📄 2006年下半年软件设计师上午试卷.htm
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<TD align=middle width="33%" bgColor=#c0c0c0>A</TD>
<TD align=middle width="33%" bgColor=#c0c0c0>B</TD>
<TD align=middle width="34%" bgColor=#c0c0c0>C</TD></TR>
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<TD align=middle width="33%">3<BR>4</TD>
<TD align=middle width="33%">7<BR>5</TD>
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width="34%">11<BR>6</TD></TR></TBODY></TABLE></CENTER></TD></TR>
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<TD align=middle width="33%" bgColor=#c0c0c0>A</TD>
<TD align=middle width="33%" bgColor=#c0c0c0>B</TD>
<TD align=middle width="34%" bgColor=#c0c0c0>C</TD></TR>
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<TD align=middle width="33%">7<BR>10</TD>
<TD align=middle width="33%">8<BR>11</TD>
<TD align=middle width="34%">9<BR>12</TD></TR></TBODY></TABLE></CENTER></TD>
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<TD align=middle width="33%" bgColor=#c0c0c0>A</TD>
<TD align=middle width="33%" bgColor=#c0c0c0>B</TD>
<TD align=middle width="34%" bgColor=#c0c0c0>C</TD></TR>
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<TD align=middle width="33%">5<BR>6</TD>
<TD align=middle width="33%">9<BR>10</TD>
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width="34%">13<BR>14</TD></TR></TBODY></TABLE></CENTER></TD></TR></TBODY></TABLE>
<P> ● 某企业职工和部门的关系模式如下所示,其中部门负责人也是一个职工。职工和部门关系的外键分别是
<U>(48)</U>。<BR> 职工(职工号,姓名,年龄,月工资,部门号,电话,办公室)<BR> 部门(部门号,部门名,负责人代码,任职时间)<BR> 查询每个部门中月工资最高的“职工号”的
SQL 查询语句如下:<BR> Select 职工号 from 职工 as E<BR> where 月工资=(Select
Max(月工资)from 职工 as M <U>(49)</U>)。<BR> (48)A. 职工号和部门号 B.
部门号和负责人代码<BR> C. 职工号和负责人代码 D. 部门号和职工号<BR> (49)A. where M.职工号=E.职工号 B.
where M.职工号=E.负责人代码<BR> C. where M.部门号=部门号 D. where M.部门号=E.部门号</P>
<P> ● 操作序列 T1、T2、T3 对数据 A、B、C 并发操作如下所示,T1 与 T2 间并发操作<U>(50)</U>, T2 与 T3
间并发操作<U>(51)</U> 。</P>
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<TABLE cellSpacing=0 width="80%" border=1>
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<TR>
<TD align=middle width="25%">时间</TD>
<TD align=middle width="25%">T1</TD>
<TD align=middle width="25%">T2</TD>
<TD align=middle width="25%">T3</TD></TR>
<TR>
<TD width="25%">
<P
align=center>t1<BR>t2<BR>t3<BR>t4<BR>t5<BR>t6<BR>t7<BR>t8<BR>t9<BR>t10<BR>t11<BR>t12</P></TD>
<TD width="25%">
<P
align=center>读A=50<BR>读B=200<BR>X1=A+B<BR><BR><BR><BR><BR>读A=50<BR>读B=200<BR>X1=A+B<BR>验算不对<BR></P></TD>
<TD width="25%">
<P
align=center><BR><BR><BR><BR>读B=200<BR>B=B-100<BR>写B<BR><BR><BR><BR><BR></P></TD>
<TD width="25%">
<P
align=center><BR><BR><BR>读B=200<BR><BR><BR><BR><BR><BR><BR>B=B+50<BR>写B</P></TD></TR></TBODY></TABLE></CENTER></DIV>
<P> (50)A. 不存在问题 B. 将丢失修改C. 不能重复读 D. 将读“脏”数据<BR> (51)A. 不存在问题 B. 将丢失修改C. 不能重复读
D. 将读“脏”数据</P>
<P> ● 结点数目为 n 的二叉查找树(二叉排序树)的最小高度为 <U>(52)</U> 、最大高度为<U>(53)</U>。<BR> (52)A.
n B.<IMG height=31 src="2006年下半年软件设计师上午试卷.files/gc06_2s04.gif" width=20
align=middle border=0> C. [log<SUB>2</SUB>n] D.
