题目一

【问题描述】

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#include <iostream.h>

#include <math.h>

class Base{

public://为了简化,数字域取公有

  double result,a,b,step;//Intevalue积分值，a积分下限，b积分上限

  int n;

  virtual double fun(double x)      (1)       ;//被积函数声明为纯虚函数

  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{//辛普生法

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;
	//分别输入员工的基本工资、奖金和津贴
	cin >> c1 >> m1;
	pc = &c1;           		
	pc->Pay();      //调用Pay()函数输出该普通员工工资，如：5000
	pc = &m1;          		
	pc->Pay();      //调用Pay()函数输出该经理工资，如：8000
	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;
//定义基类Shape
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();	
	//输入各图形数据，空白字符隔开
	for (int i = 0; i < 3; i++)
		pt[i]->Input();
	//输出各图形周长和面积，周长和面积值之间逗号隔开
	for (int i = 0; i < 3; i++)	
		pt[i]->Output();	
	return 0;
}