题目一
【问题描述】
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| #include <iostream.h>
#include <math.h>
class Base{
public:
double result,a,b,step;
int n;
virtual double fun(double x) (1) ;
virtual void Integerate(){
cout<<"这里是积分函数"<<endl;
}
Base(double ra=0,double rb=0,int nn=2000){
a=ra; b=rb; n=nn; result=0;
}
void Print(){
cout.precision(15);
cout<<"积分值="<<result<<endl;
}
};
class Simpson:public Base{
public:
void Integerate(){
int i;
step=(b-a)/n;
result= (2) ;
for(i=1;i<n;i+=2) result= (3) ;
for(i=2;i<n;i+=2) result= (4) ;
result= (5) ;
}
Simpson(double ra,double rb,int nn): (6) {}
};
class sinS:public Simpson{
public:
sinS(double ra,double rb,int nn): (7) {}
double fun(double x){return sin(x);}
};
class expS: public sinS {
public:
expS(double ra,double rb,int nn): (8) {}
double fun(double x){return exp(x);}
};
void main(){
Base *bp;
sinS ss(0.0,3.1415926535/2.0,100);
bp= (9) ;
bp->Integerate();
bp->Print();
expS es(0.0,1.0,100);
bp= (10) ;
bp->Integerate();
bp->Print();
}
|
【答案】
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| (1) 【 正确答案: 1=0】
(2)【 正确答案: fun(a)+fun(b)】
(3)【 正确答案: result+4fun(istep+a)】
(4)【 正确答案: result+2fun(istep+a)】
(5)【 正确答案: result*step/3】
(6)【 正确答案: Base(ra,rb,nn)】
(7)【 正确答案: Simpson(ra,rb,nn)】
(8)【 正确答案: sinS(ra,rb,nn)】
(9)【 正确答案: &ss】
(10)【 正确答案: &es】
|
题目二
【问题描述】
设公司有普通员工和经理两类人员。主函数已给出，编程设计普通员工类CommonWorker和经理类Manager，经理类除具有普通员工类的属性外。普通员工类的收入由基本工资和奖金构成，经理类的收入除基本工资和奖金外，还包括职务津贴。实现求两类员工工资的Pay函数。
【代码】
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| #include <iostream>
#include<iostream>
using namespace std;
class CommonWorker
{
public:
virtual void Pay()
{
earning = wage + reward;
cout << earning<<endl;
}
friend istream& operator>>(istream& inStream, CommonWorker& data)
{
inStream >> data.wage;
inStream >> data.reward;
return inStream;
}
protected:
double wage;
double reward;
double earning;
};
class Manager :public CommonWorker
{
public:
virtual void Pay()
{
earning = wage + reward + place;
cout << earning<<endl;
}
friend istream& operator>>(istream& instream, Manager& data)
{
instream >> data.wage;
instream >> data.reward;
instream >> data.place;
return instream;
}
protected:
double place;
};
int main()
{
CommonWorker c1, * pc;
Manager m1;
cin >> c1 >> m1;
pc = &c1;
pc->Pay();
pc = &m1;
pc->Pay();
return 0;
}
|
题目三
【问题描述】
根据给出的程序片段编程，使程序正确运行。由基类Shape派生出圆类Circle（数据成员：半径）、正方形类Square（数据成员：边长）、三角形类Triangle（数据成员：三条边边长），3个派生类都有输入和显示信息函数Input、Output，计算面积的函数Area，计算周长的函数Perim。
【代码】
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| #include <iostream>
#include<iomanip>
#include <cmath>
using namespace std;
const double PI = 3.14159;
class Shape
{
public:
virtual double Area() const
{ return 0; }
virtual double Perim() const
{ return 0; }
virtual void Input() {}
virtual void Output() { }
static void SetFormat()
{ cout << setiosflags(ios::fixed) << setprecision(2); }
protected:
Shape(){}
};
class Circle:public Shape
{
private:
double mr;
public:
virtual void Input()
{
double r;
cin >> r;
this->mr = r;
}
Circle(double r = 0)
{
mr = r;
}
virtual void Output()
{
cout << PI * 2 * mr << "," << PI * mr * mr<<endl;
}
};
class Square :public Shape
{
private:
double ma;
public:
Square(double a = 0, double b = 0)
{
ma = a;
}
virtual void Input()
{
double a;
cin >> a;
this->ma = a;
}
virtual void Output()
{
cout<<ma*4<<","<<ma*ma<<endl;
}
};
class Triangle :public Shape
{
private:
double ma, mb, mc;
public:
Triangle(double a = 0, double b = 0, double c = 0)
{
ma = a; mb = b; mc = c;
}
virtual void Input()
{
double a, b, c;
cin >> a >> b >> c;
ma = a; mb = b; mc = c;
}
virtual void Output()
{
cout << ma + mb + mc << ",";
double p, s;
p = (ma + mb + mc) / 2;
s = sqrt(p * (p - ma) * (p - mb) * (p - mc));
cout << s<<endl;
}
};
int main()
{
Circle circle;
Square square;
Triangle triangle;
Shape* pt[3] = { &circle,&square,&triangle };
Shape::SetFormat();
for (int i = 0; i < 3; i++)
pt[i]->Input();
for (int i = 0; i < 3; i++)
pt[i]->Output();
return 0;
}
|
