18秋浙大远程-Web程序设计在线作业

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16春浙大远程Web程序设计离线作业

浙江大学远程教育学院《Web程序设计》课程作业姓名：学号：年级：15年春学习中心：—————————————————————————————HTML部分一、选择题：1.以下哪个选项能显示成两行文字a) The first line and <BR> second lineb) The first line and \n second linec) The first line andsecond lined) The first line and second line答：a) The first line and <BR> second line2. <A href="mailto:"> post your resume </A>a) mailto:是一种协议，是email发送的地址b) mailto: 是邮件被发送的地址c) mailto: 是显示在web网页上的超链接d) 以上都不对答：a) mailto:是一种协议，是email发送的地址3. <STYLE>标记可以出现在a) <HEAD>…</HEAD>中b) <BODY>…</BODY>中c) a)和b)都正确d) 以上都不正确答：a) <HEAD>…</HEAD>中4. 以下代码会生成什么样的Frame结构<html><frameset border=”5” cols=”*,20%”><frameset rows=”20%,*”><frame src=”top.html” name=”topFrame” scrolling=”NO”/><frame src=”left.html” name=”leftFrame”/></frameset><frame src=”right.html” name=”rightFrame” scrolling=”NO”/> </frameset></html>a)b)d)答：c) 不过形状有点问题，应该是5. 以下HTML代码在浏览器中的显示结果正确的是<TABLE WIDTH="100%" BORDER=1 BGCOLOR=gray><CAPTION> Sample table </CAPTION><TR ALIGN=CENTER ><TD COLSPAN=2>Cell</TD> <TD> Cell </TD></TR><TR ALIGN=CENTER ><TD ROWSPAN=2>Cell</TD> <TD> Cell </TD> <TD> Cell </TD></TR> <TR ALIGN=CENTER ><TD> Cell </TD> <TD> Cell </TD></TR> </TABLE>a)Sample TableSample Table二、简答题：1、简述Web的工作原理答：Web的工作原理就是用户通过浏览器来得到服务器上的某项服务的过程。

秋浙大远程建筑施工技术在线作业及答案(专)

您的本次作业分数为：100 分单选题1. 【第1 章】下列哪种施工机械可不排水施工：A 铲运机B 抓铲挖掘机C 推土机D 装载机正确答案:B 单选题2. 【第1 章】四方棱柱体四个角点不可能浮现的标高情况：A 全填全挖B 两点挖两点填C 三点挖一点填D 三点为零线一点挖(填)正确答案:D 单选题3. 【第1 章】下列那一种不是影响黏土压实的施工方法：A 碾压B 振动压实C 夯实D 强夯正确答案:D 单选题4. 【第1 章】施工高度的含义指：A 设计标高B 自然地面标高C 设计标高减去自然标高D 场地中心标高减去自然标高正确答案:C 单选题5. 【第1 章】场地设计标高是指：A 平均标高B 场地中心标高C 加权标高D 场地内标高值最大正确答案:B 单选题6. 【第1 章】土方边坡自上而下分层开挖时，上台阶应比下台阶开挖深进不少于：A 10mB 20mC 30mD 40m正确答案:C 单选题7. 【第1 章】水泥土搅拌桩施工时通常采用的机械是：A 回转钻机B 浅水钻机C 搅拌桩机D 四杆钻机正确答案:C 单选题8.【第1 章】基坑(槽) 路堤的土方量可采用下列方法计算：A 三角棱柱体法B 等截面计算法C 拟柱体体积计算法D 四方棱柱体法正确答案:C 单选题9. 【第1 章】下列哪种不是影响边坡坡度的因素：A 施工工期B 气候条件C 开挖宽度D 开挖方法正确答案:C 单选题10. 【第1 章】拉铲挖掘机的开挖方式有：A 沟端开挖B 沟底开挖C 沟侧开挖D A+C正确答案:D 单选题11.【第1 章】最佳设计平面是在满足设计规划、生产工艺和运输、排水等要求的前提下，尽量使：A 土方量最小B 土方填挖平衡C 最小运输功D A+B正确答案:D 单选题12. 【第1 章】哪一种挖土机适合开挖停机面1、5m 以下的土方：A 铲运机B 抓铲挖土机C 正铲挖土机D 拉铲挖土机正确答案:C 单选题13.【第1 章】当填、挖土方量相差不大时，可采用什么方法计算土方量：A 工艺要求B 排水要求C 运输要求D 适当降低或者提高场地设计标高值正确答案:D 单选题14.【第1 章】下列哪一种土方机械适合500m 范围内的移挖作填：A 推土机B 铲运机C 装载机D 拉铲挖掘机正确答案:B 单选题15. 【第1 章】下列哪种不是刚板桩施工时的打桩方法：A 单独打入法B 整体打入法C 围檩插桩法D 分段复打桩正确答案:B 单选题16. 【第1 章】基槽支护结构的形式宜采有：A 横撑式支撑B 搅拌桩C 挡土板D 非重力式支护结构正确答案:A 单选题17.【第1 章】当场地地形变化比较平缓时，宜采用什么方法确定最佳设计平面：A 最小二乘法B 方格网法C 相关平差法D A+C正确答案:B 单选题18. 【第1 章】场地设计标高调整通常不考虑的因素有：A 土的最终可容性B 工程余土C 场外弃土D 当地的最高洪水位正确答案:D 判断题19.【第1 章】埋设井点的程序为：埋设井点管、排放总管、用弯联管接井点管与总管。

大工18秋《Web技术》在线作业3(满分题)

------------------------------------------------------------------------------------------------------------------------------ (单选题) 1: ()用于检查用户的输入数据是否在指定的上限与下限之间。

A: RangeValidatorB: CompareValidatorC: RegularExpressionValidatorD: CustomValidator正确答案:(单选题) 2: 当跨页传递数据时,哪个属性标识源页面是跨页提交()。

A: IspostBackB: IsCrossBackC: PreviousBackD: IsCrossPagePostBack正确答案:(单选题) 3: C#编写的网页后台代码被保存在扩展名为()文件中。

A: .aspxB: .cC: .csD: .asp正确答案:(单选题) 4: 在网页中要制作树状导航菜单,应使用的控件()。

A: TreeViewB: MenuC: SiteMapPathD: SiteMap正确答案:(单选题) 5: DataList通过哪个属性来控制每一行显示的记录条数()。

A: DataKeyFieldB: RepeatDirectionC: RepeatColumnsD: RepeatLayOut正确答案:(单选题) 6: 下面()不是页面之间传递值的方式。

A: 使用私有变量B: 使用Session变量C: 使用Server.TransferD: 使用Cookie正确答案:(单选题) 7: 数据集DataSet与SQL数据源之间的桥梁是()。

A: SqlConnectionB: SqlDataAdapterC: SqlCommandD: SqlTransaction正确答案:(单选题) 8: 如RadioButton控件被选中,其()属性值为true。

