Experiment 5 constructor and destructor

A. Point( Class and construction )

Title Description

The following is the class definition of a point on a plane , Please implement all its methods outside the class , And generate points to test it .

Input

Number of groups of test data t

The first set of test data points p1 Of x coordinate The first set of test data points p1 Of y coordinate The first set of test data points p2 Of x coordinate The first set of test data points p2 Of y coordinate

…

Output

Output p1 To p2 Distance of

stay C++ in , The reference code for outputting the specified precision is as follows ：

#include

#include // Must contain this header file

using namespace std;

void main( )

{ double a =3.141596;

cout<<fixed<<setprecision(3)<<a<<endl; // Output after decimal point 3 position

}

sample input

2
1 2 3 4
-1 0.5 -2 5

sample output

Distance of Point(1.00,2.00) to Point(3.00,4.00) is 2.83
Distance of Point(-1.00,0.50) to Point(-2.00,5.00) is 4.61

Reference code

#include<iostream>
#include<iomanip>
#include<cmath> 
using namespace std;
class Point
{
    
	double x,y;
	public:
		Point();// Default constructor  
		Point(double x_value,double y_value);// There are parameter constructors  
		double getX();// return x Value  
		double getY();// return y Value  
		void setX(double x_value);
		void setY(double y_value);
		double distanceToAnotherPoint(Point p); 
};
Point::Point()
{
    
	x=0;
	y=0;
} 
Point::Point(double x_value,double y_value)
{
    
	x=x_value;
	y=y_value;
} 
void Point::setX(double x_value)
{
    
	x=x_value;	
} 
void Point::setY(double y_value)
{
    
	y=y_value;	
} 
double Point::getX()
{
    
	return x;
} 
double Point::getY()
{
    
	return y;
} 
double Point::distanceToAnotherPoint(Point p)
{
    
	double len_x;
	double len_y;
	double len;
	len_x=x-p.x;
	len_y=y-p.y;
	len=sqrt(len_x*len_x+len_y*len_y);
	return len;
	
} 
int main()
{
    
	int t;// Number of test groups 
	Point p1,p2;
	double x1,y1,x2,y2,dis;
	cin>>t;
	while(t--)
	{
    
		cin>>x1>>y1>>x2>>y2;
		// Set the coordinates of the two points  
		p1.setX(x1); 
		p1.setY(y1);
		p2.setX(x2);
		p2.setY(y2);
		dis=p1.distanceToAnotherPoint(p2);
		cout<<"Distance of Point("<<fixed<<setprecision(2)<<p1.getX()<<','<<fixed<<setprecision(2)<<p1.getY()<<')';
		cout<<" to Point("<<fixed<<setprecision(2)<<p2.getX()<<','<<fixed<<setprecision(2)<<p2.getY()<<')';
		cout<<" is "<<fixed<<setprecision(2)<<dis<<endl;
		
	}	
}

B. Date( Class and construction )

Title Description

The following is the definition of a date class , Please implement all its methods outside the class , And generate the object test in the main function .

Be careful , When judging tomorrow's date , To join the cross month 、 straddle old and new years 、 Leap year judgment

for example 9. month 30 Tomorrow is 10 month 1 Japan ,12 month 31 Tomorrow is the second year 1 month 1 Japan

2 month 28 Tomorrow of the day should distinguish whether it is a leap year , Leap years are 2 month 29 Japan , Non leap years are 3 month 1 Japan

Input

Number of groups of test data t

Year of the first set of test data month Japan

…

The first date is required to be initialized with a constructor , The initialization of the second date adopts setDate Method , The third date uses the constructor , The fourth date adopts setDate Method , And so on .

Output

Output today's date

Output tomorrow's date

sample input

4
2012 1 3
2012 2 28
2012 3 31
2012 4 30

sample output

Today is 2012/01/03
Tomorrow is 2012/01/04
Today is 2012/02/28
Tomorrow is 2012/02/29
Today is 2012/03/31
Tomorrow is 2012/04/01
Today is 2012/04/30
Tomorrow is 2012/05/01

Reference code

#include<iostream>
#include<iomanip>
using namespace std;
class Date
{
    
	int year,month,day;
	public:
		Date();// Default constructor 
		Date(int y,int m,int d);
		int getYear();
		int getMonth();
		int getDay();
		void setDate(int y,int m,int d);
		void print();
		void addOneDay(); 
};
Date::Date()
{
    
	year=1900,month=1,day=1;
} 
Date::Date(int y,int m,int d)
{
    
	year=y;
	month=m;
	day=d;
} 
int Date::getDay()
{
    
	return day;
}
int Date::getMonth()
{
    
	return month;
}
int Date::getYear()
{
    
	return year;
}
void Date::setDate(int y,int m,int d)
{
    
	year=y;
	month=m;
	day=d;
} 
void Date::print()
{
    
	cout<<setfill('0')<<setw(4)<<year<<'/'<<setw(2)<<month<<'/'<<setw(2)<<day<<endl;
}
void Date::addOneDay()
{
     
