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軟件設(shè)計(jì)師每日一練試題內(nèi)容(2023/12/30)
試題1
在IE瀏覽器中,安全級(jí)別最高的區(qū)域設(shè)置是()。
A.Internet
B.本地Intranet
C.可信站點(diǎn)
D.受限站點(diǎn)
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試題參考答案:D
試題解析與討論:www.xcpkj.com/st/246123503.html
試題2
設(shè)一個(gè)包含n個(gè)頂點(diǎn)、e條弧的簡(jiǎn)單有向圖采用鄰接矩陣存儲(chǔ)結(jié)構(gòu)(即矩陣元素A[i][j]團(tuán)等于1或0,分別表示頂點(diǎn)i與頂點(diǎn)j之間有弧或無(wú)?。?則該矩陣購(gòu)非零元素?cái)?shù)目為( )。
A. e
B.2e
C.n-e
D.n+e
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試題參考答案:A
試題解析與討論:www.xcpkj.com/st/2529212674.html
試題3
分布式數(shù)據(jù)庫(kù)系統(tǒng)中, ( )是指用戶無(wú)需知道數(shù)據(jù)存放的物理位置。
A.分片透明
B.復(fù)制透明
C.邏輯透明
D.位置透明
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試題參考答案:D
試題解析與討論:www.xcpkj.com/st/2711015997.html
試題4
why have formal documents
first, writing the decisions down is essential.Only when one writes do the gaps appear and the (1) protrude(突出).The act of writing turns out to require hundreds of mini-decisions.and it is the existence of these that distinguishes clear.exact policies from fuzzy ones.
Second.the documents will communicate the decisions to others. The manager will be continually amazed that poliaes he took for common knowledge are totally unknown by some member of his team . Since his fundamentaljob is to keep everybody going in the (2)directon. his chief daily task will be communication, not decision-making,and his documents will immensely (3) this load.
Finally.a manager,s documents give him a data base and checklist. By reviewing them (4)he sees where he is. and he sees what changes of emphasis or shifts in direction are needed.
The task of the manager is to develop a plan and then to realize it. But only the written plan is precise and communicable. Such a plan consists of documents on what.when. how much.where.and who.This small set of critical documents (5) much of the manager.s work.if their comprehensive and critical nature is recognized in the beginning. the manager can approach them as friendly tools rather than annoying busywork. He will set his direction much more crisply and quickly by doing so.
(1)A.inconsistencies
B.consistencies
C.steadiness
D.adaptability
(2)A.other
B.different
C.another
D.same
(3)A.extend
B.broaden
C.lighten
D.release
(4)A.periodically
B.occasionally
C.infrequently
D.rarely
(5)A.decides
B.encapsulates
C.realizes
D.recognizes
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試題參考答案:A、D、C、A、B
試題解析與討論:www.xcpkj.com/st/255139341.html
試題5
計(jì)算機(jī)感染特洛伊木馬后的典型現(xiàn)象是 ( ) 。
A、程序異常退出
B、有未知程序試圖建立網(wǎng)絡(luò)連接
C、郵箱被垃圾郵件填滿
D、Windows 系統(tǒng)黑屏
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試題參考答案:B
試題解析與討論:www.xcpkj.com/st/81572809.html
試題6
語(yǔ)法制導(dǎo)翻譯是一種()方法。查看答案
試題參考答案:C
試題解析與討論:www.xcpkj.com/st/3961427007.html
試題7
現(xiàn)要對(duì)n個(gè)實(shí)數(shù)(僅包含正實(shí)數(shù)和負(fù)實(shí)數(shù))組成的數(shù)組A進(jìn)行重新排列,使得其中所有的負(fù)實(shí)數(shù)都位于正實(shí)數(shù)之前。求解該問(wèn)題的算法的偽代碼如下所示,則該算法的時(shí)間和空間復(fù)雜度分別為()。
i=0; j=n-1;
while i<jdo
while A[i]<0 do
i=i+1;
while A[j]>0 do
j=j-1;
if i<j do
交換A[i]和A[j];
A. A
B. B
C. C
D. D
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試題參考答案:C
試題解析與討論:www.xcpkj.com/st/2478310746.html
試題8
將高級(jí)語(yǔ)言程序翻譯為機(jī)器語(yǔ)言程序的過(guò)程中,常引入中間代碼,其好處是()
A、有利于進(jìn)行反編譯處理
B、有利于進(jìn)行與機(jī)器無(wú)關(guān)的優(yōu)化處理
C、盡早發(fā)現(xiàn)語(yǔ)法錯(cuò)誤
D、可以簡(jiǎn)化語(yǔ)法和語(yǔ)義分析
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試題參考答案:B
試題解析與討論:www.xcpkj.com/st/270978631.html
試題9
函數(shù)t()、f()的定義如下所示,若調(diào)用函數(shù)t時(shí)傳遞給x的值為5,并且調(diào)用函數(shù)F()時(shí),第一個(gè)參數(shù)采用傳值(call by value)方式,第二個(gè)參數(shù)采用傳引用(call by reference)方式,則函數(shù)t的返回值為( ) 。
A 33
B. 22
C. 11
D.負(fù)數(shù)
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試題參考答案:A
試題解析與討論:www.xcpkj.com/st/252838217.html
試題10
繼承是父類(lèi)和子類(lèi)之間共享數(shù)據(jù)和方法的機(jī)制。以下關(guān)于繼承的敘述中,不正確的是(1)。有關(guān)下圖中doIt()方法的敘述中,正確的是(2)。
(1)A.一個(gè)父類(lèi)可以有多個(gè)子類(lèi),這些子類(lèi)都是父類(lèi)的特例
B.父類(lèi)描述了這些子類(lèi)的公共屬性和操作
C.子類(lèi)可以繼承它的父類(lèi)(或祖先類(lèi))中的屬性和操作而不必自己定義
D.子類(lèi)中可以定義自己的新操作而不能定義和父類(lèi)同名的操作
(2)A.dolt()必須由Thing3實(shí)現(xiàn),同時(shí)可能用Thing4實(shí)現(xiàn)
B.dolt()必須由Thing5實(shí)現(xiàn)
C.dolt()必須由Thing2、Thing3、Thing4和Thing5實(shí)現(xiàn)
D.dolt()已經(jīng)由Thing1實(shí)現(xiàn),因此無(wú)須其他類(lèi)實(shí)現(xiàn)
查看答案
試題參考答案:D、A
試題解析與討論:www.xcpkj.com/st/248912514.html
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