試題1

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試題2

設(shè)用 2K×4 位的存儲(chǔ)器芯片組成 16K×8 位的存儲(chǔ)器（地址單元為 0000H～3FFFH，每個(gè)芯片的地址空間連續(xù)），則地址單元 0B1FH 所在芯片的最小地址編號(hào)為  （ ）  。
A、0000H
B、0800 H
C、2000 H
D、2800

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試題3

若某個(gè)計(jì)算機(jī)系統(tǒng)中，內(nèi)存地址與I/O地址統(tǒng)一編址，訪問內(nèi)存單元和I/O設(shè)備是靠（ ）采區(qū)分的。
A、數(shù)據(jù)總線上輸出的數(shù)據(jù) 
B、不同的地址代碼 
C、內(nèi)存與I/O設(shè)備使用不同的地址總線 
D、不同的指令 

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試題4

在進(jìn)行面向?qū)ο笤O(shè)計(jì)時(shí)，采用設(shè)計(jì)模式能夠（ ）。
A、復(fù)用相似問題的相同解決方案
B、改善代碼的平臺(tái)可移植性
C、改善代碼的可理解性
D、增強(qiáng)軟件的易安裝性

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試題5

算術(shù)表達(dá)式“(a-b)*(c+d)”的后綴式是( )。
A．a(chǎn)b-cd+*
B．a(chǎn)bcd-*+
C．a(chǎn)b-*cd+
D．a(chǎn)b-c+d*

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試題6

Ravi, like many project （1） , had studied the waterfall model of software development as the primary software life-cycle （2）. He was all set to use it for an upcoming project, his first assignment. However, Ravi found that the waterfall model could not be used because the customer wanted the software delivered in stages, something that implied that the system had to be delivered and built in （3） and not as （4）.
The situation in many other projects is not very different. The real world rarely presents a problem in which a standard process, or the process used in a previous project, is the best choice. To be the most suitable, an existing process must be （5） to the new problem.
A development process, even after tailoring, generally cannot handle change requests. To accommodate change requests without losing control of the project, you must supplement the development process with a requirement change management process.
（1） A. customers
B. managers 
C. users  
D. administrators
（2）A. activity   
B. procedure 
C. process   
D. progress
（3） A. parts 
B. modules  
C. software  
D. a whole
（4） A. parts
B. modules  
C. software  
D. a whole
（5） A. modified 
B. used 
C. suited
D. tailored

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試題7

RUP 在每個(gè)階段都有主要目標(biāo)，并在結(jié)束時(shí)產(chǎn)生一些制品。在 （ ） 結(jié)束時(shí)產(chǎn)生“在適當(dāng)?shù)钠脚_(tái)上集成的軟件產(chǎn)品” 。
A、初期階段
B、精化階段
C、構(gòu)建階段
D、移交階段

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試題8

設(shè)有關(guān)系模式R（A₁,A₂,A₃,A₄,A₅,A₆），其中：函數(shù)依賴集F={A₁→A₂，A₁A₃→A₄,A₅A₆→A₁,A₂A₅→A₆,A₃A₅→A₆},則（1 ）是關(guān)系模式R的一個(gè)主鍵，R規(guī)范化成都最高達(dá)到（2 ）。
（1）A.A₁A₄
B.A₂A₄
C.A₃A₅
D.A₄A₅
（2）A. 1NF
B. 2NF
C. 3NF
D. BCNF

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試題9

Why is(1)fun? What delights may its practitioner espect his reward? First is the
sheer joy of making things.As the child delights in his mud pie,so the adult enjoys
building things,especially things of his own design.Secong is the pleasure of making
things that are useful to other people.Third is the fascinanon of fashioning complex
puzzle-like objects  of interlocking moving  parts and watching them  work in subtle
eyeles,playing out the consequences of principies built in from the beginning.Fourth
is the joy of always learning,which springs from the(2)nature of the task.In one way
or    another    the    problem    is    ever    new,and    its    solver    learns
something:sometimes(3),sometimes  theoretical,and  sometimes both.Finally,there  is
the delight of working in such a tractable medium.The(4),like the poet,works only
slightly removed from nure thought-stuff.Few media of ereation are so flexible,so easy
to polish and rework,so readily capable of realizing grand conceptual structures.
Yet the program(5),unlike the poet’s words,is real in the sense that it moves and
works,producing  visible  outputs  separate  from  the  comstrct  itself.It  prints
results,draws pictures,produces sounds,moves arms.Progamming then is fun because it
gratifies creative longings built deep within us and delights sensibities we hav e in
common with all men.
（1）A、programming
B、composing
C、working
D、writing
（2）A、repeating
B、basic
C、non-repeating
D、advance
（3）A、semantic
B、practical
C、lexical
D、syntactical
（4）A、poet
B、architect
C、doctor
D、programmer
（5）A、construct
B、code
C、size
D、scale

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試題10

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