浙江大学远程教育2018面向对象程序设计在线作业答案

A a[1][0]
B a[1][1]
C a[2][0]
D a[2][1]
正确答案:A
单选题
11.【第2章】关于new运算符的下列描述中，（）是错误的。
A 它可以用来动态创建对象和对象数组
B 使用它创建的对象或对象数组可以使用运算符delete删除
C 使用它创建对象时要调用构造函数
D 使用它创建对象数组时必须指定初始值
单选题
25.【第2章】表达式：36.5-(12+7%4)/2.0的运算结果是（）类型。
A double
B int
C float
D char
正确答案:A
单选题
26.【第2章】下列描述中，（）是错误的。
A 内联函数主要解决程序的运行效率问题
B 内联函数的定义必须出现在内联函数第一次被调用之前
C 内联函数中可以包括各种语句
A printf("%s",*words[3]+1)
B printf("%s",*words[4]+1)
C printf("%s",*(words+3)+1)
D printf("%s",(*words+4)+1)
正确答案:C
单选题
23.【第2章】下列属于合法的C语言赋值语句的是（）。
A a=b=58
D 一个类中能定义多个析构函数。
正确答案:C
单选题
46.【第3章】类的友元允许访问该类的（）。
A 只能是数据成员
B 私有成员
C 只能是成员函数
D 只能是公有成员
正确答案:B
单选题

浙大远程教育Web程序设计离线作业参考答案

《Web程序设计》课程作业HTML部分一、选择题：1.以下哪个选项能显示成两行文字a)a) The first line and <BR> second lineb) The first line and \n second linec) The first line andsecond lined) The first line and second line2.<Ahref="mailto:*****************.cn">postyourresume</A>a)a) mailto:是一种协议，*****************.cn是email发送的地址b)mailto:*****************.cn是邮件被发送的地址c)mailto:*****************.cn是显示在web网页上的超链接d) 以上都不对3. <STYLE>标记可以出现在a)a) <HEAD>…</HEAD>中b) <BODY>…</BODY>中c) a)和b)都正确d) 以上都不正确4. 以下代码会生成什么样的Frame结构c)<html><frameset border=”5” cols=”*,20%”><frameset rows=”20%,*”><frame src=”top.html” name=”topFrame” scrolling=”NO”/><frame src=”left.html” name=”leftFrame”/></frameset><frame src=”right.html” name=”rightFrame” scrolling=”NO”/></frameset></html>a)b)d)5. 以下HTML代码在浏览器中的显示结果正确的是a)<TABLE WIDTH="100%" BORDER=1 BGCOLOR=gray><CAPTION> Sample table </CAPTION><TR ALIGN=CENTER ><TD COLSPAN=2>Cell</TD> <TD> Cell </TD></TR><TR ALIGN=CENTER ><TD ROWSPAN=2>Cell</TD> <TD> Cell </TD> <TD> Cell </TD></TR> <TR ALIGN=CENTER ><TD> Cell </TD> <TD> Cell </TD></TR></TABLE>a)Sample TableSample Table二、简答题：1、简述Web的工作原理答：web本意是蜘蛛网和网的意思。

东大18春学期《WEB程序设计》在线作业3

(单选题) 1: 类MyClass 中，下列哪条语句定义了一个只读的属性Count？A: private int Count;B: private int count; public int Count{ get{return count;} }C: public readonly int Count;D: public readonly int Count { get{ return count;} set{count = value;} }(单选题) 2: 下列语句创建了多少个string 对象？string[,] strArray = new string[3][4];A: 0B: 3C: 4D: 12(单选题) 3: web页面在关闭的时候触发的事件是：（）。

A: Page_LoadB: ClickC: ChangeD: Page_Unload(单选题) 4: 为了在程序中使用ODBC .NET 数据提供程序,应在源程序工程中添加对程序集______的引用.A: System.Data.dllB: System.Data.SQL.dllC: System.Data.OleDb.dllD: System.Data.Odbc.dll(单选题) 5: 某Command 对象cmd 将被用来执行以下SQL 语句，以向数据源中插入新记录：insert into Customers values(1000,"tom")请问，语句cmd.ExecuteNonQuery( );的返回值可能为( )A: 1B: 2C: 1000D: "tom"(单选题) 6: 以下语句获取表中第1行第1列（FirstName列）的数据，其中正确的语句是（）。

A: myTable.Rows[0]["FirstName"]B: myTable.Rows[1]["FirstName"]C: myTable.Rows[0][1]D: myTable.Rows[1][0](单选题) 7: 数据库连接语句：strConnString="Provider=SQLOLEDB;Data Source=(local);Initial Catalog=pubs;User ID=sa"中的Data Source=(local)是指（）。

东大16秋学期《WEB程序设计》在线作业1

16秋学期《WEB程序设计》在线作业1试卷总分：100 测试时间：--、单选题（共10 道试题，共50 分。

）1. 可以在开发环境的________窗口中对Windows操作系统的各种服务进行管理：A. 服务器管理器B. 类视图C. 属性窗口D. 解决方案资源管理器满分: 5 分2. 正则表达式“.{1,}[区,市,省]{1}.{1,}[区,市].{1,}[街,路]{1}[0-9]{1,}号.[公寓,小区]{1}[0-9]{1,}幢[0-9]{5}室”验证正确的是（）。

A. 浙江省杭州市下沙路256号富康公寓16幢18601室B. 上海市徐家汇区交大路245号高教村8幢306室C. 宁夏回族自治区吴忠市余名大街265号西湖小区8幢302室D. 浙江省杭州市西湖大道126号金星大厦16层1601室满分: 5 分3. 某Command 对象cmd 将被用来执行以下SQL 语句，以向数据源中插入新记录：insert into Customers values(1000,“tom”)请问，语句cmd.ExecuteNonQuery( );的返回值可能为( )A. 1B. 2C. 1000D. “tom”满分: 5 分4. 下列表达式表示必须输入3个以上由大小写字母、数字、符号“_”（下划线）、“－”（中划线）以及“．”组成的字符串，位数不限。

（）。

A. [A-Za-z0-9_\-\.]{3,}B. [A-Za-z0-9_/-/.]{3,}C. [A~Za~z0~9_\-\.]{3,}D. [A~Za~z0~9_/-/.]{3,}满分: 5 分5. 以下类MyClass 的属性count 属于_____属性.class MyClass{int i;int count{get{ return i; }}}A. 只读B. 只写C. 可读写D. 不可读不可写满分: 5 分6. web页面在载入的时候触发的事件是：（）。

18年4月清考《Web程序设计》考核作业

东北大学继续教育学院Web程序设计试卷（作业考核线上） A 卷学习中心：院校学号：姓名：（共7 页）一、选择题（请将所有正确答案写到答题卡中，每小题3分，共60分）1 . 目前,Microsoft .NET Framework 的发行包中包含以下.NET 数据提供程序A. SQL Server .NET 数据提供程序B. OLE DB .NET 数据提供程序C. ODBC .NET 数据提供程序D. XML .NET 数据提供程序2 . 用鼠标右击一个控件时出现的菜单一般称为：A. 主菜单B. 菜单项C. 快捷菜单D. 子菜单3 . 如果要设置TextBox为只读的，应设置：A. ReadOnly="true"B. ReadOnly="false"C. Enabled="true"D. Enabled="false"4 . 为创建在SQL Server中执行Select 语句的Command 对象,可先建立到SQL Server数据库的连接,然后使用连接对象的_______方法创建SqlCommand 对象.A. OpenB. OpenSQLC. CreateCommandD. CreateSQL5 . 网页中的表单是（）。

