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軟件設計師每日一練試題（2026/3/18）

來源：信管網 2026年03月19日【所有評論】

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軟件設計師每日一練試題內容（2026/3/18）

試題1
線性表采用鏈表存儲結構的特點中不包括（ )。
A．所需空間大小與表長成正比
B．可隨機訪問表中的任—元素
C．插入和刪除操作不需要移動元素
D．無須事先估計存儲空間大小
查看答案
試題參考答案：B
試題解析與討論：www.lyhuahuisp.com/exam/ExamDay.aspx?t1=4&day=2026/3/18
試題2
You are developing a sevver-side enterprise application. It must support a variety of different clients including desktop browsers , mobile browsers and native mobile applications. The application might also expose an API for 3rd parties to customer. It might also（）with other applications via either web services or a message broker. The application handles requests（HTTP requests and messages）by executing business logic；accessing a databse；exchanging messages with other systems；and returning a HTML /JSON/XML （） . There are logical components corresponding to different functional areas of the application.
What's the application' s deployment architecture?
Define an architecture that structures the application as a set of （），collaborating services. This approach corresponds to the Y-axis of the ScaleCube. Each service is；
Flighly maintainable and testable-enables rapid and frequent development and deployment.
Loosely coupled with other services-enables a team to work independently（the majority of time on their servicers）without being impouted by changes to other services and without affecting other services.
（） deployable-enable a team to deploy their services without having to cortdinate with other teams.
Capable of being developed by a small team-essential for high productivity by avoiding the high communication head of large teams.
Services （） using either synchronous protocols such as HTTP/REST or a synchronous protocols such as AMQP. Services can be developed and deployed independently of one another. Each service has its own database in order to be decoupled from other services. Data consistency between services is maintained using some particular pattern.
（1）A、integrate
B、Coordinate
C、cooperate
D、Communicate
（2）A、request
B、response
C、text
D、File
（3）A、loosely coupled
B、loosely cohesion
C、High coupled
D、Highly cohesion
（4）A、Dependently
B、Independently
C、Coordinately
D、Integratedly
（5）A、interoprate
B、coordinate
C、communicate
D、depend
查看答案
試題參考答案：A、B、A、B、C
試題解析與討論：www.lyhuahuisp.com/exam/ExamDay.aspx?t1=4&day=2026/3/18
試題3
Low-code and no code software development solutions have emerged as viable and convenient alternatives to the traditional development process.
Low-code is a rapid application development (RAD) approach that enables automated code generation through(71) building blocks like drag-and-drop and pull-down menu interfaces.This (72) allows low-code users to focus on the diferentiator rather than the common denominator of programming. Low-code is a balanced middle ground between manual coding and no-code as its users can still add code over auto-generated code.While in low-code there is some hand holding done by developers in the form of scripting or manual coding， no-code has a completely (73) approach, with 100% dependence on visual tools.
A low-code application platform (LCAP)——also called a low-code development platform (LCDP)——contains an integrated development environment (IDE) with (74) features like APIs, code templates, reusable plug-in modules and graphical connectors to automate a significant percentage of the application development process.LCAPs are typically available as cloud-based Platform-as-a-Service (PaaS) solutions.
A low-code platform works on the principle or lowering complexity by using visual to and techniques like process modeling. where users employ visual tools to define workflobusiness rules, user interfaces and the like.Behind the scenes, the complete workflovautomatically converted into code. LCAPs are used predominantly by professional developeautomate the generic aspects of coding to redirect effort on the last mile of (75).
(71)
A．visual
B．component-based
C．object-oriented
D．structural
(72)
A．block
B．automation
C．function
D．method
(73)
A．modern
B．hands-off
C．generic
D．labor-free
(74)
A．reusable
B．built-in
C．existed
D．well-known
(75)
A．delivery
B．automation
C．development
D．success
查看答案
試題參考答案：A、B、B、B、C
試題解析與討論：www.lyhuahuisp.com/exam/ExamDay.aspx?t1=4&day=2026/3/18
試題4
良好的啟發(fā)式設計原則上不包括（）。
A.提高模塊獨立性
B.模塊規(guī)模越小越好
C.模塊作用域在其控制域之內
D.降低模塊接口復雜性
查看答案
試題參考答案：B
試題解析與討論：www.lyhuahuisp.com/exam/ExamDay.aspx?t1=4&day=2026/3/18
試題5

在發(fā)布一訂閱(Publish-Subscribe)消息模型中，訂閱者訂閱一個主題后，當該主題有新消息到達時，所有訂閱者都會收到通知。( )設計模式最適合這一模型。
A．適配器(Adapter)
B．通知(Notifier)
C．狀態(tài)(State)
D．觀察者(Observer)
查看答案
試題參考答案：D
試題解析與討論：www.lyhuahuisp.com/exam/ExamDay.aspx?t1=4&day=2026/3/18
試題6

假定某企業(yè)2014年5月的員工工資如下表所示：

(54)
A.部門，AVG（應發(fā)工資）AS平均工資
B.姓名，AVG（應發(fā)工資）AS平均工資
C.部門，平均工資ASAVG（應發(fā)工資）
D.姓名，平均工資ASAVG（應發(fā)工資）
(55)
A.ORDERBY姓名
B.ORDERBY部門
C.GROUPBY姓名
D.GROUPBY部門
(56)
A.WHERECOUNT(姓名)>2
B.WHERECOUNT(DISTINCT(部門)）2
C.HAVINGCOUNT(姓名)>2
D.HAVINGCOUNT(DISTINCT(部門)）>2
查看答案
試題參考答案：A、D、C
試題解析與討論：www.lyhuahuisp.com/exam/ExamDay.aspx?t1=4&day=2026/3/18
試題7

如果兩名以上的申請人分別就同樣的發(fā)明創(chuàng)造申請專利，專利權應授予（）。
A、最先發(fā)明的人
B、最先申請的人
C、所有申請人
D、協商后的申請人
查看答案
試題參考答案：B
試題解析與討論：www.lyhuahuisp.com/exam/ExamDay.aspx?t1=4&day=2026/3/18
試題8

下面關于漏洞掃描系統(tǒng)的敘述，錯誤的是（）。
A、漏洞掃描系統(tǒng)是一種自動檢測目標主機安全弱點的程序
B、黑客利用漏洞掃描系統(tǒng)可以發(fā)現目標主機的安全漏洞
C、漏洞掃描系統(tǒng)可以用于發(fā)現網絡入侵者
D、漏洞掃描系統(tǒng)的實現依賴于系統(tǒng)漏洞庫的完善
查看答案
試題參考答案：C
試題解析與討論：www.lyhuahuisp.com/exam/ExamDay.aspx?t1=4&day=2026/3/18
試題9

（）的鄰接矩陣是一個對稱矩陣。
A、無向圖
B、AOV 網
C、AOE 網
D、有向圖
查看答案
試題參考答案：A
試題解析與討論：www.lyhuahuisp.com/exam/ExamDay.aspx?t1=4&day=2026/3/18
試題10
以下關于管道——過濾器軟件體系結構風格優(yōu)點的敘述中，不正確的是( )
A．構件具有良好的高內聚、低耦合的特點
B．支持軟件復用
C．支持并行執(zhí)行
D．適合交互處理應用
查看答案
試題參考答案：D
試題解析與討論：www.lyhuahuisp.com/exam/ExamDay.aspx?t1=4&day=2026/3/18