2014年宁波大学计算机网络考研复试试题(回忆版)

2014考研统考考研统考计算机基础综合真题解析计算机基础综合真题解析【回忆版】一、单项选择题单项选择题：：第1～40小题小题，，每小题2分，共80分。

下列每题给出的四个选项中下列每题给出的四个选项中，，只有一个选项是最符合题目要求的只有一个选项是最符合题目要求的。

2、假设栈初始为空，将中缀表达式a/b-(c*d+e*f)/g 转化为等价后缀表达式过程中，当扫描到f 时，栈中的元素依次为：A 、+（*-B 、+（-*C 、/+(*-*D 、/+-*涉及考点涉及考点：：考察中缀和后缀表达式的转化，并考察栈这种数据结构4、如下图二叉树进行中序线索化，则元素X 的左、右线索指向的元素为A 、 ecB 、 eaC 、 dcD 、 ba涉及考点涉及考点：：中序线索化二叉树，找出左右线索5、森林F 转化为对应二叉树T ，则F 的叶结点个数是（）A 、T 的叶结点个数B 、T 中度为1的结点个数C 、T 的左孩子指向为空的个数D 、T 的右孩子指向为空的个数涉及考点涉及考点：：森林转化为二叉树做法：第一，断开除最左孩子的孩子节点，第二，连接孩子节点中各兄弟节点，第三，将树顺时针旋转45度第四，同理处理其他树。

第五，将所有树按照先后顺序依次作为右子树连接。

6、5个元素有4种编码方案，下列不是前缀编码的是A 、01,0000,0001,001,1B 、011,000,001,010,1C 、000,001,010,011,100D 、0,100,110,1110,1100涉及考点涉及考点：：字符的前缀编码8、用哈希（散列）方法处理冲突（碰撞）时可能发生堆积（聚集)现象，则下列会直接受到堆积现象影响的是A 、存储效率B 、散列函数C 、载运因子D 、平均查找长度涉及考点涉及考点：：哈希（三列）方法处理冲突堆积现象影响的因素9、存一棵具有15个关键词的4阶B 树，则含有关键词的结点可能有A 、5B 、6C 、10D 、15涉及考点涉及考点：：B 树10、用希尔排序法，对一列数据序列排序时，若第一次排序结果为：9,1,4,13,7,8,20,23,15，则该排序可能的间隔是：A 、2B 、3C 、4D 、5涉及考点涉及考点：：希尔排序法中的间隔11、下列最不可能是快速排序第二轮的结果是A 、2,3,5,4,6,7,9B 、2,7,5,6,4,3,9C 、3,2,5,4,7,6,9D 、4,2,3,5,7,6,9涉及考点涉及考点：：快速排序法12、程序P 在装置M 执行时间为20秒，编译优化后，P 执行的指令数是以前的70%，但CPI 为以前的1.2倍，则现在P 在M 上的执行时间为A 、8.4秒B 、11.7秒C 、14.0秒D 、16.8秒涉及考点涉及考点：：cpu 计算时间的计算方法。

计算机学科专业基础综合真题2014年(总分：137.00，做题时间：90分钟)一、{{B}}单项选择题{{/B}}(总题数：40，分数：80.00)1.下列程序段的时间复杂度是count=0;for(k=1; k＜=n; k*=2)for(j=1; j＜=n; j++)count++;∙ A.O(log2n)∙ B.O(n)∙ C.O(nlog2n)∙ D.O(n2)（分数：2.00）A.B.C. √D.解析：[解析] 题目中给出了一个2层的嵌套循环，里层循环的时间复杂度是O(n)，外层循环的时间复杂度是O(log2n)。

对于嵌套循环，其整体复杂度是两层循环的复杂度的乘积，因此总体的时间复杂度是D(nlog2n)。

2.假设栈初始为空，将中缀表达式a/b+(c*d-e*f)/g转换为等价的后缀表达式的过程中，当扫描到f时，栈中的元素依次是∙ A.+(*-∙ B.+(-*∙ C./+(*-*∙ D./+-*（分数：2.00）A.B. √C.D.解析：[解析] 后缀表达式为ab/cd*ef*-g/+。

