試題1

給定一組長度為 n 的無序序列，將其存儲在一維數組 a[0..n-1]中。現采用如下方法找出其中的最大元素和最小元素：比較 a[0]和 a[n-1]，若 a[0]較大，則將二者的值進行交換；再比較 a[1]和 a[n-2]，若 a[1]較大，則交換二者的值；然后依次比較 a[2]和 a[n-3]、a[3]和 a[n-4]、…，使得每一對元素中的較小者被交換到低下標端。重復上述方法，在數組的前 n/2 個元素中查找最小元素，在后 n/2 個元素查找最大元素，從而得到整個序列的最小元素和最大元素。上述方法采用的算法設計策略是  （ ）  。
A、動態(tài)規(guī)劃法
B、貪心法
C、分治法
D、回溯法

查看答案

試題2

中央處理器CPU中的控制器是由些基本的硬件部件構成的。（ ）不是構成控制器的部件。
A、時序部件和微操作形成部件 
B、程序計數器 
C、外設接口部件 
D、指令寄存器和指令譯碼器 

查看答案

試題3

在13個元素構成的有序表A[1..13]中進行折半查找（或稱為二分查找，向下取整）。那么以下敘述中，錯誤的是（ ）。
A．無論要查找哪個元素，都是先與A[7]進行比較
B．若要查找的元素等于A[9]，則分別需與A[7]、A[11]、A[9]進行比較
C．無論要查找的元素是否在A[]中，最多與表中的4個元素比較即可
D．若待查找的元素不在A[]中，最少需要與表中的3個元素進行比較

查看答案

試題4

設有關系模式R（A₁,A₂,A₃,A₄,A₅,A₆），其中：函數依賴集F={A₁→A₂，A₁A₃→A₄,A₅A₆→A₁,A₂A₅→A₆,A₃A₅→A₆},則（1 ）是關系模式R的一個主鍵，R規(guī)范化成都最高達到（2 ）。
（1）A.A₁A₄
B.A₂A₄
C.A₃A₅
D.A₄A₅
（2）A. 1NF
B. 2NF
C. 3NF
D. BCNF

查看答案

試題5

查看答案

試題6

某二叉樹的先序遍歷序列為c a b f e d g，中序遍歷序列為a b c d e f g，則該二叉樹是(  )。
A.完全二叉樹
B.最優(yōu)二叉樹
C.平衡二叉樹
D.滿二叉樹

查看答案

試題7

查看答案

試題8

在開發(fā)一個系統(tǒng)時，如果用戶對系統(tǒng)的目標是不很清楚，難以定義需求，這時最好使用（ ）。
A、原型法
B、瀑布模型
C、V-模型
D、螺旋模型

查看答案

試題9

there is nothing in this world constant but inconstancy.-SWIFT
Project after project designs a set of algorithms and then plunges intoconstruction of customer-deliverable software on a schedule that demands delivery of the first thing built.
In most projects, the first system built is (1) usable. It may be too slow, too big , awkward to use, or all three. There is no (2 ) but to start again, smarting but smarter and build a redesigned version in which these problems are solved. The discard and (3) may be done in one lump, or it may be done piece-by-piece. But all large-system experience shows that it will be done. Where a new system concept or new technology is used, one has to build a syste m to throw away, for even the best
planning is not so omniscient (全知的) as to get it fight the first time.
the management question, therefore, is not whether to build a pilot system and throw it away. You will do that. The only question is whether to plan in advance to build a (4).or to promise to deliver the throwaway to customers. Seen this way, the answer is much clearer. Delivering that throwaway to customers buys time, but it does so only at the (5) of agony (極大痛苦) for the user, distraction for the builders while they do the redesign, and a bad..reputation for the product that the best redesign will find hard to live down.
Henceplanto throw one away; you will,anyhow.
The management question, therefore, is not whether to build a pilot system and throw it away. You will do that. The only question is whether to plan in advance to build a throwaway, or to promise to deliver the throwaway to customers. Seen this way, the answer is much clearer. Delivering that throwaway to customers buys time, but it does so only at the cost of agony for the user, distraction for the builders while they do the redesign, and a bad reputation for the product that the best redesign will find hard to live down.
(1)A.almost
B.often
C.usually
D.barely
(2)A.alternative
B.need
C.possibility
D.solution
(3)A.design
B.redesign
C.plan
D.build
(4) A.throwaway
B.system
C.software
D.product
(5) A.worth
B. value
C. cost
D. invaluable

查看答案

試題10

查看答案