试题1

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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试题2

系统中有R类资源m个，现有n个进程互斥使用。若每个进程对R资源的最大需求为w，那么当m、n、w分别取下表中的值时，对于表中的①～⑥种情况，  （1） 可能会发生死锁。若将这些情况的m分别加上  （2）  ，则系统不会发生死锁。

（1）A．①②⑤
B．③④⑤
C．②④⑤
D．②④⑥
（2）A．1、1和1
B．1、1和2
C．1、1和3
D．1、2和1

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试题3

在软件开发过程中，系统测试阶段的测试目标来自于（）阶段，
A．需求分析
B．概要设计
C．详细设计
D．软件实现

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试题4

下图所示平衡二叉树（树中任一结点的左右子树高度之差不超过1）中，结点A的右子树AR高度为h，结点B的左子树BL高度为h，结点C的左子树CL、右子树CR高度都为h-1。若在CR中插入一个结点并使得CR的高度增加1，则该二叉树 （ ） 。

A、以B为根的子二叉树变为不平衡
B、以C为根的子二叉树变为不平衡
C、以A为根的子二叉树变为不平衡
D、仍然是平衡二叉树

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试题5

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试题6

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试题7

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试题8

与多模光纤相比较，单模光纤具有（ ）等特点。
A、较高的传输率、较长的传输距离、较高的成本
B、较低的传输率、较短的传输距离、较高的成本
C、较高的传输率、较短的传输距离、较低的成本
D、较低的传输率、较长的传输距离、较低的成本

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试题9

某个计算机中心有28台微机，每台微机有24个应用，每个应用占用1个端口地址，则这个计算机中心所有应用的地址总数为（ ）。
A、24 
B、28 
C、52 
D、672 

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试题10

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