試題1

以下關(guān)于總線的敘述中，不正確的是（）
A．并行總線適合近距離高速數(shù)據(jù)傳輸．
B．串行總線適合長(zhǎng)距離數(shù)據(jù)傳輸
C．單總線結(jié)構(gòu)在一個(gè)總線上適應(yīng)不同種類的設(shè)備，設(shè)計(jì)簡(jiǎn)單且性能很高
D.專用總線在設(shè)計(jì)上可以與連接設(shè)備實(shí)現(xiàn)最佳匹配

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試題2

（ ）是多媒體內(nèi)容描述接口標(biāo)準(zhǔn)。
A、MPEG-1 
B、MPEG-2 
C、MPEG-4 
D、MPEG-7 

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試題3

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試題4

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試題5

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試題6

In a world where it seems we already have too much to do, and too many things to think about, it seems the last thing we need is something new that we have to learn.
But use cases do solve a problem with requirements: with （1 ） declarative requirements it's hard to describle steps and sequences of events.
Use cases, stated simply, allow description of sequences of events that, taken together, lead to a system doing something useful. As simple as this sounds, this is important. When confronted only with a pile of requiements, it's often（2 ）to make sense of what the authors of the  requirements really wanted the system to do. In the preceding example, use cases reduce the ambiguity of the requirements by specifying exactly when and under what conditions certain behavior occurs; as such, the sequence of the behaviors can be regarded as a requirement. Use cases are particularly well suited to capture approaches. Although this may sound simple, the fact is that (3 ) requirement capture approaches, with their emphasis on declarative requirements and "shall" statements, completely fail to capture fail to capture the (4 ) of the system's behavior. Use cases are a simple yet powerful way to express the behavior of the system in way that all stakeholders can easily understand.
But, like anything, use cases come with their own problems, and as useful as they are, they can be (5 ). The result is something that is as bad, if not worse, that the original problem. Therein it's important to utilize use cases effectively without creating a greater problem than the one you started with.
（1）A.plenty
B.loose
C.extra
D.strict
（2）A.impossible
B.possible
C.sensible
D.practical
（3）A.modern
B.conventional
C.different
D.formal
（4）A.statics
B.nature
C.dynamics
D.originals
（5）A.misapplied
B.applied
C.used
D.powerful

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試題7

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試題8

設(shè)求解某問(wèn)題的遞歸算法如下：
F(int n){
if n==1 {
Move(1)
}else{
F(n-1);
Move(n);
F(n-1);
}
}
求解該算法的計(jì)算時(shí)間時(shí),僅考慮算法Move所做的計(jì)算為主要計(jì)算,且Move 為常數(shù)級(jí)算法。則算法F的計(jì)算時(shí)間T(n)的遞推關(guān)系式為（1） ；設(shè)算法Move的計(jì)算時(shí)間為k，當(dāng)n=4 時(shí)，算法F的計(jì)算時(shí)間為（2） 。
（1）A、T(n)=T(n-1)+1
B、T(n)=2T(n-1)
C、T(n)=2T(n-1)+1
D、T(n)=2T(n+1)+1
（2）A、14k
B、15k
C、16k
D、17k

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試題9

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試題10

如下圖所示，模塊A和模塊B都訪問(wèn)相同的全局變量和數(shù)據(jù)結(jié)構(gòu)，則這兩個(gè)模塊之間的耦合類型為（ ）耦合。



A. 公共
B．控制
C．標(biāo)記
D．?dāng)?shù)據(jù)

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