[log<SUB>2</SUB>(n+1)]<BR> (53)A. n B.<IMG height=31
src="2006年下半年软件设计师上午试卷.files/gc06_2s04.gif" width=20 align=middle border=0>
C. [log<SUB>2</SUB>n] D. [log<SUB>2</SUB>(n+1)]</P>
<P> ● 某双向链表中的结点如下图所示,删除 t 所指结点的操作为 <U>(54)</U> 。</P>
<P align=center><IMG height=119 src="2006年下半年软件设计师上午试卷.files/gc06_2s05.gif"
width=363 border=0></P>
<P> (54)A. t->prior->next = t->next; t->next->prior =
t->prior;<BR> B. t->prior->prior = t->prior; t->next->next
= t->next;<BR> C. t->prior->next = t->prior;
t->next->prior = t->next;<BR> D. t->prior->prior =
t->next; t->next->prior = t->prior;
<P> ● 对于二维数组 a[0..4,1..5],设每个元素占 1 个存储单元,且以列为主序存储,则元素 a[2,2]相对于数组空间起始地址的偏移量是
<U>(55)</U>。<BR> (55)A. 5 B. 7 C. 10 D. 15
<P> ● 对于n个元素的关键字序列{k1 , k 2 ,..., k n } ,当且仅当满足关系<IMG height=24
src="2006年下半年软件设计师上午试卷.files/gc06_2s07.gif" width=295 align=middle
border=0>称其为小根堆,反之则为大根堆。以下序列中,<U>(56)</U>
不符合堆的定义。<BR> (56)A.(4,10,15,72,39,23,18) B.(58,27,36,12,8,23,9)<BR> C.(4,10,18,72,39,23,15) D.(58,36,27,12,8,23,9)
<P> ● 求单源点最短路径的迪杰斯特拉(Dijkstra)算法是按 <U>(57)</U> 的顺序求源点到各顶点的最短路径的。<BR> (57)A.
路径长度递减 B. 路径长度递增 C. 顶点编号递减 D. 顶点编号递增
<P> ● <U>(58)</U> 算法策略与递归技术的联系最弱。<BR> (58)A. 动态规划 B. 贪心 C. 回溯 D. 分治
<P> ● 对于具有 n 个元素的一个数据序列,若只需得到其中第 k 个元素之前的部分排序, 最好采用 (59) ,使用分治(Divide and
Conquer)策略的是 <U>(60)</U> 算法。<BR> (59)A. 希尔排序 B. 直接插入排序 C. 快速排序 D.
堆排序<BR> (60)A. 冒泡排序 B. 插入排序 C. 快速排序 D. 堆排序
<P> ● ARP 协 议的 作 用 是 <U>(61)</U> ,ARP 报文封装在 <U>(62)</U> 中传送。<BR> (61)A. 由 IP
地址查找对应的 MAC 地址 B. 由 MAC 地址查找对应的 IP 地址<BR> C. 由 IP 地址查找对应的端口号 D. 由 MAC
地址查找对应的端口号<BR> (62)A. 以太帧 B. IP 数据报 C. UDP 报文 D. TCP 报文
<P> ● 802.11 标准定义的分布式协调功能采用了 <U>(63)</U> 协议。<BR> (63)A. CSMA/CD B.
CSMA/CA C. CDMA/CD D. CDMA/CA
<P> ● 设有两个子网 202.118.133.0/24 和 202.118.130.0/24,如果进行路由汇聚,得到的网 络地址是 <U>(64)</U>
。<BR> (64)A. 202.118.128.0/21 B. 202.118.128.0/22<BR> C.
202.118.130.0/22 D. 202.118.132.0/20
<P> ● 路由器收到一个数据包,其目标地址为 195.26.17.4,该地址属于 <U>(65)</U>子网。<BR> (65)A.
195.26.0.0/21 B. 195.26.16.0/20<BR> C. 195.26.8.0/22 D.
195.26.20.0/22
<P> ● NAC’s(Network Access Control) role is to restrict network access to only
compliant endpoints and (66) users. However, NAC is not a complete LAN
<U>(67)</U> solution; additional proactive and <U>(68) </U>security measures
must be implemented. Nevis is the first and only comprehensive LAN security
solution that combines deep security processing of every packet at 10Gbps,
ensuring a high level of security plus application availability and performance.
Nevis integrates NAC as the first line of LAN security <U>(69)</U> . In addition
to NAC, enterprises need to implement role-based network access control as well
as critical proactive security measures— real-time, multilevel
<U>(70)</U>inspection and microsecond threat containment.<BR> (66)A.
automated B. distinguished C. authenticated D. destructed<BR> (67)A.
crisis B. security C. favorable D. excellent<BR> (68)A.
constructive B. reductive C. reactive D. productive<BR> (69)A.
defense B. intrusion C. inbreak D. protection<BR> (70)A. port B.
connection C. threat D. insurance
<P> ● Virtualization is an approach to IT that pools and shares <U>(71)</U> so
that utilization is optimized and supplies automatically meet demand.
Traditional IT environments are often silos, where both technology and human
<U>(72)</U>are aligned around an application or business function. With a
virtualized <U>(73)</U> , people, processes, and technology are focused on
meeting service levels, <U>(</U> <U>74</U> <U>)</U> is allocated dynamically,
resources are optimized, and the entire infrastructure is simplified and
flexible. We offer a broad spectrum of virtualization <U>(75)</U> that allows
customers to choose the most appropriate path and optimization focus for their
IT infrastructure resources.<BR> (71)A. advantages B. resources C.
benefits D. precedents<BR> (72)A. profits B. costs C. resources
D. powers<BR> (73)A. system B. infrastructure C. hardware D.
link<BR> (74)A. content B. position C. power D.
capacity<BR> (75)A. solutions B. networks C. interfaces D.
connections</P>
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