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信息安全工程師每日一練試題(2024/3/6)

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信息安全工程師每日一練試題內(nèi)容(2024/3/6)

  • 試題1

    從安全屬性對各種網(wǎng)絡(luò)攻擊進(jìn)行分類,阻斷攻擊是針對()的攻擊
    A.機(jī)密性
    B.可用性
    C.完整性
    D.真實性

    查看答案

    試題參考答案:B

    試題解析與討論:www.jazzmuze.com/st/3271121583.html

  • 試題2

    Linux系統(tǒng)的運(yùn)行日志存儲的目錄是(  )。
    A./var/log
    B./usr/log
    C./etc/log
    D./tmp/log

    查看答案

    試題參考答案:A

    試題解析與討論:www.jazzmuze.com/st/3894728881.html

  • 試題3

    以下關(guān)于VPN的敘述中,正確的是()
    A、VPN指的是用戶通過公用網(wǎng)絡(luò)建立的臨時的、安全的連接
    B、VPN指的是用戶自己租用線路,和公共網(wǎng)絡(luò)物理上完全隔離的、安全的線路
    C、VPN不能做到信息認(rèn)證和身份認(rèn)證
    D、VPN只能提供身份認(rèn)證,不能提供數(shù)據(jù)加密的功能

    查看答案

    試題參考答案:A

    試題解析與討論:www.jazzmuze.com/st/285048692.html

  • 試題4

    計算機(jī)網(wǎng)絡(luò)為了實現(xiàn)資源共享,采用協(xié)議分層設(shè)計思想,每層網(wǎng)絡(luò)協(xié)議都有地址信息,如網(wǎng)卡(MAC)地址、IP地址、端口地址和域名地址,以下有關(guān)上述地址轉(zhuǎn)換的描述錯誤的是 (   )。
    A.DHCP協(xié)議可以完成IP地址和端口地址的轉(zhuǎn)換
    B.DNS協(xié)議可以實現(xiàn)域名地址和IP地址之間的轉(zhuǎn)換
    C.ARP協(xié)議可以實現(xiàn)MAC地址和IP地址之間的轉(zhuǎn)換
    D.域名地址和端口地址無法轉(zhuǎn)換

    查看答案

    試題參考答案:A

    試題解析與討論:www.jazzmuze.com/st/522428592.html

  • 試題5

    外部網(wǎng)關(guān)協(xié)議BGP是不同自治系統(tǒng)的路由器之間交換路由信息的協(xié)議,BGP-4使用四種報文:打開報文、更新報文、?;顖笪暮屯ㄖ獔笪摹F渲杏脕泶_認(rèn)打開報文和周期性地證實鄰站關(guān)系的是(  )。
    A.打開報文
    B.更新報文
    C.?;顖笪?br />D.通知報文

    查看答案

    試題參考答案:C

    試題解析與討論:www.jazzmuze.com/st/411084710.html

  • 試題6

    一個密碼系統(tǒng)如果用E表示加密運(yùn)算,D表示解密運(yùn)算,M表示明文,C表示密文,則下面描述必然成立的是(  )。
    A. E(E(M))=C
    B. D(E(M))=M
    C. D(E(M))=C
    D. D(D(M))=M

    查看答案

    試題參考答案:B

    試題解析與討論:www.jazzmuze.com/st/389293064.html

  • 試題7

    安全策略表達(dá)模型是一種對安全需求與安全策略的抽象概念模型,一般分為自主訪問控制模型和強(qiáng)制訪問控制模型。以下屬于自主訪問控制模型的是()
    A. BLP模型
    B. HRU模型
    C. BN模型
    D.基于角色的訪問控制模型