// Put the days of each month into the array  
	int mon[13]={
    0,31,28,31,30,31,30,31,31,30,31,30,31};
	int Leap;
	if(year%4==0&&year%100!=0||year%400==0) Leap=1;
	else Leap=0;
	if(Leap) mon[2]=29;
	day++;
	if(day>mon[month]) 
	{
    
		month++;
		day=1; 
	}
	if(month>12)
	{
    
		year++;
		month=1;
	}
	
	
}
int main()
{
    
	int t,ye,mo,da;
	cin>>t;
	Date p1;
	while(t--)
	{
    
		cin>>ye>>mo>>da;
		p1.setDate(ye,mo,da);
		cout<<"Today is ";
		p1.print() ;
		p1.addOneDay();
		cout<<"Tomorrow is ";
		p1.print() ; 
	}
}

C. Fraction class （ Class and construction ）

Title Description

Complete the implementation of the following score classes :

class CFraction
{
private:
int fz, fm;
public:
CFraction(int fz_val, int fm_val) ;
CFraction add(const CFraction &r);
CFraction sub(const CFraction &r);
CFraction mul(const CFraction &r);
CFraction div(const CFraction &r);
int getGCD(); // Find the maximum common divisor of the numerator and denominator of the object
void print();
};

Find two numbers a、b The greatest common divisor of can be divided by rolling , Also known as Euclid algorithm , The steps are :

1. In exchange for a, b send a > b;
2. use a except b Get the remainder r, if r=0, be b For the greatest common divisor , sign out .
3. if r Not for 0, Then use b Instead of a, r Instead of b, here a,b It's smaller than the last one , The problem has shrunk ;
4. Continue to the first 2 Step .

Input

Number of groups of test data t

The first score of the first group

The second score of the first group

The first score in the second group

The second score of the second group

…

Output

The sum of the two scores in the first group

The difference between the two scores in the first group

The product of the first set of two scores

The quotient of the first set of two scores

The sum of the two scores in the second group

The difference between the two scores in the second group

The second set is the product of two scores

The quotient of two scores in the second group

…

sample input

3
1/2
2/3
3/4
5/8
21/23
8/13

sample output

7/6
-1/6
1/3
3/4

11/8
1/8
15/32
6/5

457/299
89/299
168/299
273/184

Reference code

#include<iostream>
#include<iomanip>
#include<cmath> 
using namespace std;
class CFraction
{
    

	int fz, fm;// Molecular denominator  
	public:
	CFraction(int fz_val, int fm_val) ;
	CFraction add(const CFraction &r);// Add  
	CFraction sub(const CFraction &r);// reduce  
	CFraction mul(const CFraction &r);// ride  
	CFraction div(const CFraction &r);// except  
	int getGCD();   //  Find the maximum common divisor of the numerator and denominator of the object 
	void print();
	CFraction(){
    };
};
CFraction::CFraction(int fz_val, int fm_val)
{
    
	fz=fz_val;
	fm=fm_val;
}  
CFraction CFraction::add(const CFraction &r)
{
    
	CFraction temp(fz*r.fm+r.fz *fm,fm*r.fm);
	int gcd=temp.getGCD();
	temp.fz=temp.fz/gcd;
	temp.fm=temp.fm/gcd;
	return temp;
}
CFraction CFraction::sub(const CFraction &r)
{
    
	CFraction temp((fz*r.fm-r.fz *fm),fm*r.fm);
	int gcd=temp.getGCD();
	temp.fz=temp.fz/gcd;
	temp.fm=temp.fm/gcd;
	return temp;
}
CFraction CFraction::mul(const CFraction &r)
{
    
	CFraction temp(fz*r.fz,fm*r.fm);
	int gcd=temp.getGCD();
	temp.fz=temp.fz/gcd;
	temp.fm=temp.fm/gcd;
	return temp;
}
CFraction CFraction::div(const CFraction &r)
{
    
