实验3

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Problem A: 平面上的点——Point类 (IV)

Problem A: 平面上的点——Point类 (IV)

Time Limit: 1 Sec  Memory Limit: 4 MB
Submit: 4107  Solved: 1975
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Description

在数学上，平面直角坐标系上的点用X轴和Y轴上的两个坐标值唯一确定。现在我们封装一个“Point类”来实现平面上的点的操作。

根据“append.cc”，完成Point类的构造方法和show()、showCounter()、showSumOfPoint()方法；实现showPoint()函数。

接口描述：
showPoint()函数：按输出格式输出Point对象，调用Point::show()方法实现。
Point::show()方法：按输出格式输出Point对象。
Point::showCounter()方法：按格式输出当前程序中Point对象的计数。
Point::showSumOfPoint()方法：按格式输出程序运行至当前存在过的Point对象总数。

Input

输入多行，每行为一组坐标“x,y”，表示点的x坐标和y坐标，x和y的值都在double数据范围内。

Output

对每个Point对象，调用show()方法输出其值，或者用showPoint()函数来输出（通过参数传入的）Point对象的值：X坐标在前，Y坐标在后，Y坐标前面多输出一个空格。每个坐标的输出精度为最长16位。调用用showCounter()方法和showSumOfPoint()输出Point对象的计数统计，输出格式见sample。

C语言的输入输出被禁用。

Sample Input

1,2 3,3 2,1

Sample Output

Point : (1, 2) Current : 2 points. Point : (3, 3) Current : 2 points. Point : (2, 1) Current : 2 points. In total : 4 points. Current : 3 points. Point : (0, 0) Point : (1, 1) Point : (0, 0) In total : 6 points.

HINT

对象计数通过静态成员来实现

Append Code

append.cc,

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한국어<   中文  فارسی  English  ไทย
All Copyright Reserved 2010-2011 SDUSTOJ TEAM
GPL2.0 2003-2011 HUSTOJ Project TEAM
Anything about the Problems, Please Contact Admin:admin

#include<iostream>
#include<iomanip>
using namespace std;
class Point{
private:
    double x,y;
    static int sum,num;
public:
    Point():x(0),y(0){num++;sum++;}
    Point(double a):x(a),y(1){num++;sum++;}
    Point(double a,double b):x(a),y(b){num++;sum++;}
    Point(const Point&p){x=p.x;y=p.y;num++;sum++;}
    ~Point(){num--;}
    void show(){cout<<setprecision(16)<<"Point : ("<<x<<", "<<y<<")"<<endl;}
    static void showCounter(){cout<<setprecision(16)<<"Current : "<<num<<" points."<<endl;}
    static void showSumOfPoint(){cout<<setprecision(16)<<"In total : "<<sum<<" points."<<endl;}
};
void showPoint(Point &a,Point &b,Point &c){a.show();b.show();c.show();}
int Point::sum=0;
int Point::num=0;
int main()
{
    char c;
    double a, b;
    Point q;
    while(std::cin>>a>>c>>b)
    {
        Point p(a, b);
        p.show();
        p.showCounter();
    }
    q.showSumOfPoint();
    Point q1(q), q2(1);
    Point::showCounter();
    showPoint(q1, q2, q);
    Point::showSumOfPoint();
}

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Problem B: 平面上的点——Point类 (V)

Problem B: 平面上的点——Point类 (V)

Time Limit: 1 Sec  Memory Limit: 4 MB
Submit: 5067  Solved: 1715
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Description

在数学上，平面直角坐标系上的点用X轴和Y轴上的两个坐标值唯一确定。现在我们封装一个“Point类”来实现平面上的点的操作。

根据“append.cc”，完成Point类的构造方法和接口描述中的方法。

接口描述：
showPoint()函数：按输出格式输出Point对象。
Point::show()方法：按输出格式输出Point对象。
Point::showSumOfPoint()方法：按格式输出程序运行至当前存在过的Point对象总数。
Point::x()方法：取x坐标。
Point::y()方法：取y坐标。
Point::x(double)方法：传参数设置x坐标并返回。
Point::y(double)方法：传参数设置y坐标并返回。
Point::getX()方法：取x坐标。
Point::getY()方法：取y坐标。
Point::setX()方法：传参数设置x坐标并返回。
Point::setY()方法：传参数设置y坐标并返回。
Point::setPoint(double,double)方法：设置Point对象的x坐标（第一个参数）和y坐标（第二个参数）并返回本对象