根据中缀表达式a/b+(c*d-e*f)/g转换为等价的后缀表达式的过程，字母不需要入栈，只有扫描到符号时才需要入栈。

最先入栈的是“/”，当扫描完b时出栈。

接下来入栈的是“+”和“(”，然后扫描c，后面的“*”要入栈，再扫描d，然后“*”出栈。

接下来“-”入栈，扫描e，接下来的“*”入栈，接下来就扫描到f了。

此时没有出栈的有“+，(，-，*”。

3.循环队列存放在一维数组A[0..M-1]中，end1指向队头元素，end2指向队尾元素的后一个位置。

假设队列两端均可进行人队和出队操作，队列中最多能容纳M-1个元素，初始时为空。

下列判断队空和队满的条件中，正确的是∙ A.队空：end1==end2；队满：end1==(end2+1)mod M∙ B.队空：end1==end2；队满：end2==(end1+1)mod(M-1)∙ C.队空：end2==(end1+1)mod M；队满：end1==(end2+1)mod M∙ D.队空：end1=(end2+1)mod M；队满：end2==(end1+1)mod(M-1)（分数：2.00）A. √B.C.D.解析：[解析] 对于循环链表来说，队列空的条件是队头指针和队尾指针指向同一个位置，即end1==end2；队列满的条件是队尾指针指向队头指针的前一个位置，即end1==(end2+1)mod M。

1 . 用预处理指令#define 声明一个常数，用以表明1年中有多少秒（忽略闰年问题） #define SECONDS_PER_YEAR (60 * 60 * 24 * 365)UL2. 嵌入式系统中经常要用到无限循环，你怎么样用C编写死循环呢？while(1){}或者for(;;){}3. 用变量a给出下面的定义 a) 一个整型数（An integer）b)一个指向整型数的指针（ A pointer to an integer）c)一个指向指针的的指针，它指向的指针是指向一个整型数（ A pointer to a pointer to an intege）r d)一个有10个整型数的数组（ An array of 10 integers）e) 一个有10个指针的数组，该指针是指向一个整型数的。

（An array of 10 pointers to integers） f) 一个指向有10个整型数数组的指针（ A pointer to an array of 10 integers）g) 一个指向函数的指针，该函数有一个整型参数并返回一个整型数（A pointer to a function that takes an integer as an argument and returns an integer）h) 一个有10个指针的数组，该指针指向一个函数，该函数有一个整型参数并返回一个整型数（ An array of ten pointers to functions that take an integer argument and return an integer ）答案是：a) int a; // An integerb) int *a; // A pointer to an integerc) int **a; // A pointer to a pointer to an integer d) int a[10]; // An array of 10 integerse) int *a[10]; // An array of 10 pointers to integers f) int (*a)[10]; // A pointer to an array of 10 int egersg) int (*a)(int); // A pointer to a function a that takes an integer argument and returns an integer h) int (*a[10])(int); // An array of 10 pointers to functions that take an integer argument and return a n integer4.关键字static的作用是什么？这个简单的问题很少有人能回答完全。

2014年宁波大学计算机网络考研复试试题（回忆版）
一、填空题（每空2分，共30分）
1. 三种数据交换方式：（电路交换）、（报文交换）、（分组交换）
2. 网络类型划分：（WAN—广域网）、（LAN—局域网）、（MAN—城域网）
3. 计算机网络的层次结构，TCP/IP模型，从下往上分为四层：（网络接口层）
（网际层）（传输层）（应用层）
4. AM、FM、DM对应的中文：（调幅），（调频），(调相)
5. 双绞线分为3类线（10Mbps）, 5类线（100Mbps）
6. TDMA,FDMA,CDMA对应的中文：（时分多址复用）（频分多址复用）（码分多
址复用）
二、写出对应的英文缩写（每个1分）
1.HTTP
2.SMTP
3.DNS
4.SNMP
5.ARP
6.RIP
7.ICMP
8.BGP
9.FTP
10.OSPF
三、请说出M A C地址、I P地址、域名三者的区别。

（10分）
四、请写出集线器、交换机、路由器三者所属的层次及各自的功能。

（10分）
五、什么是C S M A/C D?为何C S M A/C D以太网有最小帧长和网络有效长度上限？（10分）
六、请介绍一个你熟悉的网络协议，300字以上，写出其功能，实现原理，及应用的特点，最好图文结合。