	CFraction temp(fz*r.fm,r.fz *fm);
	int gcd=temp.getGCD();
	temp.fz=temp.fz/gcd;
	temp.fm=temp.fm/gcd;
	return temp;
}
int CFraction::getGCD()
{
    
	int r;
	int a,b;
	if(fabs(fz)>fabs(fm))
	{
    
		a=fabs(fz);
		b=fabs(fm);
	}
	else 
	{
    
		a=fabs(fm);
		b=fabs(fz);
	}

	while(a%b!=0)// Find the greatest common divisor by the division of rolling  
	{
    
		r=a%b;
		a=b;
		b=r;
	}
	return b;
}
void CFraction::print()
{
    
	// If there are negative numbers , Notice that the minus sign comes first 
	if(fz<0&&fm>0||fz>0&&fm<0) cout<<'-';
	cout<<fabs(fz)<<'/'<<fabs(fm)<<endl;
}
int main()
{
    
	int t;
	cin>>t;
	int fz1,fm1,fz2,fm2;
	CFraction p3;
	while(t--)
	{
    
		char ch;
		cin>>fz1>>ch>>fm1>>fz2>>ch>>fm2;
		CFraction p1(fz1,fm1);
		CFraction p2(fz2,fm2);
		p3=p1.add(p2);// To add  
		p3.print();
		p3=p1.sub(p2); // Subtraction  
		p3.print();
		p3=p1.mul(p2);// Multiplication  
		p3.print();
		p3=p1.div(p2);// division  
		p3.print();
		cout<<endl;		
	} 
}

D. Point_Array( class + structure + An array of objects ）

Title Description

The above is an exercise we have practiced , Please make the following modifications on the basis of the original code ：1、 Add self written destructor ;2、 take getDisTo The parameter of the method is modified to getDisTo(const Point &p);3、 Modify the corresponding constructor according to the output below .

Then, in the main function, it is established according to the number of user inputs Point Array , Find the distance between the two points with the largest distance in the array .

Input

Number of groups of test data t

Number of points in the first group

The first point is x coordinate y coordinate

On the second point x coordinate y coordinate

…

Output

Output the two points with the largest distance in the first group and their distance

…

stay C++ in , The reference code for outputting the specified precision is as follows ：

#include

#include // Must contain this header file

using namespace std;

void main( )

{ double a =3.141596;

cout<<fixed<<setprecision(3)<<a<<endl; // Output after decimal point 3 position

sample input

2
4
0 0
5 0
5 5
2 10
3
-1 -8
0 9
5 0

sample output

Constructor.
Constructor.
Constructor.
Constructor.
The longeset distance is 10.44,between p[1] and p[3].
Distructor.
Distructor.
Distructor.
Distructor.
Constructor.
Constructor.
Constructor.
The longeset distance is 17.03,between p[0] and p[1].
Distructor.
Distructor.
Distructor.

Reference code

#include<iostream>
#include<iomanip>
#include<cmath> 
using namespace std;
class Point
{
    
	double x,y;
	public:
		Point();// Default constructor  
		~Point()
		{
    
			cout<<"Distructor."<<endl;
		}
		Point(double x_value,double y_value);// There are parameter constructors  
		double getX();// return x Value  
		double getY();// return y Value  
		void setXY(double x1,double y1)// Set the... In the array x,y 
		{
    
			x=x1;
			y=y1;
		}
		void setX(double x_value);
		void setY(double y_value);
		double getDisTo(const Point &p); 
};
Point::Point()
{
    
	cout<<"Constructor."<<endl;
	x=0;
	y=0;
} 
Point::Point(double x_value,double y_value)// There are parameter constructors  
{
    
	x=x_value;
	y=y_value;
} 
double Point::getX()// return x Value  
{
    
	return x;
} 
double Point::getY()// return y Value  
{
    
	return y;
} 
void Point::setX(double x_value)// Set up x 
{
    
	x=x_value;	
} 
void Point::setY(double y_value)// Set up y 
{
    
	y=y_value;	
} 
double Point::getDisTo(const Point &p)
{
    
	double len_x;
	double len_y;
	double len;
	len_x=x-p.x;
	len_y=y-p.y;
	len=sqrt(len_x*len_x+len_y*len_y);
	return len;
	
} 
int main()
{
    
	int t;// Number of test groups 
	int n;// Number of points  
	double x,y,dis;
	cin>>t;
	while(t--)
	{
    
		cin>>n;
		Point *p=new Point[n]; 
		for(int i=0;i<n;i++)
		{
    
			cin>>x>>y;
			p[i].setXY(x,y);
		}
		double max=-99999;
		int tag1,tag2;
		for(int i=0;i<n-1;i++)
		{
    
			for(int j=i+1;j<n;j++)
			{
    
				if(p[i].getDisTo(p[j])>max)
				{
    
					max=p[i].getDisTo(p[j]);
					tag1=i;
					tag2=j;// Record which two points generate the maximum value  
				}
			}
		}
		if(tag1>tag2)
		{
    
			int tem;
			tem=tag1;
			tag1=tag2;
			tag2=tem;
		}
		cout<<"The longeset distance is "<<fixed<<setprecision(2)<<max<<",between p["<<tag1<<"] and p["<<tag2<<"]."<<endl;
		delete []p;
	}	
}

