第1题:
有以下程序 #include <iostream> using namespace std; class Complex { public: Complex (double r=0, double i =0 :re(r) ,im (i) {} double real() const {return re;} double imag() const { return im;} Complex operator + (Complex c} const {return Complex(re+c.re, im+c.im);} privane: double re,im; }; int main { Complex a =Complex (1,1)+Complex(5); cout<<a.real()<<'+'<<a.imag() << 'i' <<endl return 0; } 程序执行后的输出结果是
A.6+6i
B.6+1i
C.1+6i
D.1+1i
第2题:
下面是复数类complex的定义,其中作为友元函数重载的运算符“--”的功能是将参数对象的实部减1,然后返回对该对象的引用;请补充完整。
class complex
{
private:
int real;
int imag;
public:
complex(int r=0,int i=0):real(r),imag(i){}
void show ()
{
cout<<real<<(imag<0?"-":"+")<<imag<<'i';
}
______;
};
complex& operator -- (complex &c)
{
c.real --;
return c;
}
第3题:
有以下程序: #include<iostream> using namespace std; class Complex { public: Complex(dOuble r=0,dOuble i=0):re(r),im(i){} doublereal()const{return re;} doubleimag()const{return im;} Complex operator+(Complex c)const {return Complex
A.6+6i
B.6+1i
C.1+6i
D.1+1i
第4题:
第5题:
CRM就是“One-To-One”一对一营销。
第6题:
junctional complex
第7题:
Inheritance represents an is-a relationship.
Inheritance represents a has-a relationship.
Interfaces must be used when creating a has-a relationship.
Instance variables can be used when creating a has-a relationship.
第8题:
The relationship of individual playing cards to a 52-card deck
The relationship of a group of related species of animals to the genus to which the species belong
The relationship of a compact disk to the songs contained on that disk
The relationship of a transmission to the car in which the transmission has been installed
The relationship of an updated version of software to the original version of that software
第9题:
as the representative of
in the interest of
in the name of
in the process of
第10题:
unconditionally
unbearably
unapproachably
unavoidably
第11题:
第12题:
对
错
第13题:
有如下程序: #include <iostream> using namespace std; class Complex { double re, im, public: Complex(double r, double i): re(r), im(i) {} double real() const {return re;} double image() const {return im,} Complex& operator +=(Complex a) { re +=a.re; im +=a.im; return *this; } }; ostream& operator << (ostream& s, const Complex& z) { return s<<'('<<z.real()<<','<<z.image()<<')'; } int main() { Complex x(1,-2), y(2,3); cout << (x+=y) << endl; return 0; } 执行这个程序的输出结果是( )。
A.(1,-2)
B.(2,3)
C.(3,5)
D.(3,1)
第14题:
有下列程序: #include<iostream> using namespace std; class Complex { double re,im; public: Complex(double r,double i):re(r),im(i){} double real()const{retum re;} double image()const{return im;} Complex& operator+=(Complex A) { r
A.(1,-2)
B.(2,3)
C.(3,5)
D.(3,1)
第15题:
A.Customer Relationship Management Services
B.Customer Request Management
C.Customer Relationship Manager
D.Co-operation Relationship Management
第16题:
Which of the following is a benefit of having a Solution Assurance?()
第17题:
C.RM就是“One-To-One”一对一营销。
第18题:
students don’t have to travel away from home to take a test
the content of distance learning program is the same as that of the full time program in colleges and universities
the good distance learning programs include courses that are not taught in average universities
the relationship between students and professors are one-to-one
第19题:
ISM Code
SOLAS 74 Convention
Safety Management System
MARPOL Convention
第20题:
对
错
第21题:
EIGRPv6 will form a neighbor relationship.
EIGRPv6 will not form a neighbor relationship.
EIGRPv6 will form a neighbor relationship, but with the log MSG:// EIGRPv6 neighbor not on a commonsub net.
EIGRPv6 will form a neighbor relationship, but routes learned from that neighbor will not be installed inthe routing table.
第22题:
第23题: