单选题73 A degree B request C person D masterAABBCCDD
单选题
73 A degree B request C person D master
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解析：
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By evening, he had decided to make his request ______. (A) at hand(B) in hand(C) by person(D) in person
能将程序补充完整的选项是( )。class Person{ private int a; public int change(int m){ return m; }}public class Teacher extends Person{ public int b; public static void main(String arg[]) { Person p = new Person(); Teacher t = new Teacher(); int i; ______ }} B.A．i=mB．i=bC． i=p.aD．i=p. change(50)
某井下入φ73mm油管2829m,地层出砂后将油管卡住,用清水压井处理事故,倒出油管1350m。再下反扣φ73mm钻杆以上提力F=210kN倒扣,预计倒扣位置在多少米?清水中φ73mm钻杆每米质量为12.4kg/m,φ73mm油管每米质量为8.3kg/m。(取小数点后一位,g=10m/S??2??) 。??
某井井深2100m,用φ73mm油管冲砂,当冲砂至2000m时,发生油管折断事故,落鱼长1000m,鱼顶深1000m,φ73mm钻杆打捞发生卡钻事故,卡后上提管柱F1=200kN时,钻具伸长了1.0m，求卡点深度(φ73mm油管K2=2450,φ73mm钻杆K1=3800)。
能将程序补充完整的选项是______。 class Person{ private int a； phblic int change(int m){return m；} } public class Teacher extends Person{ public int b； public static void main(String arg[ ]){ Person p=new Person( )； Teacher t=new Teacher( )； int i; ______； } }A．i=mB．i=bC．i=p.aD．i=p.change(50)
Do you think you are the suitable person for this position?().A. I'm not sure if I will be here for the task.B. I don't think Xiao Ming is the right person to count on.C. Yes. I'm hard working and I think I am suitable for this position.
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使用VC6打开考生文件夹下的工程test28_3。此工程包含一个test28_3.cpp，其中定义了类Person，但该类的定义并不完整。请按要求完成下列操作，将程序补充完整。(1)定义类Person的私有数据成员forename、surname 和 money，forename 和 surname都是char型的指针数据，money是double型的数据。请在注释“//**1**”之后添加适当的语句。(2)完成类Person的带三个参数的构造函数Person(char *f, char *s，double m)，分别为forename和surname申请新的空间来存储参数f和s指针指向的内容，注意空间的大小，最后把参数m的值赋给money，请在注释“//**2**”之后添加适当的语句。(3)完成类Person的析构函数的定义，把forename和surname指向的空间释放，请在注释“//**3**”之后添加适当的语句。(4)完成类Person的成员函数display的定义，使其以格式“forname surname has money”的形式输出内容，请在注释“//**4*。”之后添加适当的语句。注意：除在指定位置添加语句之外，请不要改动程序中的其他内容。输出结果如下：Richard Berk has 1000.56源程序文件test28_3.cpp清单如下：include ＜iostream.h＞include ＜string.h＞class Person{private://** 1 **double money;public:Person(char *f, char *s, double m);～Person();void display();};Person::Person(char *f, char *s, double m){//** 2 **strcpy(forename, f);surname = new char[strlen(s)+1];strcpy(surname, s);money=m;}Person: :-Person (){//** 3 **}void Person:: display ( ){//** 4 **}void main ( ){Person p("Richard", "Berk", 1000.56);p.display ( );}
下列程序运行后的输出结果是( )。#include#includeusing namespace std;class Person{public:Person(string n):name(n) { coutprivate:int year,month,day;};class Student:public Person{public:Student(string n,int y,int m,int d,char c):birthday(y,m,d),sex(c),Person(n) { coutA. SB.PSC.DPSD.PDS
●以下的SQL 99语句，Student与Person之间的关系是 (62) 。CREATE TYPE Person(name char(20)，address varchar(50))；CREATE TYPE Student(under Person(degree char(20)department char(20)）；(62)A．类型继承B．类型引用C．表继承D．无任何关系
下列程序段的执行结果为 a=95 If a ＞ 60 Then degree=1 If a ＞ 70 Then degree=2 If a ＞ 80 Then degree=3 If a ＞ 90 Then degree=4 Print"degree="；degreeA．degree=1B．degree=2C．degree=3D．degree=4
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