    查看答案

    試題參考答案:B

    試題解析與討論:www.jazzmuze.com/st/3272020263.html

  • 試題8

    Kerberos 是一個網(wǎng)絡(luò)認(rèn)證協(xié)議,其目標(biāo)是使用密鑰加密為客戶端/服務(wù)器應(yīng)用程序提供強(qiáng)身份認(rèn)證。以下關(guān)于 Kerberos 的說法中,錯誤的是( )。
    A.通常將認(rèn)證服務(wù)器AS 和票據(jù)發(fā)放服務(wù)器TGS統(tǒng)稱為 KDC
    B.票據(jù)(Ticket)主要包括客戶和目的服務(wù)方 Principal、客戶方IP 地址、時間戳、Ticket 生存期和會話密鑰
    C.Kerberos利用對稱密碼技術(shù),使用可信第三方為應(yīng)用服務(wù)器提供認(rèn)證服務(wù)
    D.認(rèn)證服務(wù)器AS為申請服務(wù)的用戶授予票據(jù)

    查看答案

    試題參考答案:D

    試題解析與討論:www.jazzmuze.com/st/5737026640.html

  • 試題9

    數(shù)字簽名是對以數(shù)字形式儲存的消息就行某種處理,產(chǎn)生一種類似于傳統(tǒng)手書簽名功效的消息處理過程,一個數(shù)字簽名體制通常包括兩個部分,()
    A.施加簽名和驗證簽名
    B.數(shù)字證書和身份認(rèn)證
    C.身份消息加密和解密
    D.數(shù)字證書和消息摘要

    查看答案

    試題參考答案:A

    試題解析與討論:www.jazzmuze.com/st/3270223694.html

  • 試題10

    Trust is typically interpreted as a subjective belief in the reliability, honesty and  security  of an entity on which we depend ( )our welfare .In online environments we depend on a wide spectrun of things , ranging from computer hardware,software and data to people and organizations. A security solution always assumes certain entities function according to specific policies.To trust is precisely to make this sort of assumptions , hence , a trusted entity is the same as an entity that is assumed to function according to  policy . A consequence of this is that a trust component of a system must work correctly in order   for the security of that system to hold, meaning that when a trusted(  )fails , then the sytems and applications that depend on it can(  )be considered secure.An often cited articulation of this principle is:" a trusted system or component is one that can break your security policy” ( which happens when the trust system fails ). The same applies to a trusted party such as a service provider ( SP for short )that is , it must operate according to the agreed or assumed   policy in order to ensure the expected level of securty and quality of services . A paradoxical   conclusion to be drawn from this analysis is that security assurance may decrease when increasing the number of trusted components and parties that a service infrastructure depends on . This is because the security of an infrastructure consisting of many.
    Trusted components typically follows the principle of the weakest link , that is ,in many situations the the overall security can only be as strong as the least reliable or least secure of all the trusted components. We cannot avoid using trusted security components,but the fewer the better. This is important to understand when designing the  identity management architectures,that is, fewer the trusted parties in an identity management model , stronger the security that can be achieved by it.
    The transfer of the social constructs of identity and trust into digital and computational concepts helps in designing and implementing large scale online markets and communities,and also plays an important role in the converging mobile and Internet environments.Identity management (denoted Idm hereafter ) is about recognizing and verifying the correctness of identitied in online environment .Trust management becomes a component of (  )whenever different parties rely on each other for identity provision and authentication . IdM and Trust management therefore depend on each other in complex ways because the correctness of the identity itself must be trusted for the quality and reliability of the corresponding entity to be trusted.IdM is also an essential concept when defining  authorisation policies in personalised services.
    Establishing trust always has a cost, so that having  complex trust requirement typically leads to high overhead in establishing the required trust. To reduce costs there will be incentives for stakeholders to “cut corners”regarding trust requirements ,which could lead to inadequate security . The challenge is to design IdM systems with relatively simple trust requirements.Cryptographic mechanisms are often a core component of IdM solutions,for example,for entity and data authentication.With cryptography,it is often possible to propagate trust from where it initially exists to where it is needed .The establishment of initial(  )usually takes place in the physical world,and the subsequent propagation of trust happens online,often in an automated manner.
    (71)A.with
    B. on
    C. of
    D. for
    (72)A.entity
    B.person
    C.component
    D.thing
    (73)A. No longer
    B. never
    C. always
    D.often
    (74)A. SP
    B. IdM
    C.Internet
    D.entity
    (75)A.trust
    B.cost
    C.IdM
    D. solution

    查看答案

    試題參考答案:D、C、A、B、A

    試題解析與討論:www.jazzmuze.com/st/389944612.html

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