Input

输入多行，每行为一组坐标“x,y”，表示点的x坐标和y坐标，x和y的值都在double数据范围内。

Output

用ShowPoint()函数来输出（通过参数传入的）Point对象的值或坐标值：X坐标在前，Y坐标在后，Y坐标前面多输出一个空格。每个坐标的输出精度为最长16位。
对每个Point对象，调用show()方法输出其值，输出格式与ShowPoint()函数略有不同：“Point[i] :”，i表示这是程序运行过程中第i个被创建的Point对象。
调用showSumOfPoint()输出Point对象的计数统计，输出格式见sample。

C语言的输入输出被禁用。

Sample Input

1,2 3,3 2,1

Sample Output

Point : (1, 2) Point : (3, 3) Point : (2, 1) Point : (1, 1) Point : (4, -3) ==========gorgeous separator========== Point[1] : (1, 0) Point[2] : (3, 3) Point[3] : (0, 0) Point[4] : (4, -3) Point[64] : (1, 0) Point[64] : (1, 0) ==========gorgeous separator========== In total : 66 points.

HINT

传递和返回引用是不构造新对象的。给函数正确的返回值。

Append Code

append.cc,

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한국어<   中文  فارسی  English  ไทย
All Copyright Reserved 2010-2011 SDUSTOJ TEAM
GPL2.0 2003-2011 HUSTOJ Project TEAM
Anything about the Problems, Please Contact Admin:admin

#include<iostream>
#include<iomanip>
using namespace std;
class Point{
private:
    double m,n;
    int id;
    static int sum;
public:
    double x(){return m;}
    double y(){return n;}
    double x(double a){m=a;return m;}
    double y(double b){n=b;return n;}
    double getX(){return m;}
    double getY(){return n;}
    double setX(double a){m=a;return m;}
    double setY(double b){n=b;return n;}
    Point& setPoint(double a,double b){m=a;n=b;return *this;}
    Point():m(0),n(0){sum++;id=sum;}
    Point(double a):m(a),n(a){sum++;id=sum;}
    Point(double a,double b):m(a),n(b){sum++;id=sum;}
    Point(const Point&p){m=p.m;n=p.n;sum++;id=sum;}
    ~Point(){}
    void show(){cout<<setprecision(16)<<"Point["<<id<<"] : ("<<m<<", "<<n<<")"<<endl;}
    static void showSumOfPoint(){cout<<setprecision(16)<<"In total : "<<sum<<" points."<<endl;}
};
int Point::sum=0;
void ShowPoint(Point p)
{
    cout<<std::setprecision(16)<<"Point : ("<<p.x()<<", "<<p.y()<<")"<<endl;
}

void ShowPoint(double x, double y)
{
    Point p(x, y);
    cout<<std::setprecision(16)<<"Point : ("<<p.x()<<", "<<p.y()<<")"<<endl;
}

void ShowPoint(Point &p, double x, double y)
{
    cout<<std::setprecision(16)<<"Point : ("<<p.x(x)<<", "<<p.x(y)<<")"<<endl;
}

int main()
{
    int l(0);
    char c;
    double a, b;
    Point pt[60];
    while(std::cin>>a>>c>>b)
    {
        if(a == b)
            ShowPoint(pt[l].setPoint(a, b));
        if(a > b)
            ShowPoint(a, b);
        if(a < b)
            ShowPoint(pt[l], a, b);
        l++;
    }
    Point p(a), q(b);
    ShowPoint(q);
    double x(0), y(0);
    for(int i = 0; i < l; i++)
        x += pt[i].getX(), y -= pt[i].getY();
    ShowPoint(pt[l].setX(x), pt[l].setY(y));
    cout<<"==========gorgeous separator=========="<<endl;
    for(int i = 0; i <= l; i++)
        pt[i].show();
    q.setPoint(q.x() - p.x() + a, q.y() - p.y() + b).show();
    q.show();
    cout<<"==========gorgeous separator=========="<<endl;
    p.showSumOfPoint();
}

