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軟件設(shè)計師每日一練試題內(nèi)容(2025/4/10)
試題1
隊列的特點是先進先出,若用循環(huán)單鏈表表示隊列,則()。查看答案
試題參考答案:A
試題解析與討論:www.jazzmuze.com/exam/ExamDay.aspx?t1=4&day=2025/4/10
試題2
CPU訪問存儲器時,被訪問數(shù)據(jù)一般聚集在一個較小的連續(xù)存儲區(qū)域中。若一個存儲單元已被訪問,則其鄰近的存儲單元有可能還要被訪問,該特性被稱為( )查看答案
試題參考答案:C
試題解析與討論:www.jazzmuze.com/exam/ExamDay.aspx?t1=4&day=2025/4/10
試題3
美國卡內(nèi)基—梅隆大學(xué)SEI提出的CMM模型將軟件過程的成熟度分為5個等級,以下選項中,屬于可管理級的特征是( )。
A、工作無序,項目進行過程中經(jīng)常放棄當初的計劃
B、建立了項目級的管理制度
C、建立了企業(yè)級的管理制度
D、軟件過程中活動的生產(chǎn)率和質(zhì)量是可度量的
查看答案
試題參考答案:D
試題解析與討論:www.jazzmuze.com/exam/ExamDay.aspx?t1=4&day=2025/4/10
試題4
Teams are required for most engineering projects. Although some small hardware or software products can be developed by individuals, the scale and complexity of modem systems is such, and the demand for short schedules so great, that it is no longer (1) for one person to do most engineering jobs. Systems development is a team (2) ,and the effectiveness of the team largely determines the (3) of the engineering. Development teams often behave much like baseball or basketball teams. Even though they may have multiple specialties, allthe members work toward (4) .However,on systems maintenance and enhancement teams, the engineers often work relatively independently, much like wrestling and track teams.
A team is (5) just a group of people who happen to work together. Teamwork takes practice and it involves special skills. Teams require common processes; they need
agreed-upon goals; and they need effective guidance and leadership. The methods for guiding and leading such teams are well known, but they are not obvious.
(1)A.convenient
B.existing
C.practical
D.real
(2)A.a(chǎn)ctivity
B.job
C.process
D.a(chǎn)pplication
(3)A.size
B.quality
C.scale
D.complexity
(4)A.multiple objectives
B.different objectives
C.a(chǎn) single objective
D.independent objectives
(5)A.relatively
B./
C.only
D.more than
查看答案
試題參考答案:C、A、B、C、D
試題解析與討論:www.jazzmuze.com/exam/ExamDay.aspx?t1=4&day=2025/4/10
試題5
結(jié)構(gòu)化開發(fā)方法中,數(shù)據(jù)流圖是 ( ) 階段產(chǎn)生的成果。
A、需求分析
B、總體設(shè)計
C、詳細設(shè)計
D、程序編碼
查看答案
試題參考答案:A
試題解析與討論:www.jazzmuze.com/exam/ExamDay.aspx?t1=4&day=2025/4/10
試題6
已知一個線性表(16, 25, 35, 43, 51, 62, 87, 93),采用散列函數(shù) H(Key)=Key mod 7將元素散列到表長為 9 的散列表中。若采用線性探測的開放定址法解決沖突(順序地探查可用存儲單元),則構(gòu)造的哈希表為 (1) ,在該散列表上進行等概率成功查找的平均查找長度為 (2) (為確定記錄在查找表中的位置,需和給定關(guān)鍵字值進行比較的次數(shù)的期望值稱為查找算法在查找成功時的平均查找長度)。
(1)A. A
B. B
C. C
D. D
(2)A、(5*1+2+3+6) / 8
B、(5*1+2+3+6) / 9
C、(8*1) / 8
D、(8*1) / 9
查看答案
試題參考答案:C、A
試題解析與討論:www.jazzmuze.com/exam/ExamDay.aspx?t1=4&day=2025/4/10
試題7
函數(shù)main()、f()的定義如下所示。調(diào)用函數(shù)f()時,第一個參數(shù)采用傳值(call by value)方式,第二個參數(shù)采用傳引用(call by referen)方式,則函數(shù)main()執(zhí)行后輸出的值為(50)。
查看答案
試題參考答案:D
試題解析與討論:www.jazzmuze.com/exam/ExamDay.aspx?t1=4&day=2025/4/10
試題8
建立一個供應(yīng)商、零件數(shù)據(jù)庫。其中“供應(yīng)商”表S(Sno,Sname,Zip,City)分別表示:供應(yīng)商代碼、供應(yīng)商名、供應(yīng)商郵編、供應(yīng)商所在城市,其函數(shù)依賴為:Sno→(Sname,Zip,City ),Zip→City。“供應(yīng)商”表S屬于 ( ) 。
A、1NF
B、2NF
C、3NF
D、BCNF
查看答案
試題參考答案:B
試題解析與討論:www.jazzmuze.com/exam/ExamDay.aspx?t1=4&day=2025/4/10
試題9
原碼表示法和補碼表示法是計算機中用于表示數(shù)據(jù)的兩種編碼方法,在計算機系統(tǒng)中常采用補碼來表示和運算數(shù)據(jù),原因是采用補碼可以()。
A.保證運算過程與手工運算方法保持一致
B.簡化計算機運算部件的設(shè)計
C.提高數(shù)據(jù)的運算速度
D.提高數(shù)據(jù)的運算精度
查看答案
試題參考答案:B
試題解析與討論:www.jazzmuze.com/exam/ExamDay.aspx?t1=4&day=2025/4/10
試題10
面向?qū)ο蟪绦蛟O(shè)計語言為 ( ) 提供支持。
A、面向?qū)ο笥美O(shè)計階段
B、面向?qū)ο蠓治鲭A段
C、面向?qū)ο笮枨蠓治鲭A段
D、面向?qū)ο髮崿F(xiàn)階段
查看答案
試題參考答案:D
試題解析與討論:www.jazzmuze.com/exam/ExamDay.aspx?t1=4&day=2025/4/10
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