試題1

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試題2

（ ）  不能保證求得 0-1 背包問(wèn)題的最優(yōu)解。
A、分支限界法
B、貪心算法
C、回溯法
D、動(dòng)態(tài)規(guī)劃策略

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試題3

一個(gè)類是  （1）。在定義類時(shí)，將屬性聲明為private的目的是  （2）。
（1） A．一組對(duì)象的封裝
B．表示一組對(duì)象的層次關(guān)系
C．一組對(duì)象的實(shí)例
D．一組對(duì)象的抽象定義
（2）A．實(shí)現(xiàn)數(shù)據(jù)隱藏，以免意外更改
B．操作符重載
C．實(shí)現(xiàn)屬性值不可更改
D．實(shí)現(xiàn)屬性值對(duì)類的所有對(duì)象共享

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試題4

兩個(gè)同符號(hào)的數(shù)相加或異符號(hào)的數(shù)相減，所得結(jié)果的符號(hào)位SF和進(jìn)位標(biāo)志CF進(jìn)（ ）運(yùn)算為1時(shí)，表示運(yùn)算的結(jié)果產(chǎn)生溢出。
A、與
B、或
C、與非
D、異或

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試題5

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6

在面向?qū)ο筌浖_發(fā)過(guò)程中，采用設(shè)計(jì)模式（ ）。
A、允許在非面向?qū)ο蟪绦蛟O(shè)計(jì)語(yǔ)言中使用面向?qū)ο蟮母拍? 
B、以復(fù)用成功的設(shè)計(jì)和體系結(jié)構(gòu) 
C、以減少設(shè)計(jì)過(guò)程創(chuàng)建的類的個(gè)數(shù)  
D、以保證程序的運(yùn)行速度達(dá)到最優(yōu)值

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試題7

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試題8

設(shè)某語(yǔ)言的語(yǔ)法規(guī)則用上下文無(wú)關(guān)文法 G=(N,T,P,S)表示，其中 N 是非終結(jié)符號(hào)的集合，T是終結(jié)符號(hào)的集合，P 是產(chǎn)生式集合，S 是開始符號(hào)，令 V=N∪T，那么符合該語(yǔ)言的句子是（ ） 。
A、從 S 出發(fā)推導(dǎo)的、僅包含 T 中符號(hào)的符號(hào)串
B、從 N 中符號(hào)出發(fā)推導(dǎo)的、僅包含 T 中符號(hào)的符號(hào)串
C、從 S 出發(fā)推導(dǎo)的、包含 V 中符號(hào)的符號(hào)串
D、從 N 中符號(hào)出發(fā)推導(dǎo)的、包含 V 中符號(hào)的符號(hào)串

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試題9

冗余技術(shù)通常分為 4 類，其中  （ ）  按照工作方法可以分為靜態(tài)、動(dòng)態(tài)和混合冗余。
A、時(shí)間冗余
B、信息冗余
C、結(jié)構(gòu)冗余
D、冗余附加技術(shù)

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試題10

模塊A直接訪問(wèn)模塊B的內(nèi)部數(shù)據(jù)，則模塊A和模塊B的耦合類型為（）。
A．?dāng)?shù)據(jù)耦合
B．標(biāo)記耦合
C．公共耦合
D．內(nèi)容耦合

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