A. 一个容器类对象B. 一个图像对象C. 一个子网页D. 一个对话框6 . SQL Server 的Windows 身份验证机制是指,当网络用户尝试连接到SQL Server 数据库时,A. Windows 获取用户输入的用户和密码,并提交给SQL Server 进行身份验证,并决定用户的数据库访问权限B. SQL Server 根据用户输入的用户和密码,提交给Windows 进行身份验证,并决定用户的数据库访问权限C. SQL Server 根据已在Windows 网络中登录的用户的网络安全属性,对用户身份进行验证,并决定用户的数据库访问权限D. 登录到本地Windows 的用户均可无限制访问SQL Server 数据库7 . 分析下列程序：public class class4{private string _sData = "";public string sData{set{_sData = value;}}}在Main 函数中，在成功创建该类的对象obj 后，下列哪些语句是合法的？A. obj.sData = "It is funny!";B. Console.WriteLine(obj.sData);C. obj._sData = 100;D. obj.set(obj.sData);8 . 如果需要确保用户输入大于30的值，应该使用（）验证控件。

浙大远程操作系统原理在线作业及答案

操作系统原理在线作业1.对磁盘进行移臂调度时，既考虑了减少寻找时间，又不频繁改变动臂的移动方向的调度算法是（ C ）。

A 先来先服务B 最短寻找时间优先C 电梯调度D 优先级高者优先2.下列进程调度算法中，综合考虑进程等待时间和执行时间的是（ D ）。

A 时间片轮转调度算法B 短进程优先调度算法C 先来先服务调度算法D 高响应比优先调度算法O系统有三种常用方式来与主机交换数据，它们是程序轮询方式、中断方式和ＤＭＡ方式，其中ＤＭＡ方式主要由硬件来实现，此时高速外设和内存之间进行数据交换（ B ）。

A 不通过ＣＰＵ的控制，不利用系统总线B 不通过ＣＰＵ的控制，利用系统总线C 通过ＣＰＵ的控制，不利用系统总线D 通过ＣＰＵ的控制，利用系统总线4. 某进程由于需要从磁盘上读入数据而处于等待状态.当系统完成了所需的读盘操作后,此时该进程的状态将（ D ）。

A 从就绪变为运行B 从运行变为就绪C 从运行变为阻塞D 从等待变为就绪5.在段页式存储管理系统中时，每次从主存中取指令或取操作数，至少要访问（ C ）主存。

A 1次B 2次C 3次D 4次6.设某进程的页访问串为：1、3、1、2、4，工作集为3块，问：按FIFO页面替换算法，当访问4号页面时，应淘汰（ C ）号页面。

A 1B 2C 3D 47.假设一个正在运行的进程对信号量S进行了P操作后，信号量S的值变为-1，此时该进程将（ A ）。

A 转为等待状态B 转为就绪状态C 继续运行D 终止8. 下列选项中，降低进程优先级的合理时机是（ A ）。

A进程的时间片用完 B进程刚完成I/O，进入就绪队列 C进程长期处于就绪队列中 D进程从就绪态转为运行态9.两个进程合作完成一个任务，在并发执行中，一个进程要等待其合作伙伴发来信息，或者建立某个条件后再向前执行，这种关系是进程间的（ A ）关系。

A 同步B 互斥C 竞争D 合作10. 当被阻塞进程所等待的事件出现时，如所需数据到达或者等待的I/O操作已完成，则调用唤醒原语操作，将等待该事件的进程唤醒。

秋浙大远程面向对象程序设计的离线作业

秋-浙大远程-面向对象程序设计的-离线作业————————————————————————————————作者: ————————————————————————————————日期:浙江大学远程教育学院《面向对象程序设计》课程作业姓名：学号：年级：学习中心：—————————————————————————————第2章【2.3】测试下面的注释(它在C++风格的单行注释中套入了类似于C的注释)是否有效。

//ｔhis iｓa strａnge ／*waｙto do a commｅnt＊/答:此注释有效,单行注释中可以嵌套 / * …… * /方式的注释。

【2.4】以下这个简短的C＋+程序不可能编译通过，为什么?#include＜iostreaｍ>ｕsing namｅspaｃe stｄ;ｓuｍ(ｉｎt ａ，iｎt b)ｉｎt ｍain(){ｉnt ａ,b，c；cout<<＂Enter twｏnuｍbers：";cｉｎ>>a>＞b;c=sum(a，ｂ)；cｏuｔ<＜"suｍis："＜<ｃ;reｔｕrn 0;}sum(intａ,inｔb){ return a+b;}答：不可能通过编译.在usｉｎｇｎａmespacｅｓtｄ；后面加上ｓuｍ(int a,int b) 就可以通过。

【2.5】回答问题。

(1) 以下两个函数原型是否等价：ｆlｏat ｆun(inｔa,fｌoａtｂ,char *ｃ)；float fuｎ(ｉnｔ，ｆloat,char* );（2) 以下两个函数的第一行是否等价：float fun(int ａ，flｏat b,ｃhaｒ* ｃ）;floａt fｕn(iｎt,float,chaｒ＊);答:(１）这两个函数原型是等价的，函数原型中的参数名可以缺省。

(2)这两个函数的第1行是不等价的，因为这个函数的第1行中必须包含参数名。

浙大远程管理信息系统在线作业答案

浙大远程管理信息系统在线作业答案The document was finally revised on 2021窗体顶端您的本次作业分数为：100分1.【第1章】对管理信息系统进行综合，我们可以了解到，管理信息系统是由多个功能子系统组成的，这些功能子系统又可以分为业务处理、运行控制、管理控制和（）几个主要的信息处理部分。

A 财务管理B 信息管理C 人力资源管理D 战略管理正确答案:D2.【第1章】DSS是以下哪个术语的简称？A 决策支持系统B 群体决策支持系统C 智能决策支持系统D 管理决策系统正确答案:A3.【第1章】管理信息系统的应用离不开一定的环境和条件，这里所说的“环境”具体指的是（）。

A 组织所处的自然环境B 组织所处的社会环境C 组织内外各种因素的综合D 组织所处的自然环境和社会环境的综合正确答案:C4.【第1章】按照不同级别管理者对管理信息的需要，通常把管理信息分为以下三级（）。

A 公司级、工厂级、车间级B 工厂级、车间级、工段级C 厂级、处级、科级D 战略级、战术级、作业级正确答案:D5.【第1章】从管理决策问题的性质来看，在运行控制层上的决策大多属于（）的问题。

A 结构化B 半结构化C 非结构化D 以上都有正确答案:A6.【第1章】EDPS是以下哪个术语的简称？A 电子数据处理系统B 单项数据处理阶段C 综合数据处理阶段D 管理信息系统正确答案:A7.【第1章】（）反映了某个企业、组织或部门所涉及的数据本身的内容，同时也反映了数据之间的联系。

