x y x ,)(21x y p p i i ∆-∆+=+; (ii) 0<i p 时,取点),1(),(11i i i i y x y x +=++,y p p i i ∆+=+21. 可得计算过程如下表所示:
扫描转换结果如右上图所示.
视点位置为E =(0,6.25,10)
5. (30分) 构造世界坐标系中场景如下:
(a) 有一点光源, 其位置为O =(0,6.25,−10), 入射光光强I e =0.5; 环境光I a =0.2I e .
(b) 视点位置为E =(0,6.25,10)；
(c)
曲面的环境反射、漫反射和高光反射系数分别为1: k a
d s 0.5, 高光反射衰减速指数n =2.
(1) 写出曲面上参数位置(u,v) 处的法向量的计算过程；
(2) 用Phong 光照模型的近似计算公式计算视点位置观察到的曲面上点求B (u =0.5,v =0.5)的颜色(光强)值。
提示1：二次Bézier 曲面
B(u,v)=∑∑P ij 2j=02i=0B i,2(u)B j,2(v), B i,2(u )=C 2i u i (1−u)
2−i
. 的矩阵表示
B (u,v )=(u 2,u,1)MPM T (v 2,v,1)T , M =(1−21−220100
).
MPM T =((0,20,0)(0,−20,0)(0,0,0)
(0,−20,0)(0,20,0)
(10,0,0)(0,0,0)(0,0,−10)(−5,−5,5)
)
提示2：Phong光照模型的近似计算公式I=k a I a+k d I e N∙L+k s I e(N∙H)n, 其中H是光源射线与视点射线的平分线方向.
解：(1)根据矩阵表示有
∂B(u,v)
ðu
=(2u,1,0)MPM T(v2,v,1)T
∂B(u,v)
ðv
=(u2,u,1)MPM T(2v,1,0)T
从而该点处的法向为
N=∂B(u,v)
ðu ×∂B(u,v)
ðv
，
最后归一化得: N←N/‖N‖.
(3)Q= B(0.5,0.5)=(0.25,0.5,1)((0,20,0)(0,−20,0)(0,0,0)
(0,−20,0)(0,20,0)(10,0,0) (0,0,0)(0,0,−10)(−5,−5,5))(
0.25
0.5
1
)
=(0,−3.75,0)
∂B(u,v)
ðu |
(u,v)=(0.5,0.5)
=(10,0,0)
∂B(u,v)
ðv |
(u,v)=(0.5,0.5)
=(0,0,−5)
Q点处的法向量N=(10,0,0)×(0,0,−5)=(0,50,0),归一化后N=(0,1,0)光线方向:L=O−Q=(0,6.25,−10)−(0,−3.75,0)=(0,10,−10),归一化
L=(0,√2
2
,−
√2
2
)
视线方向:V=E−Q=(0,6.25,10)−(0,−3.75,0)=(0,10,10), 归一化
V=(0,√2
2
,
√2
2
)
Algorithm Design and Analysis
H=L+V
2
=(0,√2,0), 归一化H=(0,1,0).
N←N/‖N‖
因此该点(Q)处的光强值
I =k a I a+k d I e N∙L+k s I e(N∙H)n
=0.01+0.5(0,1,0)∙(0,√2
2,−√2
2
)+0.5×0.5((0,1,0)∙(0,1,0))2
=0.6135.
Algorithm Design and Analysis。