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Problem C: 平面上的点——Point类 (VI)

Problem C: 平面上的点——Point类 (VI)

Time Limit: 1 Sec  Memory Limit: 4 MB
Submit: 4838  Solved: 1847
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Description

在数学上，平面直角坐标系上的点用X轴和Y轴上的两个坐标值唯一确定。现在我们封装一个“Point类”来实现平面上的点的操作。

根据“append.cc”，完成Point类的构造方法和接口描述中的方法和函数。

接口描述：
showPoint()函数：按输出格式输出Point对象。
Point::show()方法：按输出格式输出Point对象。
Point::showSumOfPoint()方法：按格式输出程序运行至当前存在过的Point对象总数。
Point::x()方法：取x坐标。
Point::y()方法：取y坐标。
Point::x(double)方法：传参数设置x坐标并返回。
Point::y(double)方法：传参数设置y坐标并返回。
Point::setPoint(double,double)方法：设置Point对象的x坐标（第一个参数）和y坐标（第二个参数）并返回本对象。
Point::isEqual()方法：判断传入的参数与对象的坐标是否相同，相同返回true。
Point::copy()方法：传参数复制给对象。
Point::inverse()方法，有两个版本：不传参数则将对象自身的x坐标和y坐标互换；若将Point对象做参数传入，则将传入对象的坐标交换复制给对象自身，不修改参数的值。

Input

输入多行，每行为一组坐标“x,y”，表示点的x坐标和y坐标，x和y的值都在double数据范围内。

Output

用ShowPoint()函数来输出（通过参数传入的）Point对象的值或坐标值：X坐标在前，Y坐标在后，Y坐标前面多输出一个空格。每个坐标的输出精度为最长16位。
对每个Point对象，调用show()方法输出其值，输出格式与ShowPoint()函数略有不同：“Point[i] :”，i表示这是程序运行过程中第i个被创建的Point对象。
调用showSumOfPoint()输出Point对象的计数统计，输出格式见sample。

C语言的输入输出被禁用。

Sample Input

1,2 3,3 2,1

Sample Output

Point[3] : (1, 2) Point[1] : (2, 1) Point[4] : (3, 3) Point[1] : (3, 3) Point[5] : (1, 2) Point[1] : (1, 2) Point[2] : (0, 0) ==========gorgeous separator========== Point[2] : (-7, 5) Point[3] : (1, 2) Point[4] : (3, 3) Point[5] : (1, 2) Point[6] : (-7, 5) ==========gorgeous separator========== Point[63] : (3, 3) Point : (3, 3) Point : (3, 3) Point : (3, 3) In total : 64 points.

HINT

给函数正确的返回值。注意常量对象调用的函数。

Append Code

append.cc,

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한국어<   中文  فارسی  English  ไทย
All Copyright Reserved 2010-2011 SDUSTOJ TEAM
GPL2.0 2003-2011 HUSTOJ Project TEAM
Anything about the Problems, Please Contact Admin:admin