（30分）。

一、单选题，每题仅一个正确答案，每题1.5分，共30分：[ ]1. Which is not a network communication links?D.Subway[ ]2.Which is not the types of protocol data unit in TCP/IP?C.datetime[ ]3.In which state the delays of router become small?A. La/R ~ 0[ ]4.When congest occur in network, which delay become large?B.queuing delay[ ]5.Application protocol define the types of messages exchanged, which are:A.request & response messages[ ]6.Which version of hypertext transfer protocol are used in default mode now?C.HTTP 1.1[ ]7.The Http client works on which port:C.TCP port above 1024[ ]8.To indicates end of message in Http, we can use:A.Carriage return and line feed[ ]9.Which is not an email protocol:C.Emule[ ]10. TCP retransmission can not be caused by:D. ACK[ ]11. Which of the following measures the time from one segment transmission to ACK receipt:A. sample RTT[ ]12. Which event will not be in TCP Receiver:C.Arrival of out-of-order segment higher-than-expect seq. # .No gap detected.[ ]13.TCP Flow Control is a speed-matching service for matching:A.the sender rate to the receiving application’s drain rate[ ]14.Which one is not the mechanism of TCP Congestion Control:C.FIFO[ ]15.In RDT protocol, which can increase the use of physical resources:C.Pipeline[ ]16. Which one is not the Key Network-Layer Functions in today’s internet:C. Connection setup[ ]17. Which of the following IP address belongs to category C address?A. 210.98.3.4[ ]work address translation implement in NAT router, it must use: C．NAT translation table[ ]19. ICMP do not error reporting about:D. unknown username[ ]20. In Ethernet’s CSMA/CD Jam Signal make sure:A.other transmitters are aware of collision二、多选题，每题有2-3个正确答案，请在正确项前打钩(全钩不给分)，每题2分，共20分：1. The communication services be provided to applications, It can be:[ ]A.connectionless[ ]B.connection-oriented2.Which are application architectures model in the internet?[ ]A.client/server model[ ]B.peer-peer model[ ]C.browser/server model3. User Agent is an interface with user “above” and network “below”, such as:[ ]A.Web: browser[ ]B.E-mail: mail reader[ ]D.Streaming audio/video: media player4.Ftp protocol can work in which model that listed:[ ]A.PASV model[ ]C.Port model5. The problems with centralized directory of P2P maybe:[ ]A.Single point of failure[ ]C.Performance bottleneck[ ]D.Copyright infringement6.Which services are not available in TCP:[ ]A.delay guarantees[ ]C.bandwidth guarantees7. In TCP congestion control, there are those State in TCP Sender:[ ]A. Congestion Avoidance[ ]C. Slow Start[ ]D. Timeout8.In TCP three way handshake, it will be:[ ]A.client host sends TCP SYN segment to server[ ]C.server host receives SYN, replies with SYN/ACK segment[ ]D.client receives SYN/ACK, replies with ACK segment9.Those are the routing in the Internet:[ ]A.RIP[ ]B.OSPF[ ]D.BGP10.Those are the Random Access Protocols：[ ]B.CSMA[ ]C.CSMA/CD三、名称缩写选择，请在匹配项上填写缩写编号，如果在下列18个选项中没有，可以在19－20项中自主添加选项。