A 数据库B 数据文件（表）C 文件系统D 数据结构正确答案:A8.【第1章】局域网与使用调制解调器进行计算机通信的远程网相比，它的信息传送速度要（）。

A 高得多B 低得多C 差不多D 无法比较正确答案:A9.【第1章】数据库是比文件系统更高级的一种数据组织方式。

正确错误正确答案: 对10.【第1章】DSS是在人和计算机交互的过程中帮助决策者探索可能的方案，为管理者提供决策所需的信息。

16秋浙大《程序设计基础(VB)》在线作业

浙江大学17春16秋浙大《程序设计基础(VB)》在线作业一、单选题（共35 道试题，共70 分。

）1. Visual Basic规定，不同类型的数据占用存储空间的长度是不同的。

下列各组数据类型中，满足占用存储空间从小到大顺序排列的是（）。

A. Byte，Integer，Long，DoubleB. Byte，Integer，Double，BooleanC. Boolean，Byte，Integer，DoubleD. Boolean，Byte，Integer，Long正确答案：2. Visual Basic 是一种面向对象的程序设计语言，构成对象的三要素是( ) 。

A. 属性、事件、方法B. 控件、属性、事件C. 窗体、控件、过程D. 窗体、控件、模块正确答案：3. 设菜单中有一个菜单项为"Open"。

若要为该菜单命令设计访问键，即按下Alt及字母O 时，能够执行"Open"命令，则在菜单编辑器中设置“Open"命令的方式是（）。

A. 把Caption属性设置为&OpenB. 把Caption属性设置为O&penC. 把Name属性设置为&OpenD. 把Name属性设置为O&pen正确答案：4. 实现Windows应用程序之间信息共享的技术是（）。

A. 开放式数据链接功能B. 对象链接与嵌入C. 动态链接库D. 以上三项都是正确答案：5. Visual Basic根据计算机访问文件的方式将文件分成三类，其中不包括什么。

A. 顺序文件B. Unix文件C. 二进制文件D. 随机文件正确答案：6. 不能正确表示条件“两个整型变量A和B之一为0，但不能同时为0”的布尔表达式是（）。

A. A*B=0 AND A<>BB. （A=0 OR B=0）AND A<>BC. A=0 AND B<>0 OR A<>0 AND B=0D. A*B=0 AND （A=0 OR B=0）正确答案：7. OPTION EXPLICIT语句不可以放在（）。

奥鹏大工18秋《Web技术》大作业及要求

大工作业答案《Web技术》大作业及要求网络教育学院《Web技术》课程设计题目：奥鹏作业答案：层次：专业：年级：年春/秋季学号：学生：辅导教师：完成日期：年月日大工作业答案《Web技术》课程设计注意：从以下5个题目中任选其一作答。

题目一：设计一个在线考试系统总则：利用相关的web平台和数据库设计一个在线答疑系统。

（具体工具平台及语言可以自己根据自己的习惯选用）要求：（1）建立一个完整的文件夹，所有系统内容都要包含在建立的这个文件夹内，系统各类元素文件在总文件夹中以子文件夹形式分类清楚。

（如图片都放在子文件夹image里，html页都放在pages子文件夹内，数据库文件放在DB 文件夹内）（2）系统用户分为：学生用户、教师用户和管理员用户。

不同的用户有用不同的权限，各自完成各自的管理功能，不同的用户看到不同的系统功能。

（3）用SQL Server/MySql创建后台数据库，并完善数据库的结构和数据。

（4）普通用户的主要功能模块：登录与注册、答疑模块等。

（5）管理员的主要功能模块：用户信息的管理、答疑室的管理（添加、删除、修改、查询）。

（6）数据库至少包含：用户信息表、答疑室表等。

（7）附带一份课程设计功能说明书，也放在总文件夹中。

功能书包含：介绍各模块功能、数据库设计说明、各模块核心程序、总结及体会。

（8）将所完成的相关文件压缩成一个文件，文件名为[姓名奥鹏卡号奥鹏作业答案]（如戴卫东101410013979浙江台州奥鹏奥鹏作业答案[1]VIP ）作业提交：大作业上交时文件名写法为：[姓名奥鹏卡号奥鹏作业答案]（如：戴卫东101410013979浙江台州奥鹏奥鹏作业答案[1]VIP）以附件形式上交离线作业（附件的大小限制在10M以内），选择已完成的作业（注意命名），点提交即可。

如下图所示。

注意事项：独立完成作业，不准抄袭其他人或者请人代做，如有雷同作业，成绩以零分计！题目二：资源共享平台总则：利用先关的平台和数据库设计一个资源共享平台。

东大18秋学期《WEB程序设计》在线作业3

------------------------------------------------------------------------------------------------------------------------------ (单选题) 1: 已知ds1,ds2 分别代表两个不同的DataSet 对象.其中ds1 已包含名为"Customer"的DataTable 对象,且该DataTable 对象被变量dt_Customer 引用.已知dt_Customer表中有100 条记录,则执行下列语句后,新的数据表new_dt_Customer 中包含______条记录. DataTable new_dt_Customer = dt_Customer.Copy();A: 0B: 100C: 200D: 300正确答案：(单选题) 2: 数据集ds 中两数据表(父表:Customer;子表:CartItems)之间有如图外键约束. 父表Customers 中有一行数据的CustomerID=100,子表CartItems 中有20 行数据的CustomerID=100.若从父表中删除该行,则:A: 子表不发生任何变化B: 子表中20 个相关行的CustomerID 列的值变为DBNullC: 子表中20 个相关行被删除D: 引发Exception正确答案：(单选题) 3: 下列表达式表示必须输入3个以上由大小写字母、数字、符号“_”（下划线）、“－”（中划线）以及“．”组成的字符串，位数不限。

（）。

A: [A-Za-z0-9_\-\.]{3,}B: [A-Za-z0-9_/-/.]{3,}C: [A~Za~z0~9_\-\.]{3,}D: [A~Za~z0~9_/-/.]{3,}正确答案：(单选题) 4: 在编写页面时，若希望在每次页面载入时执行某个操作，则应将相应代码写入_______事件处理程序中。