#include<iostream>
#include<iomanip>
using namespace std;
class Point
{
private:
    double x1;
    double y1;
    int order;
    static int coun ;
public :
    Point ():x1(0),y1(0){coun ++; order=coun;}
    Point(double a , double b):x1(a) , y1(b){coun ++;order=coun;}
    Point(double a ):x1(a),y1(a){coun ++;order=coun;}
    ///Point(const Point & p){ coun++;aCoun++;x1 = p.x1;y1 = p.y1;order = (aCoun+1);}
    ///~Point(){coun--;}
    void show()const
    {
        std::cout << "Point["<<order<<"] : ("<<x1<<", " <<y1<<")"<<endl;
    }
   /*static void showCounter()
    {
        cout << "Current : "<<coun <<" points."<<endl;
    }*/
    static void showSumOfPoint(){cout<<setprecision(16)<<"In total : "<<coun<<" points."<<endl;}
    double x()const{return x1;}
    double y()const{return y1;}
    double x(double x){x1 = x;return x1;}
    double y(double y){y1 = y;return y1;}
    double getX(){return x1;}
    double getY(){return y1;}
    double setX(double x){return x1=x;}
    double setY(double y){return y1=y;}
    Point & setPoint(double x ,double y)
    {
        x1 = x;
        y1 = y;
        return *this;
    }
    bool isEqual( Point & p)const
    {
        if(p.x1 == x1 && p.y1 == y1 )
            return true;
        else
            return false;
    }
    Point & copy(Point & p)
    {
        x1 = p.x1;
        y1 = p.y1;
        return *this;
    }
   Point & inverse()
   {
       double temp;
         temp = x1;
         x1 = y1;
         y1 = temp;
         return*this;
   }
    Point& inverse(Point &p){x1=p.y1;y1=p.x1;return *this;}

};
 int Point::coun =0;

void ShowPoint(const Point &p)
{
    cout << "Point : ("<<std::setprecision(16)<<p.x()<<", " <<p.y() <<")"<<endl;
}
void ShowPoint(double x, double y)
{
 cout << "Point : ("<<std::setprecision(16)<<x<<", " <<y <<")"<<endl;
}
void ShowPoint(Point &p, double x, double y)
{
    cout<<std::setprecision(16)<<"Point : ("<<p.x(x)<<", "<<p.x(y)<<")"<<endl;
}

int main()
{
    int l(0);
    char c;
    double a, b;
    Point p, q, pt[60];
    while(std::cin>>a>>c>>b)
    {
        if(a == b)
            p.copy(pt[l].setPoint(a, b));
        if(a > b)
            p.copy(pt[l].setPoint(a, b).inverse());
        if(a < b)
            p.inverse(pt[l].setPoint(a, b));
        if(a < 0)
            q.copy(p).inverse();
        if(b < 0)
            q.inverse(p).copy(pt[l]);
        pt[l++].show();
        p.show();
    }
    q.show();
    cout<<"==========gorgeous separator=========="<<endl;
    double x(0), y(0);
    for(int i = 0; i < l; i++)
        x += pt[i].x(), y -= pt[i].y();
    pt[l].x(y), pt[l].y(x);
    q.copy(pt[l]).show();
    for(int i = 0; i <= l; i++)
        pt[i].show();
    cout<<"==========gorgeous separator=========="<<endl;
    const Point const_point(3, 3);
    const_point.show();
    for(int i = 0; i <= l; i++)
    {
        if(const_point.isEqual(pt[i]))
        {
            ShowPoint(const_point);
            ShowPoint(const_point.x(), const_point.y());
            ShowPoint(Point(const_point.x(), const_point.y()));
        }
    }
    const_point.showSumOfPoint();
}

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Problem D: 平面上的点和线——Point类、Line类 (I)

Problem D: 平面上的点和线——Point类、Line类 (I)

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 2814  Solved: 1374
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Description

在数学上，平面直角坐标系上的点用X轴和Y轴上的两个坐标值唯一确定，两点确定一条线段。现在我们封装一个“Point类”和“Line类”来实现平面上的点的操作。

 

根据“append.cc”，完成Point类和Line类的构造方法和show()方法。

 