学号 姓名I.Choose the single correct answer from following choose. (1.5*20=30) [Correct checked:1.5; Other wise:0]1.Which physical media has high-speed operation and low error rate:[] A.Fiber optic cable[] B.Coaxial cable [] C.Twisted pair [] D.Radio2.If no free buffers in router, the arriving packets will be:[] A.dropped[] B.queued [] C.returned [] D.marked3.Which can provides delay measurement from source to router along end-end Internet path towards destination ：[] A.Ping [] B.Traceroute [] C.Ipconfig [] D.Nslookup4.In TCP/IP , which layer can make routing of datagrams from source to destination:[] A.Applicaion[] B.Transport [] work[] D.Data Link5.Web page consists of ( ) which includes several referenced objects ：[] A.referenced HTML-file [] B.host HTML-file [] C.path HTML-file [] D.base HTML-file6.What is the default persistent model in HTTP/1.1:[] A.Nonpersistent HTTP[] B.Persistent without pipelining [] C.Persistent with pipelining [] D.Nonpersistent with pipelining7.Web server maintains no information about past client requests, so HTTP is:[] A.stateful [] B.stateless [] C.satisfied [] D.unsatisfied8.FTP client browses remote directory by sending commands over:[] A.connection-less [] B.free connection [] C.data connection [] D.control connection9.Which can satisfy client request without involving origin server:[] A.Web caches[] B.Write caches [] C.TCP buffer [] D.Router buffer10.UDP socket identified by:[] A.two-tuple (source IP address, source port number)[] B.two-tuple (dest IP address, dest port number) [] C.two-tuple (source IP address, dest port number) [] D.two-tuple (dest IP address, source port number)11.In GBN,when receiver receive a out-of-order packet,then discard and re-ACK the packet with:[] A. highest in-order sequence # [] B. lowest in-order sequence # [] C. highest in-order port # [] D. lowest in-order port #12.In RDT Approachs, which is designed for performance:[] A. checksum [] B. pipeline [] C. sequence # [] D. ACK or NAK13.Queued datagram at front of router’s queue prevents others in queue from moving forward is:[] A.Head-of-the-Line (HOL) blocking [] B.Hops-of-the-Line (HOL) blocking [] C.Head-of-the-List (HOL) blocking [] D.Hops-of-the-List (HOL) blocking14.What’s a network ? From IP address perspective they can physically reach each other without intervening router and the device interfaces with:[] A. same IP address[] B. same TCP port #[] C. same network part of IP address [] D. same host part of IP address学号 姓名15. Large IP datagram divided (“fragmented”) within network,it will be reassembled:[] A. only at last router [] B. only at final destination [] C. only at next router[] D. maybe at next router16.Which is not a common Intra-AS routing protocols:[] A.RIP: Routing Information Protocol [] B.OSPF: Open Shortest Path First[] C.IGRP: Interior Gateway Routing Protocol[] D.ICMP: Interior Control Message Protocol17.Which is not a MAC Random Access protocol in Ethernet:[] A.SCMA [] B.SCMA/CA [] C.SCMA/CD[] D.Slotted SCMA18.In DHCP client-server scenario, which message has DHCP-options field:[] A. host broadcasts “DHCP discover” [] B. DHCP server responds with “DHCP offer” [] C. host requests IP address: “DHCP request”[] D. DHCP server sends addr ess: “DHCP ack”19.How to determine MAC address of host B, If knowing host B’s IP address?[] A.ARP [] B.RARP [] C.RAP[] D.RIP20. Which device can break subnet into LAN segments:[] A.IP mask [] B.NA T [] C.Router [] D.SwitchII.Choose the multiple correct answer from following choose. (2*10=20) [All correct checked:2; Part correct checked:1; No checked:0; Full checked:0] 1.Which is the part of network structure:[] work edge [] work core [] work user [] D.access networks2.How to connect end systems to edge router?[] A.Microsoft access networks [] B.Residential access networks [] C.Institutional access networks[] D.Mobile access networks3.What kind of transport service does an application need?[] A.Data loss [] B.Timing [] C.Bandwidth[] D.Security4.Electronic Mail three phases of transfer is:[] A.handshaking (greeting) [] B.transfer of messages[] C.opens the 2nd TCP connection[] D.close5.In TCP Connection Management, initialize TCP variables include ：[] A.sequence # [] B.buffers[] C.Sender MTU[] D.RcvWindow6.How does sender perceive congestion?[] A.timeout [] B.3 duplicate ACKs [] C.3 duplicate data[] D.slow start7.TCP Congestion Control use three mechanisms:[] A.additive increase and multiplicative decrease [] B.slow start[] C.Conservative after timeout events[] D.additive decrease and multiplicative increase8.What are the Key Network-Layer Functions:[] A.forwarding [] B.routing [] C.connection setup [] D.flow control学号 姓名9.Link Layer Services include:[] A.Reliable delivery between adjacent nodes [] B.Flow Control between adjacent nodes [] C.Connection Manage[] D.Error Detection and Correction10.MAC Protocol ’s taxonomy, three broad classes is:[] A.Channel Partitioning [] B.Random Access [] C.Taking turns [] D.AdvertisementIII. Fill the blank from options. (1.5*16=24)1).The network protocols define 1 , 2 of message sent and received among network entities, and 3 on message transmission, receipt.(options: delay / format / policy / order / request / replay / actions taken / price / interface )2).InTCP Congestion Control, after 3 duplicate ACKs CongWin is cut in ___4___ and window then grows____5____. But after timeout event, CongWin instead set to___6___, window then grows____7____, when it up to a ___8_____ again, then grows linearly.