試題1

以下關于總線的敘述中，不正確的是（）
A．并行總線適合近距離高速數據傳輸．
B．串行總線適合長距離數據傳輸
C．單總線結構在一個總線上適應不同種類的設備，設計簡單且性能很高
D.專用總線在設計上可以與連接設備實現最佳匹配

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試題2

（ ）是多媒體內容描述接口標準。
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

設求解某問題的遞歸算法如下：
F(int n){
if n==1 {
Move(1)
}else{
F(n-1);
Move(n);
F(n-1);
}
}
求解該算法的計算時間時,僅考慮算法Move所做的計算為主要計算,且Move 為常數級算法。則算法F的計算時間T(n)的遞推關系式為（1） ；設算法Move的計算時間為k，當n=4 時，算法F的計算時間為（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都訪問相同的全局變量和數據結構，則這兩個模塊之間的耦合類型為（ ）耦合。



A. 公共
B．控制
C．標記
D．數據

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