接口描述：

Point::show()方法：按格式输出Point对象。

Line::show()方法：按格式输出Line对象。

Input

输入的第一行为N，表示后面有N行测试样例。

每行为两组坐标“x,y”，分别表示线段起点和终点的x坐标和y坐标，两组坐标间用一个空格分开，x和y的值都在double数据范围内。

Output

输出为多行，每行为一条线段，起点坐标在前终点坐标在后，每个点的X坐标在前，Y坐标在后，Y坐标前面多输出一个空格，用括号包裹起来。输出格式见sample。

Sample Input

4 0,0 1,1 1,1 2,3 2,3 4,5 0,1 1,0

Sample Output

Point : (0, 0) Line : (0, 0) to (1, 1) Line : (1, 1) to (2, 3) Line : (2, 3) to (4, 5) Line : (0, 1) to (1, 0) Line : (1, -2) to (2, -1) Line : (1, -2) to (0, 0) Line : (2, -1) to (0, 0) Line : (0, 0) to (2, -1)

HINT

 

Append Code

append.cc,

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한국어<   中文  فارسی  English  ไทย
All Copyright Reserved 2010-2011 SDUSTOJ TEAM
GPL2.0 2003-2011 HUSTOJ Project TEAM
Anything about the Problems, Please Contact Admin:admin

#include<iostream>
#include<iomanip>
using namespace std;
class Point{
        friend class Line;
private:
    double x,y;
public:
    Point():x(0),y(0){};
    Point(double a,double b):x(a),y(b){}
    void show(){cout<<"Point : ("<<x<<", "<<y<<")"<<endl;}
};
class Line{
        friend class Point;
private:
    Point p1,p2;
public:
    Line(Point p,Point q):p1(p),p2(q){}
    Line(double x1,double y1,double x2,double y2):p1(x1,y1),p2(x2,y2){}
    void show(){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<")"<<endl;}
};
int main()
{
    char c;
    int num, i;
    double x1, x2, y1, y2;
    Point p(1, -2), q(2, -1), t;
    t.show();
    std::cin>>num;
    for(i = 1; i <= num; i++)
    {
        std::cin>>x1>>c>>y1>>x2>>c>>y2;
        Line line(x1, y1, x2, y2);
        line.show();
    }
    Line l1(p, q), l2(p, t), l3(q, t), l4(t, q);
    l1.show();
    l2.show();
    l3.show();
    l4.show();
}

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Problem E: 平面上的点和线——Point类、Line类 (II)

Problem E: 平面上的点和线——Point类、Line类 (II)

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 2641  Solved: 1357
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Description

在数学上，平面直角坐标系上的点用X轴和Y轴上的两个坐标值唯一确定，两点确定一条线段。现在我们封装一个“Point类”和“Line类”来实现平面上的点的操作。

 

根据“append.cc”，完成Point类和Line类的构造方法和show()方法，输出各Line对象的构造和析构次序。

 

接口描述：

Point::show()方法：按格式输出Point对象。

Line::show()方法：按格式输出Line对象。

Input

输入的第一行为N，表示后面有N行测试样例。每行为两组坐标“x,y”，分别表示线段起点和终点的x坐标和y坐标，两组坐标间用一个空格分开，x和y的值都在double数据范围内。

Output

输出为多行，每行为一条线段，起点坐标在前终点坐标在后，每个点的X坐标在前，Y坐标在后，Y坐标前面多输出一个空格，用括号包裹起来。输出格式见sample。

Sample Input

4 0,0 1,1 1,1 2,3 2,3 4,5 0,1 1,0

Sample Output

Point : (0, 0) Line : (0, 0) to (1, 1) is created. Line : (0, 0) to (1, 1) Line : (0, 0) to (1, 1) is erased. Line : (1, 1) to (2, 3) is created. Line : (1, 1) to (2, 3) Line : (1, 1) to (2, 3) is erased. Line : (2, 3) to (4, 5) is created. Line : (2, 3) to (4, 5) Line : (2, 3) to (4, 5) is erased. Line : (0, 1) to (1, 0) is created. Line : (0, 1) to (1, 0) Line : (0, 1) to (1, 0) is erased. Line : (1, -2) to (2, -1) is created. Line : (1, -2) to (0, 0) is created. Line : (2, -1) to (0, 0) is created. Line : (0, 0) to (2, -1) is created. Line : (1, -2) to (2, -1) Line : (1, -2) to (0, 0) Line : (2, -1) to (0, 0) Line : (0, 0) to (2, -1) Line : (0, 0) to (2, -1) is erased. Line : (2, -1) to (0, 0) is erased. Line : (1, -2) to (0, 0) is erased. Line : (1, -2) to (2, -1) is erased.