(options: half / double / 1 MSS / 0 MSS / linearly / exponentially / threshold / top / bottom )3).Please fill the general format of Http request message:(options: header field name / URL / field value / version / method / Cr Lf/ 200 OK )IV.Question (26)1.As follow, LAN1 connect to LAN2 via a router:In session 1, Host A send a HTTP connection to WEB server D ，if Host A initial TCP port 1025，Host D use TCP port 80；In session 2, Host A send a HTTP connection to WEB server B ，if Host A initial TCP port 1026，Host B use TCP port 80；2.Read and answer ：(17)Two of the m ost important fields in the TCP segm ent header are the sequence number field and the acknowledgment number field. These fields are a critical part of TCP's reliable data transfer service. But before discussing how these fields are used to provide reliable data transfer, let us first explain what exactly TCP puts in these fields. TCP views data as an unstructured, but ordered, stream of bytes. TCP's use of sequence numbers reflects this view in that sequence num bers are over the stream of transmitted bytes and not over the series of transmitted segm ents. The sequence学号 姓名number for a segm ent is the byte-stream number of the first byte in the segm ent. Let's look at an example. Suppose that a process in host A wants to send a stream of data to a process in host B over a TCP connection. The TCP in host A will implicitly number each byte in the data stream . Suppose that the data stream consists of a file consisting of 500,000 bytes, that the MSS is 1,000 bytes, and that the first byte of the data stream is numbered zero. As shown in Figure 3.5-3, TCP constructs 500 segm ents out of the data stream . The first segm ent gets assigned sequence number 0, the second segm ent gets assigned sequence num ber 1000, the third segment gets assigned sequence num ber 2000, and so on.. Each sequence number is inserted in the sequence num ber field in the header of the appropriate TCP segment.Figure 3.5-3: Dividing file data into TCP segm ents.Now let us consider acknowledgment numbers. These are a little trickier than sequence numbers. Recall that TCP is full duplex, so that host A m ay be receiving data from host B while it sends data to host B (as part of the sam e TCP connection). Each of the segm ents that arrive from host B have a sequence number for the data flowing from B to A. The acknowledgment number that host A puts in its segm ent is sequence number of the next byte host A is expecting from host B. It is good to look at a few examples to understand what is going on here. Suppose that host A has received all bytes num bered 0 through 535 from B and suppose that it is about to send a segm ent to host B. In other words, host A is waiting for byte 536 and all the subsequent bytes in host B's data stream . So host A puts 536 in the acknowledgment number field of the segm ent it sends to B.As another example, suppose that host A has received one segm ent from host B containing bytes 0 through 535 and another segment containing bytes 900 through 1,000. For som e reason host A has not yet received bytes 536 through 899. In this example, host A is still waiting for byte 536 (and beyond) in orde r to recreate B's data stream . Thus, A's next segm ent to B will contain 536 in the acknowledgment number field. Because TCP only acknowledges bytes up to the first missing byte in the stream , TCP is said to provide cumulative acknowledgements.This last example also brings up an important but subtle issue. Host A received the third segment (bytes 900 through 1,000) before receiving the second segm ent (bytes 536 through 899). Thus, the third segm ent arrived out of order. The subtle issue is: What does a host do when it receives out of order segm ents in a TCP connection? Interestingly, the TCP RFCs do not im pose any rules here, and leave the decision upto the people programming a TCP implem entation. There are basically two choices: either (i) the receiver immediately discards out-of-order bytes; or (ii) the receiver keeps the out-of-order bytes and waits for the m issing bytes to fill in the gaps. Clearly, the latter choice is m ore efficient in terms of network bandwidth, whereas the former choice significantly sim plifies the TCP code. Throughout the rem ainder of this introductory discussion of TCP, we focus on the form er implementation, that is, we assum e that the TCP receiver discards out-of-order segm ents.In Figure 3.5.3 we assumed that the initial sequence number was zero. In truth, both sides of a TCP connection randomly choose an initial sequence number. This is done to m inimize the possibility a segm ent that is still present in the network from an earlier, already-terminated connection between two hosts is m istaken for a valid segm ent in a later connection between these sam e two hosts (who also happen to be using the sam e port numbers as the old connection) .Question 1: Does TCP's use of sequence numbers over the series of transmitted segments ? (3) Question 2: What does the sequence number for a segment means? For example. (4)Question 3: What does the acknowledgment number means, that host A puts in its segment to host B?(3) Question 4: How does TCPs to choose an initial sequence number? (3)Question 5: What does a host do when it receives out of order segments in a TCP connection? (4)(所有答案请填在答题卡上，答在试卷上的答案一律无效)。