2018年浙江大学大学生程序设计竞赛试题

The 18th Zhejiang UniversityProgramming ContestSponsored byContest SectionApril 7, 2018Problem ListThis problem set should contain 10 (ten) problems on 18 (eighteen) numbered pages. Please inform a runner immediately if something is missing from your problem set.Prepared by Zhejiang University ACM/ICPC Problem Setter Team.Problem A.Pretty MatrixDescriptionDreamGrid’s birthday is coming.As his best friend,BaoBao is going to prepare a gift for him.As we all know,BaoBao has a lot of matrices.This time he picks an integer matrix with n rows and m columns from his collection,but he thinks it’s not pretty enough.On the one hand,he doesn’t want to be stingy,but some integers in the matrix seem to be too small.On the other hand,he knows that DreamGrid is not good at algebra and hates large numbers,but some integers in the matrix seem to be too large and are not suitable for a gift to DreamGrid.Based on the above consideration,BaoBao declares that a matrix is pretty,if the following conditions are satisﬁed:1.For every integer a ij in the matrix,a ij≥A.2.For every integer a ij in the matrix,a ij≤B.where a ij is the integer located at the i-th row and the j-th column in the matrix,and A and B are two integers chosen by BaoBao.Given the matrix BaoBao picks,along with the two integers A and B,please help BaoBao change some integers in the matrix(BaoBao can change an integer in the matrix to any integer)so that the matrix becomes a pretty matrix.As changing integers in the matrix is tiring,please tell BaoBao the minimum number of integers in the matrix he has to change to make the matrix pretty.InputThere are multiple test cases.Theﬁrst line of input is an integer T(about100),indicating the number of test cases.For each test case:Theﬁrst line contains four integers n,m,A and B.(1≤n,m≤100,1≤A,B≤105).Their meanings are described above.For the next n lines,the i-th line contains m integers a i1,a i2,...,a im(1≤a ij≤105),representing the original matrix.OutputFor each test case output one line indicating the answer.If it’s impossible to make the matrix pretty, print“No Solution”(without quotes)instead.Examplestandard input standard output23423 3222 2123 231003 2121 122No SolutionProblem B.LiblumeDescriptionDreamGrid has a matrix A consisting of lowercase English letters.You are allowed to rearrange its rows any number of times(including zero times).Please calculate the maximum possible area of a submatrix in which every row and column is a palindromic string after the rearrangements.Let’s assume that the rows of matrix A are numbered from1to n from top to bottom and the columns are numbered from1to m from left to right.A matrix cell on the intersection of the i-th row and the j-th column can be represented as(i,j).A palindromic string is a string that can be read the same way from left to right and from right to left. For example,“abacaba”,“z”,“abba”are palindromes.A submatrix of matrix A is a group of four integers d,u,l,r(1≤d≤u≤n,1≤l≤r≤m).The submatrix contains all the cells(i,j)satisfying both d≤i≤u and l≤j≤r.The area of the submatrix is the number of cells it contains.InputThere are multiple test cases.Theﬁrst line of input contains an integer T,indicating the number of test cases.For each test case:Theﬁrst line contains two integers n and m(1≤n×m≤2×105)–the number of rows and columns of the matrix.Each of the next n lines contains m characters denoting the matrix.It is guaranteed that the sum of n×m in all cases does not exceed2×106.OutputFor each test case,output an integer denoting the maximum possible area of a submatrix in which every row and column is a palindromic string after the rearrangements.Examplestandard input standard output4 22 aa aa 33 aaa aaa bbb 33 abc def ghi 33 aba bab aba 4 9 1 9Problem C.Mergeable StackDescriptionGiven n initially empty stacks,there are three types of operations:•1s v:Push the value v onto the top of the s-th stack.•2s:Pop the topmost value out of the s-th stack,and print that value.If the s-th stack is empty, pop nothing and print“EMPTY”(without quotes)instead.•3s t:Move every element in the t-th stack onto the top of the s-th stack in order.Precisely speaking,denote the original size of the s-th stack by S(s),and the original size of the t-th stack by S(t).Denote the original elements in the s-th stack from bottom to top by E(s,1),E(s,2),...,E(s,S(s)),and the original elements in the t-th stack from bottom to top by E(t,1),E(t,2),...,E(t,S(t)).After this operation,the t-th stack is emptied,and the elements in the s-th stack from bottom to top becomes E(s,1),E(s,2),...,E(s,S(s)),E(t,1),E(t,2),...,E(t,S(t)).Of course,if S(t)=0, this operation actually does nothing.There are q operations in total.Pleaseﬁnish these operations in the input order and print the answer for every operation of the second type.InputThere are multiple test cases.Theﬁrst line of the input contains an integer T,indicating the number of test cases.For each test case:Theﬁrst line contains two integers n and q(1≤n,q≤3×105),indicating the number of stacks and the number of operations.Theﬁrst integer of the following q lines will be op(1≤op≤3),indicating the type of operation.•If op=1,two integers s and v(1≤s≤n,1≤v≤109)follow,indicating an operation of theﬁrst type.•If op=2,one integer s(1≤s≤n)follows,indicating an operation of the second type.•If op=3,two integers s and t(1≤s,t≤n,s=t)follow,indicating an operation of the third type. It’s guaranteed that neither the sum of n nor the sum of q over all test cases will exceed106.OutputFor each operation of the second type output one line,indicating the answer.Examplestandard input standard output2215 1110 1111 1212 1213 312 1214 2121212121 321 22222237 312 313 321 2122232313121110 EMPTY 14 EMPTY EMPTY EMPTY EMPTY EMPTY EMPTYProblem D.RAID-ZOJDescriptionIt’s a beautiful day outside.Birds are singing.Flowers are blooming.You turn on your computer and are ready for a day’s work.But,alas,your disk F AILED and all your data goes down the drain!But don’t worry,it’s time for RAID(Redundant Array of Independent Disks)to come to your rescue!The idea behind RAID is quite simple:make copies of data on other disks.This is exactly what RAID1 does.RAID1makes a mirror for every disk,so each bit of your data is stored identically on two diﬀerent disks.If one disk fails,you can still restore your data from another disk.RAID1is nice,but isn’t clever enough,as you can only use50%of your disks to store meaningful data. RAID3is a smarter approach.Let’s say you have m data disks,each containing n bits,where the i-th bit in the j-th data disk(note that we count i from0to(n−1),and j from0to(m−1))is indicated as b i,j. RAID3only requires you to buy one more disk,called the row parity disk,to store extra data.Let the i-th bit in the row parity disk to be r i.RAID3calculates r i as follows:r i=b i,0⊕b i,1⊕···⊕b i,m−2⊕b i,m−1where⊕is the bitwise exclusive or operation(0⊕0=1⊕1=0,0⊕1=1⊕0=1).With the row parity disk in hand,if the j-th data disk fails,we can restore the data on that disk by adjusting the above formula tob i,j=r i⊕b i,0⊕b i,1⊕···⊕b i,j−2⊕b i,j−1⊕b i,j+1⊕b i,j+2⊕···⊕b i,m−2⊕b i,m−1of your disks to store meaningful data!Horray!What’s more,you can now use mm+1RAID3seems to be such a perfect approach,that we even put ourselves under the illusion that we can sit back and relax after deploying RAID5(similar to RAID3,but more reliable)on ZOJ and stop thinking about disk failures.But guess what,the careless administrator of the data center didn’t notice the warning when a disk failure occurred,and when another disk failed in May2016,everything was too late.RAID3and RAID5can both protect their users from one disk failure,but if two disks fail simultaneously (though the probability is small)or you have a careless data center administrator,they still can’t help you restore your data.This is why we invent RAID-ZOJ.RAID-ZOJ is another RAID approach hoping to guard its users against two simultaneous disk failures. To achieve this goal,apart from the row parity disk,we add a second parity disk to the system,called the diagonal parity disk.For a bit b i,j on the data disk,we deﬁne its diagonal value D(b i,j)=(i+j)mod n.Let d i be the i-th bit on the diagonal parity disk,we calculate d i as follows:⊕d i=b k,jD(b k,j)=iwhich means that d i is calculated as the bitwise exclusive or of every bit whose diagonal value equals i. To help you understand,check the sampleﬁgure showing a RAID-ZOJ system with n=3and m=4 below.Now comes theﬁnal problem:is RAID-ZOJ capable of restoring data on two failed disks?We now provide you with a RAID-ZOJ system with m data disks,each containing n bits,a row parity disk,and a diagonalparity disk.In this system,exactly two data disks are broken.Your task is to restore the data on the two failed data disks so that the restored data are consistent with the two parity disks.As there might be multiple valid answers,you just need to calculate the minimum and the maximum possible number of1s in the broken bits.InputThere are