HINT

 

Append Code

append.cc,

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한국어<   中文  فارسی  English  ไทย
All Copyright Reserved 2010-2011 SDUSTOJ TEAM
GPL2.0 2003-2011 HUSTOJ Project TEAM
Anything about the Problems, Please Contact Admin:admin

#include<iostream>
using namespace std;
class Point{
        friend class Line;
private:
    double x,y;
public:
    Point():x(0),y(0){};
    Point(double a,double b):x(a),y(b){}
    void show(){cout<<"Point : ("<<x<<", "<<y<<")"<<endl;}
};
class Line{
        friend class Point;
private:
    Point p1,p2;
public:
    Line(Point p,Point q):p1(p),p2(q){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is created."<<endl;}
    Line(double x1,double y1,double x2,double y2):p1(x1,y1),p2(x2,y2){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is created."<<endl;}
    ~Line(){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is erased."<<endl;}
    void show(){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<")"<<endl;}
};

int main()
{
    char c;
    int num, i;
    double x1, x2, y1, y2;
    Point p(1, -2), q(2, -1), t;
    t.show();
    std::cin>>num;
    for(i = 1; i <= num; i++)
    {
        std::cin>>x1>>c>>y1>>x2>>c>>y2;
        Line line(x1, y1, x2, y2);
        line.show();
    }
    Line l1(p, q), l2(p, t), l3(q, t), l4(t, q);
    l1.show();
    l2.show();
    l3.show();
    l4.show();
}

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Problem F: 平面上的点和线——Point类、Line类 (III)

Problem F: 平面上的点和线——Point类、Line类 (III)

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 3854  Solved: 1367
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Description

在数学上，平面直角坐标系上的点用X轴和Y轴上的两个坐标值唯一确定，两点确定一条线段。现在我们封装一个“Point类”和“Line类”来实现平面上的点的操作。

 

根据“append.cc”，完成Point类和Line类的构造方法和show()方法，输出各Line对象和Point对象的构造和析构次序。

 

接口描述：

Point::show()方法：按格式输出Point对象。

Line::show()方法：按格式输出Line对象。

 

Input

输入的第一行为N，表示后面有N行测试样例。每行为两组坐标“x,y”，分别表示线段起点和终点的x坐标和y坐标，两组坐标间用一个空格分开，x和y的值都在double数据范围内。

Output

输出为多行，每行为一条线段，起点坐标在前终点坐标在后，每个点的X坐标在前，Y坐标在后，Y坐标前面多输出一个空格，用括号包裹起来。输出格式见sample。

 