2014年国防科技大学计算机考研复试试题（回忆版）
一、单选题
二、大题
1.关于在页式存储管理方式中，计算读取几个给定逻辑地址中数据所用的时间，题目涉及到的知识点主要有：逻辑地址的表示方式，简单CLOCK页面置换算法，对于给定的逻辑地址先查找TLB，再查找页表，更新TLB和页表内容等等。

2.关于文件系统的几个问答题，比如文件名和文件属性是不是应该分开存储；一次性写入的情况下，用哪种物理存储方式（应该是连续存储方式），用什么表项来表示文件的大小（应该是起始块号和文件大小)。

3.关于PV操作的填空题，需要在空白处填写PV操作语句，一共有2个进程，一个发送消息进程，另外一个是接受消息的进程。

承德分公司岗位技能大赛互联网专业试题1、在Internet中，IP地址由（）位二进制数组成。

CA．16 B．24 C．32 D．642、接收功率为1mW，发送功率为10mW的通信系统的衰减是（）。

A、-10dB，B、-1dB，C、-100dB，D、-20dB3、调制解调器（MODEM）的主要功能是（）。

A、模拟信号的放大，B、数字信号的整形C、模拟信号与数字信号的转换，D、数字信号的编码4、NAT可以解决什么问题？（）A、节省Internet公有地址，B、提高路由器转发性能C、增强数据传输的安全性，D、防止计算机遭受病毒攻击5.哪种网际协议用来将IP地址映射为MAC地址？（C）A.UDPB.ICMPC.ARPD.RARP6、IP地址的两个部分是什么？（D）A.网络地址和MAC地址B．主机地址和MAC地址C．网络地址和子网掩码D.网络地址和主机地址7、ICMP的功能是什么？（B）A.传输路由更新信息B. 发送错误和控制信息C. 避免路由循环D.冲突检测HTML是指（）。

8、OSI模拟中最底层和最高层分别是（）。

CA.物理层和表示层B.网络层和应用层C.物理层和应用层 D .网络层和表示层9、华为公司GPON设备操作中，以下哪个命令是用于查询的？（ A ）A、displayB、showC、queryD、lookup10、通常中的1KB是指( ) 。

CA. 1000个字节B. 1000个二进制位C. 1024个字节D. 1024个二进制位8、（）命令可以查看到目的主机的路由信息A、PING，B、TRACERT，C、CMD，D、IPCONFIG9、交换机工作在OSI七层的哪一层？（）A、一层，B、二层，C、三层，D、三层以上标准网线的标准制作线序是（）。

A、一头是橙白橙绿白蓝蓝白绿棕白棕一头是橙白橙绿白绿蓝白蓝棕白棕B、一头是橙白橙绿白蓝蓝白绿棕白棕一头是橙白橙绿白蓝蓝白绿棕白棕C、一头是橙白橙绿白绿蓝白蓝棕白棕一头是橙白橙绿白绿蓝白蓝棕白棕D、一头是棕白橙绿白蓝蓝白绿棕白棕一头是棕白橙绿白蓝蓝白绿棕白棕2、GPON单板中GPBC能够提供____个PON口，GPBD能够提供___个PON口。