E. Point circle operation （ Structure and destruction ）

Title Description

Design a point class Point, Contains private properties x Coordinates and y coordinate , Operations include

1、 Constructors , Two conditions are required ：1. Can use class Point To create an array of objects （ Default constructor ！）;2. Capable of receiving external input x And coordinate initialization , Tips ： Constructor overload

2、 Destructor , hold x Coordinates and y Coordinates are clear 0, And output information “point clear”

3、 Set up (setXY), Accept external input , And set up x Coordinates and y coordinate

4、 obtain x coordinate , Go straight back to x value

5、 obtain y coordinate , Go straight back to y value

Design a circle class Circle, Contains private properties ： The coordinates of the center of the circle x and y、 radius r; Operations include ：

1、 Constructors , Accept external input , Set the center x coordinate 、y Coordinates and radius

2、 Destructor , Coordinate the center of the circle x and y And the radius are cleared , And the output "circle clear"

3、 contain （Contain）, Judge whether a circle contains a point , Calculate the distance from the center of the circle to this point , And then compare it with the radius , Greater than does not include , Less than or equal to includes . Tips ： Using a point object as a parameter does not necessarily conform to the output

Input

On the first line, enter the of a point x Coordinates and y coordinate , use Point Class to create a point object , And is automatically initialized by the constructor

Second line input n, use Point Class to create an array of point objects , contain n A little bit

Enter... From the third line n That's ok , Enter a dot per line x and y coordinate , Use settings (setXY) For each point x and y coordinate

Then enter three parameters on one line , Represents the coordinates of the center of a circle x and y, And the radius , Use Circle Class to create a circular object , And automatically initialize through the constructor

Output

By calling the inclusion of the circle (Contain) Method , Determine whether each point is in the circle .

According to the input order of points , Output the judgment result of one point per line , If included, output in, If not, output out

explain ： When an array of objects is created dynamically , So at the end of the program , This array is not recycled . Only add code delete []p, To reclaim the array .

This question does not require the retrieval of arrays .

sample input

5 2
3
4 7
9 9
2 4
3 3 3

sample output

in
out
out
in
circle clear
point clear

Reference code

#include<iostream>
#include<iomanip>
#include<cmath> 
using namespace std;
class Point
{
    
	int x,y;
	public:
		Point()// Default constructor  
		{
    
			x=0;
	     	y=0;
		}
		~Point()
		{
    
			x=0;
			y=0;
			cout<<"point clear"<<endl;
		}
		Point(int x_value,int y_value)// There are parameter constructors  
		{
    
			x=x_value;
			y=y_value;
		}
		int getX()// return x Value 
		{
    
			return x;
		} 
		int getY()// return y Value  
		{
    
			return y;
		}
		void setXY(int x1,int y1)// Set the... In the array x,y 
		{
    
			x=x1;
			y=y1;
		}
	
};
class Circle
{
    
	int x;int y;int r;
	public:
	Circle(int x1,int y1,int r1)
	{
    
		x=x1;
		y=y1;
		r=r1;
	}
	~Circle()
	{
    
		x=0;
		y=0;
		r=0;
		cout<<"circle clear"<<endl;
	}
	void setXY(int x1,int y1,int r1)
	{
    
		x=x1;
		y=y1;
		r=r1;
	}
	void contain(int x1,int y1)
	{
    
		if(sqrt((x1-x)*(x1-x)+(y1-y)*(y1-y))<=r) cout<<"in"<<endl;
		else cout<<"out"<<endl;
	} 
			
};
int main()
{
    

	int n;// Number of points  
	int p1_x,p1_y,x,y;// The coordinates of point 
	int cir_x,cir_y,r;// The coordinates of the center of the circle  , radius  
	cin>>p1_x>>p1_y;
	Point p1(p1_x,p1_y);
	cin>>n;
	Point *p=new Point[n]; 
	for(int i=0;i<n;i++)
	{
    
		cin>>x>>y;
		p[i].setXY(x,y);
	}
	cin>>cir_x>>cir_y>>r;
	Circle c(cir_x,cir_y,r);// initialization 
	c.contain(p1_x,p1_y);
	for(int i=0;i<n;i++)
	{
    
		x=p[i].getX();
		y=p[i].getY();
		c.contain(x,y); 
	}
	//delete []p;
}