C语言的输入输出被禁用。

Sample Input

4 0,0 1,1 1,1 2,3 2,3 4,5 0,1 1,0

Sample Output

Point : (1, -2) is created. Point : (2, -1) is created. Point : (0, 0) is created. Point : (0, 0) ========================= Point : (0, 0) is created. Point : (1, 1) is created. Line : (0, 0) to (1, 1) is created. Line : (0, 0) to (1, 1) Line : (0, 0) to (1, 1) is erased. Point : (1, 1) is erased. Point : (0, 0) is erased. ========================= Point : (1, 1) is created. Point : (2, 3) is created. Line : (1, 1) to (2, 3) is created. Line : (1, 1) to (2, 3) Line : (1, 1) to (2, 3) is erased. Point : (2, 3) is erased. Point : (1, 1) is erased. ========================= Point : (2, 3) is created. Point : (4, 5) is created. Line : (2, 3) to (4, 5) is created. Line : (2, 3) to (4, 5) Line : (2, 3) to (4, 5) is erased. Point : (4, 5) is erased. Point : (2, 3) is erased. ========================= Point : (0, 1) is created. Point : (1, 0) is created. Line : (0, 1) to (1, 0) is created. Line : (0, 1) to (1, 0) Line : (0, 1) to (1, 0) is erased. Point : (1, 0) is erased. Point : (0, 1) is erased. ========================= Point : (1, -2) is copied. Point : (2, -1) is copied. Line : (1, -2) to (2, -1) is created. Point : (1, -2) is copied. Point : (0, 0) is copied. Line : (1, -2) to (0, 0) is created. Point : (2, -1) is copied. Point : (0, 0) is copied. Line : (2, -1) to (0, 0) is created. Point : (0, 0) is copied. Point : (2, -1) is copied. Line : (0, 0) to (2, -1) is created. Line : (1, -2) to (2, -1) Line : (1, -2) to (0, 0) Line : (2, -1) to (0, 0) Line : (0, 0) to (2, -1) Line : (0, 0) to (2, -1) is erased. Point : (2, -1) is erased. Point : (0, 0) is erased. Line : (2, -1) to (0, 0) is erased. Point : (0, 0) is erased. Point : (2, -1) is erased. Line : (1, -2) to (0, 0) is erased. Point : (0, 0) is erased. Point : (1, -2) is erased. Line : (1, -2) to (2, -1) is erased. Point : (2, -1) is erased. Point : (1, -2) is erased. Point : (0, 0) is erased. Point : (2, -1) is erased. Point : (1, -2) is erased.

HINT

 

Append Code

append.cc,

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한국어<   中文  فارسی  English  ไทย
All Copyright Reserved 2010-2011 SDUSTOJ TEAM
GPL2.0 2003-2011 HUSTOJ Project TEAM
Anything about the Problems, Please Contact Admin:admin

#include<iostream>
using namespace std;
class Point{
        friend class Line;
private:
    double x,y;
public:
    Point():x(0),y(0){cout<<"Point : ("<<x<<", "<<y<<") is created."<<endl;};
    Point(double a,double b):x(a),y(b){cout<<"Point : ("<<x<<", "<<y<<") is created."<<endl;}
    Point(const Point&p){x=p.x;y=p.y;cout<<"Point : ("<<x<<", "<<y<<") is copied."<<endl;}
    ~Point(){cout<<"Point : ("<<x<<", "<<y<<") is erased."<<endl;}
    void show(){cout<<"Point : ("<<x<<", "<<y<<")"<<endl;}
};
class Line{
        friend class Point;
private:
    Point p1,p2;
public:
    Line(Point &p,Point &q):p1(p),p2(q){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is created."<<endl;}
    Line(double x1,double y1,double x2,double y2):p1(x1,y1),p2(x2,y2){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is created."<<endl;}

    ~Line(){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is erased."<<endl;}
    void show(){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<")"<<endl;}
};
int main()
{
    char c;
    int num, i;
    double x1, x2, y1, y2;
    Point p(1, -2), q(2, -1), t;
    t.show();
    std::cin>>num;
    for(i = 1; i <= num; i++)
    {
        std::cout<<"=========================
";
        std::cin>>x1>>c>>y1>>x2>>c>>y2;
        Line line(x1, y1, x2, y2);
        line.show();
    }
    std::cout<<"=========================
";
    Line l1(p, q), l2(p, t), l3(q, t), l4(t, q);
    l1.show();
    l2.show();
    l3.show();
    l4.show();
}

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Problem G: 平面上的点和线——Point类、Line类 (IV)

Problem G: 平面上的点和线——Point类、Line类 (IV)

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 2619  Solved: 1277
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Description

在数学上，平面直角坐标系上的点用X轴和Y轴上的两个坐标值唯一确定，两点确定一条线段。现在我们封装一个“Point类”和“Line类”来实现平面上的点的操作。

 

根据“append.cc”，完成Point类和Line类的构造方法和show()方法，输出各Line对象和Point对象的构造和析构次序。

 

接口描述：

Point::show()方法：按格式输出Point对象。

Line::show()方法：按格式输出Line对象。

 