multiple test cases.Theﬁrst line of input contains an integer T,indicating the number of test cases.For each test case:Theﬁrst line contains two integers n and m(1≤n×m≤106),indicating the number of bits in each data disk and the number of data disks.Each of the next n lines contains m characters denoting the bits in the data disks.The j-th character on the i-th line(don’t forget we count both i and j from0)is b i,j(b i,j∈{0,1,X}),indicating the i-th bit in the j-th data disk,where b i,j=X(ASCII code:88)means that this bit is broken and its your task to restore its value.The next line contains n characters r0,r1,...,r n−1(r i∈{0,1}),indicating the bits in the row parity disk. The next line contains n characters d0,d1,...,d n−1(d i∈{0,1}),indicating the bits in the diagonal parity disk.It’s guaranteed that there exist exactly two diﬀerent integers a and b such that1≤a,b≤m and b i,a=b i,b=X for all1≤i≤n.It’s also guaranteed that the sum of n×m over all test cases will not exceed3×106.OutputFor each test case output one line containing two integers separated by a space,indicating the minimum and the maximum possible number of1s in the broken bits.If no valid answer exists,print“No Solution”(without quotes)instead.Examplestandard input standard output434 0XX1 1XX1 0XX0 000 110 23 X1X X0X 100135 X0X01 X0X01 X0X01 101 010 12 XX111504No Solution 11NoteThe two possible answers for theﬁrst sample test case are shown as follows(the restored broken bits are shown in bold):Theﬁrst possible answer hasﬁve1s in the broken bits,while the second possible answer has one1.So the answer is“15”.The four possible answers for the second sample test case are shown as follows(the restored broken bits are shown in bold):Theﬁrst possible answer has no1in the broken bits,while the second and the third possible answer each has two1s,and the fourth possible answer has four1s.So the answer is“04”.Problem E.Crosses PuzzlesDescriptionGiven a grid with n rows and m columns,each cell may contain nothing or a cross consisting of three dotted segments and one solid segment.Let’s denote the cell located on the i-th row and the j-th columnby(i,j).When you click on a cross,it will rotate90degrees clockwise,thus changing the direction the solid segment points to.Your goal is to click on some crosses in some proper order so that the solid segment of every cross points upwards.But beware,this task is harder than you think it may be.After the rotation of a cross in(i,j)ends,the cross(i′,j′)pointed by the solid segment of(i,j)will also begin to rotate90degrees clockwise,and after the rotation of(i′,j′)ends,another cross may be triggered if pointed by the solid segment of(i′,j′)... This chain reaction comes to an end only when the solid segment of the previously-rotated cross pointsto an empty cell or points out of the map.To help you understand,please check the following pictures showing a chain reaction on a grid withn=m=3step by step.This is the initial state before the chain reaction.After clicking,(1,1)rotates90degrees clockwise Now let’s click on the cross(1,1).and its solid segment points to(1,2)(1,2)is activated and rotates90degrees clockwise.(1,3)is activated and rotates90degrees clockwise.After rotation,its solid segment points to(1,3).After rotation,its solid segment points back to(1,2).(1,2)is activated again and rotates.(2,2)is activated and rotates90degrees clockwise. After rotation,its solid segment points to(2,2).After rotation,its solid segment points to(2,3).As(2,3)is an empty cell,the chain reaction stops.Given a grid with n rows and m columns,pleaseﬁnd out a proper order to click on the crosses so that the solid segment of every cross points upwards.You can perform at most6000clicks for each test case. InputThere are multiple test cases.Theﬁrst line of the input contains an integer T(1≤T≤50),indicating the number of test cases.For each test case:Theﬁrst line contains two integers n and m(1≤n,m≤10),indicating the size of the grid.For the following n lines,the i-th line contains m integers t i,1,t i,2,...,t i,m(−1≤t i,j≤3),where t i,j indicates the initial state of the cross at intersection(i,j).•If t i,j=0,the solid segment of the cross in cell(i,j)is initially pointing upwards.•If t i,j=1,the solid segment of the cross in cell(i,j)is initially pointing leftwards.•If t i,j=2,the solid segment of the cross in cell(i,j)is initially pointing downwards.•If t i,j=3,the solid segment of the cross in cell(i,j)is initially pointing rightwards.•If t i,j=−1,cell(i,j)is an empty cell.OutputFor each test case,ﬁrst output one line containing one integer k(0≤k≤6000),indicating the number of clicks.Then k lines follow.Each line contains two integers i,j(1≤i≤n,1≤j≤m)separated by a space,indicating that you click on the cross in cell(i,j).If there’s no way to make the solid segment of every cross point upwards in6000clicks,just output“-1”(without quotes)instead.Examplestandard input standard output232 23 -10 32 22 32 018 11 12 12 22 22 31 31 32 5 11 21 21 21 12NoteWe’ve also prepared a playable example on the webpage.You can check that example for further understanding.Problem F.Schr¨o dinger’s KnapsackDescriptionDreamGrid has a magical knapsack with a size capacity of c called the Schr¨o dinger’s knapsack(or S-knapsack for short)and two types of magical items called the Schr¨o dinger’s items(or S-items for short). There are n S-items of theﬁrst type in total,and they all have a value factor of k1;While there are m S-items of the second type in total,and they all have a value factor of k2.The size of an S-item is given and is certain.For the i-th S-item of theﬁrst type,we denote its size by s1,i;For the i-th S-item of the second type,we denote its size by s2,i.But the value of an S-item remains uncertain until it is put into the S-knapsack(just like Schr¨o dinger’s cat whose state is uncertain until one opens the box).Its value is calculated by two factors:its value factor k,and the remaining size capacity r of the S-knapsack just after it is put into the S-knapsack.Knowing these two factors,the value v of an S-item can be calculated by the formula v=kr.For a normal knapsack problem,the order to put items into the knapsack does not matter,but this is not true for our Schr¨o dinger’s knapsack problem.Consider an S-knapsack with a size capacity of5,an S-item with a value factor of1and a size of2,and another S-item with a value factor of2and a size of 1.If we put theﬁrst S-item into the S-knapsackﬁrst and then put the second S-item,the total value of the S-items in the S-knapsack is1×(5−2)+2×(3−1)=7;But if we put the second S-item into the S-knapsackﬁrst,the total value will be changed to2×(5−1)+1×(4−2)=10.The order does matter in this case!Given the size of DreamGrid’s S-knapsack,the value factor of two types of S-items and the size of each S-item,please help DreamGrid determine a proper subset of S-items and a proper order to put these S-items into the S-knapsack,so that the total value of the S-items in the S-knapsack is maximized. InputTheﬁrst line of the input contains an integer T(about500),indicating the number of test cases.For each test case:Theﬁrst line contains three integers k1,k2and c(1≤k1,k2,c≤107),indicating the value factor of the ﬁrst type of S-items,the value factor of the second type of S-items,and the size capacity of the S-knapsack. The second line contains two integers n and m(1≤n,m≤2000),indicating the number of theﬁrst type of S-items,and the number of the second type of S-items.The next line contains n integers s1,1,s1,2,...,s1,n(1≤s1,i≤107),indicating the size of the S-items of theﬁrst type.The next line contains m integers s2,1,s2,2,...,s2,m(1≤s2,i≤107),indicating the size of the S-items of the second type.It’s guaranteed that there are at most10test cases with their max(n,m)larger than100.OutputFor each test case output one line containing one integer,indicating the maximum possible total value of the S-items in the S-knapsack.Examplestandard input standard output3327 2343 132 1210 34 212 3231 125 112123 45 10NoteFor theﬁrst sample test case,you canﬁrst choose the1st S-item of the second type,then choose the 3rd S-item of the second type,andﬁnally choose the2nd S-item of theﬁrst type.The total value is 2×(7−1)+2×(6−2)+3×(4−3)=23.For the second sample test case,you canﬁrst choose the4th S-item of the second type,then choose the2nd S-item of theﬁrst type,then choose the2nd S-item of the second type,then choose the 1st S-item of the second type,andﬁnally choose the1st S-item of theﬁrst type.The total value is 2×(10−1)+1×(9−1)+2×(8−2)+2×(6−3)+1×(3−2)=45.The third sample test case is explained in the description.It’s easy to prove that no larger total value can be achieved for the sample test cases.Problem G.Traﬃc LightDescriptionDreamGrid City is a city with n×m intersections arranged into a grid of n rows and m columns.The intersection on the i-th row and the j-th column can be described as(i,j),and two intersections(i1,j1) and(i2,j2)are connected by a road if|i1−i2|+|j1−j2|=1.At each intersection stands a traﬃc light.A traﬃc light can only be in one of the two states:0and1. If the traﬃc light at the intersection(i,j)is in state0,one can only move from(i,j)to(i+1,j)or (i−1,j);If the traﬃc light is in state1,one can only move from(i,j)to(i,j+1)or(i,j−1)(of course, the destination must be another intersection in the city).BaoBao lives at the intersection(s i,s j),and he wants to visit his best friend DreamGrid living at the intersection(f i,f j).After his departure,in each minute the following things will happen in order:•BaoBao moves from his current intersection to another neighboring intersection along a road.As a law-abiding citizen,BaoBao has to obey the traﬃc light rules when moving.•Every traﬃc light changes its state.If a traﬃc light is in state0,it will switch to state1;If a traﬃc light is in state1,it will switch to state0.As an energetic young man,BaoBao doesn’t want to wait for the traﬃc lights,and he must move in each minute until he arrives at DreamGrid’s house.Please tell BaoBao the shortest possible time he can move from(s i,s j)to(f i,f j)to meet his friend,or tell him that this is impossible.InputThere are multiple test cases.Theﬁrst line of the input contains an integer T,indicating the number of test cases.For each test case:Theﬁrst line contains two integers n and m(1≤n×m≤105),indicating the size of the city.For the following n lines,the i-th line contains m integers t i,1,t i,2,...,t i,m(0≤t i,j≤1),where t i,j indicates the initial state of the traﬃc light at intersection(i,j).The next line contains four integers s i,s j,f i and f j(1≤s i,f i≤n,1≤s j,f j≤m),indicating the starting intersection and the destination intersection.It’s guaranteed that the sum of n×m over all test cases will not exceed3×105.OutputFor each test case output one line containing one integer,indicating the shortest possible time(in minute) BaoBao can move from(s i,s j)to(f i,f j)without stopping.If it is impossible for BaoBao to arrive at DreamGrid’s house,print“-1”(without quotes)instead.Examplestandard input standard output423 110 010 1321 23 100 110 1312 221010 1122 1201 11113 5 -1 0NoteFor theﬁrst sample test case,BaoBao can follow this path:(1,3)→(2,3)→(2,2)→(2,1).For the second sample test case,due to the traﬃc light rules,BaoBao can’t go from(1,3)to(1,2)directly. Instead,he should follow this path:(1,3)→(2,3)→(2,2)→(2,1)→(1,1)→(1,2).For the third sample test case,it’s easy to discover that BaoBao can only go back and forth between(1,1) and(1,2).Problem H.Boolean ExpressionDescriptionGiven a valid boolean expression consisting of seven types of characters‘T’(true),‘F’(false),‘!’(not),‘&’(and),‘|’(or),‘(’(left bracket)and‘)’(right bracket),you can add any number of these seven types of characters into the expression to make it true(of course,the expression must still be valid).Please calculate the minimum number of characters you have to add into the expression to make it true. The priority of the boolean operators are:‘!’>‘&’>‘|’,just the same as their priority in most of the programming languages.To be speciﬁc,a valid boolean expression can be expressed by the following BNF,<bool>::=‘T’|‘F’|‘(’<expr>‘)’<not-expr>::=‘!’<not-expr>|<bool><and-expr>::=<and-expr>‘&’<not-expr>|<not-expr><expr>::=<expr>‘|’<and-expr>|<and-expr>where<expr>is a valid boolean expression.InputThere are multiple test cases.Theﬁrst line of the input contains an integer T,indicating the number of test cases.For each test case:Theﬁrst and only line contains a boolean expression s(1≤|s|≤5×105)without space consisting of ‘T’,‘F’,‘!’,‘&’,‘|’,‘(’and‘)’.It’s guaranteed that the given expression is valid,and the sum of|s|over all test cases will not exceed 5×106.OutputFor each test case output one line containing one integer,indicating the minimum number of characters you have to add into the expression to make it true.Examplestandard input standard output4!T&FT&!F|T&F(T&(!T|(!!F))&T) !!!!!F 2 0 1 0NoteFor theﬁrst sample test case,we can add a pair of brackets and change the expression to!(T&F).For the second sample test case,note that the priority of‘&’is higher than‘|’,so the original expression is already true.For the third sample test case,we can add a‘!’and change the expression to!(T&(!T|(!!F))&T).We kindly remind you that the stack size of the online judge system is limited to8M,so please use your stack space wisely.Problem I.HonoriﬁcsDescriptionAn honoriﬁc is a title that conveys esteem or respect for position or rank when used in addressing or referring to a person.Sometimes,the term“honoriﬁc”is used in a more speciﬁc sense to refer to an honorary academic title.It is also often conﬂated with systems of honoriﬁc speech in linguistics,which are grammatical or morphological ways of encoding the relative social status of speakers.In Japenese,“-san”(sometimes pronounced as“han”in Kansai dialect)is the most commonplace honoriﬁc and is a title of respect typically used between equals of any age.Although the closest analogs in English are the honoriﬁcs“Mr.”,“Miss”,“Ms.”,or“Mrs.”,“-san”is almost universally added to a person’s name;“-san”can be used in formal and informal contexts and for any gender.Because it is the most common honoriﬁc,it is also the most often used to convert common nouns into proper ones.In Korean,“-ssi”is the most commonly used honoriﬁc used amongst people of approximately equal speech level.It is attached at the end of the full name,such as Gim Cheolsu-ssi or simply after theﬁrst name, Cheolsu-ssi,if the speaker is more familiar with someone.Appending“-ssi”to the surname,for instance Gim-ssi,can be quite rude,as it indicates that the speaker considers himself to be of a higher social status than the person he is speaking to.In this problem,you are given a series of Japanese or Korean names,please add an honoriﬁc to each name. You should append“-san”to each Japanese name,and append“-ssi”to each Korean name.In the inputﬁle,Japanese names are given in Kunrei-shiki romanization system,and Korean names are given in revised romanization system.Each name consists of two capitalized words composed of Latin alphabet.Theﬁrst word is the surname(family name),and the second word is the forename(given name). The test data contains10000randomly generated names.train.txt is the training data for you,which can be downloaded from the webpage of this problem,and its format is described in the input section.The training data only has6000names covering60%∼70%surnames and forenames.Please pay attention to the generalization of your solution.Since there may be some rare and very confused cases even for a human,your solution only needs to achieve at least99%correct rate on the test.InputFor the formal test,there are multiple test cases.Theﬁrst line of input contains an integer T(T=10000), indicating the number of test cases.For each test case:Theﬁrst line contains two words(theﬁrst character of each word is capitalized),which is either a Japanese name or a Korean name.The length of each word will not exceed20.For train.txt,theﬁrst line contains an integer T(T=6000),indicating the number of training data. For each training data:Theﬁrst line contains two words(theﬁrst character of each word is capitalized),which is either a Japanese name or a Korean name followed by the correct honoriﬁc of this data.The length of each word(without the honoriﬁc)will not exceed20.OutputFor each test case,output the name and the honoriﬁc.You should append“-san”to each Japanese name, and append“-ssi”to each Korean name.You need to achieve99%correct rate on it at least.Examplestandard input standard output7Fuzii Mina Nakamoto Yuuta Song Junggi Hirai Momo Seonu Jeonga Gong YuBang Mina Fuzii Mina-san Nakamoto Yuuta-san Song Junggi-ssi Hirai Momo-san Seonu Jeonga-ssi Gong Yu-ssiBang Mina-ssiNoteBoth the training data and the test data are generated as follows:for each case,toss a coin with an equally likely outcome to decide whether it is a Japanese name or a Korean name.If it is a Japanese name,then a surname and a forename are picked randomly from the Japanese surname list and Japanese forename list respectively;If it is a Korean name,then a surname and a forename are picked randomly from the Korean surname list and Korean forename list respectively.In theﬁnal test data,the Japanese surname and forename list contain about480surnames and2800 forenames respectively,and the Korean surname and forename list contain about200surnames and2800 forenames respectively.In the downloadable training data,the Japanese/Korean surnames/forenames list is a random subset of the corresponding one in theﬁnal test data.These are no common surnames between the Japanese surname list and the Korean surname list,so a generated Japanese name and a generated Korean name will not be the same.However,there are a very small amount of common forenames between the Japanese forename list and the Korean forename list,so please be careful with these cases.Reference:https:///wiki/Honorifichttps:///wiki/Japanese_honorifics#Sanhttps:///wiki/Korean_honorifics#-ssi。

东师《基于Web程序设计》18秋在线作业2（满分）

东师《基于Web程序设计》18秋在线作业2（满分）(单选题) 1: 如果希望打开的记录集可以前后移动指针，并且可读可写，则应该为下⾯哪⼀句（）A: rs.open sql,connB: rs.open sql,conn,1,3C: rs.open sql,conn,,3D: rs.open sql,conn,1正确答案:(单选题) 2: 下列哪个函数可以将数值转换为字符串？A: CdateB: CintC: CStrD: CDbl正确答案:(单选题) 3: 执⾏完如下语句后，a的值为() <% Dim a a=FormatNumber(10.223344,3) %>A: 10.223344B: 10.21C: 10.223D: 10.000正确答案:(单选题) 4: /doc/f23e485db42acfc789eb172ded630b1c58ee9b06.html 框架中，服务器控件是为配合Web表单⼯作⽽专门设计的，（）是服务器控件A: JS控件B: Web控件C: ⾃定义控件D: COM控件正确答案:(单选题) 5: 执⾏完 a=5 Mod 3 语句后，a的值为A: 0B: 2C: 3D: 5正确答案:(单选题) 6: 以下标记中,()可⽤于在⽹页插⼊图像A: 标记B:标记C:标记D: 标记正确答案:(单选题) 7: Response.Redirect "login.asp"表⽰()A: 覆盖login.aspB: 关闭login.aspC: 在⼀个新窗⼝中打开login.aspD: 重定向到login.asp正确答案:(单选题) 8: 下⾯有关数据类型的描述中不正确的是( )A: 在引⽤类型中，有可能两个变量引⽤同⼀个对象。

B: bool类型中可以⽤数字1表⽰trueC: byte类型的取值范围是0~255D: 可以通过转义符⽅式输⼊字符正确答案:(单选题) 9: 在/doc/f23e485db42acfc789eb172ded630b1c58ee9b06.html 中，Application是（）类的实例A: HttpApplicationB: HttpApplicationUtilityC: HttpApplicationStateD: Page正确答案:(单选题) 10: 函数Instr(“jjshang@/doc/f23e485db42acfc789eb172ded630b1c58ee9b06.html ”,”@”)的返回的值为A: 3B: 5C: 7D: 8正确答案:(单选题) 11: 如果指针指向第1条记录，则rs.bof和rs.eof的值分别为A: True、FalseB: False、TrueC: True、TrueD: False、False正确答案:(单选题) 12: ⽤于让代码在每次/doc/f23e485db42acfc789eb172ded630b1c58ee9b06.html ⽹页加载是执⾏的事件处理程序的名称是（）A: Page_loadB: Page_startC: Page_InitD: Page_Click正确答案:(单选题) 13: 对于利⽤ Dim a(4,5) 语句定义的⼆维数组，Ubound(a,1)将返回()A: 0B: 4C: 5D: 6正确答案:(单选题) 14: 在ASP。

web程序设计大作业

web程序设计课程考试大作业姓名：李德坤学号：0121310870824专业：计算机1301班题目3：结合所学的课程内容，论述web程序设计的相关技术、应用及学习体会1.HTML的主要知识点框架是一种在一个浏览器窗口中显示多个HTML文件的网页制作技术。

框架基本结构框架分为框架集和框架两部分。

定义框架的语法格式如下。

<frameset><frame src="url1"><frame src="url2">......</frameset>标记HTML中用于描述功能的符号称为“标记”。

标记在使用时必须用尖括号“<>”括起来，比如<html>、<head>、<body>等，都是标记。

多数标记必须成对出现，以开头无斜杠的标记开始，以有斜杠的标记结束。

标记还可以嵌套使用。

属性与标记相关的特性称为属性，每个属性总是对应一个属性值，将其称为“属性/值”对。

“属性/值”对出现在开始标记的“>”之前。

在HTML文档中，标题（Heading）很重要。

标题是通过<h1>-<h6>等6对标记进行定义的。

<h1>定义最大的标题，<h6> 定义最小的标题。

font标记用于编辑网页文字的样式，主要是设置文字的字体、字号、颜色等属性。

font标记语法格式如下：<font face="" size="" color=""></font>HTML的一些元素可以定义字体信息，如粗体、斜体等。

不论是在普通文档，还是在网页文字中，合理的使用段落会使文字的显示更加美观，要表达的内容也更加清晰。

在HTML文件中，有专门的段落标记<p>。

粗体标记<b>，被<b>标记对括起的字词或句子在网页中表现为粗体。