Input

输入的第一行为N，表示后面有N行测试样例。

每行为两组坐标“x,y”，分别表示线段起点和终点的x坐标和y坐标，两组坐标间用一个空格分开，x和y的值都在double数据范围内。

Output

输出为多行，每行为一条线段，起点坐标在前终点坐标在后，每个点的X坐标在前，Y坐标在后，Y坐标前面多输出一个空格，用括号包裹起来。输出格式见sample。

 

C语言的输入输出被禁用。

Sample Input

4 0,0 1,1 1,1 2,3 2,3 4,5 0,1 1,0

Sample Output

Point : (1, -2) is created. Point : (2, -1) is created. Point : (0, 0) is created. Point : (0, 0) Point : (0, 0) is created. Point : (0, 0) is created. Line : (0, 0) to (0, 0) is created. Point : (0, 0) is created. Point : (0, 0) is created. Line : (0, 0) to (0, 0) is created. Point : (0, 0) is created. Point : (0, 0) is created. Line : (0, 0) to (0, 0) is created. Point : (0, 0) is created. Point : (0, 0) is created. Line : (0, 0) to (0, 0) is created. ========================= Line : (0, 0) to (1, 1) ========================= Line : (1, 1) to (2, 3) ========================= Line : (2, 3) to (4, 5) ========================= Line : (0, 1) to (1, 0) ========================= Point : (1, -2) is copied. Point : (2, -1) is copied. Line : (1, -2) to (2, -1) is created. Point : (1, -2) is copied. Point : (0, 0) is copied. Line : (1, -2) to (0, 0) is created. Point : (2, -1) is copied. Point : (0, 0) is copied. Line : (2, -1) to (0, 0) is created. Point : (0, 0) is copied. Point : (2, -1) is copied. Line : (0, 0) to (2, -1) is created. Line : (1, -2) to (2, -1) Line : (1, -2) to (0, 0) Line : (2, -1) to (0, 0) Line : (0, 0) to (2, -1) Line : (0, 0) to (2, -1) is erased. Point : (2, -1) is erased. Point : (0, 0) is erased. Line : (2, -1) to (0, 0) is erased. Point : (0, 0) is erased. Point : (2, -1) is erased. Line : (1, -2) to (0, 0) is erased. Point : (0, 0) is erased. Point : (1, -2) is erased. Line : (1, -2) to (2, -1) is erased. Point : (2, -1) is erased. Point : (1, -2) is erased. Line : (2, 3) to (4, 5) is erased. Point : (4, 5) is erased. Point : (2, 3) is erased. Line : (1, 1) to (2, 3) is erased. Point : (2, 3) is erased. Point : (1, 1) is erased. Line : (0, 0) to (1, 1) is erased. Point : (1, 1) is erased. Point : (0, 0) is erased. Line : (0, 0) to (0, 0) is erased. Point : (0, 0) is erased. Point : (0, 0) is erased. Point : (0, 0) is erased. Point : (2, -1) is erased. Point : (1, -2) is erased.

HINT

 

Append Code

append.cc,

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한국어<   中文  فارسی  English  ไทย
All Copyright Reserved 2010-2011 SDUSTOJ TEAM
GPL2.0 2003-2011 HUSTOJ Project TEAM
Anything about the Problems, Please Contact Admin:admin

#include<iostream>
using namespace std;

class Point{
        friend class Line;
private:
    double x,y;
public:
    Point(double a=0,double b=0):x(a),y(b){cout<<"Point : ("<<x<<", "<<y<<") is created."<<endl;}
    Point(const Point&p){x=p.x;y=p.y;cout<<"Point : ("<<x<<", "<<y<<") is copied."<<endl;}
    ~Point(){cout<<"Point : ("<<x<<", "<<y<<") is erased."<<endl;}
    void show(){cout<<"Point : ("<<x<<", "<<y<<")"<<endl;}
};
class Line{
        friend class Point;
private:
    Point p1,p2;
public:
    Line(Point &p,Point &q):p1(p),p2(q){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is created."<<endl;}
    Line(double x1=0,double y1=0,double x2=0,double y2=0):p1(x1,y1),p2(x2,y2){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is created."<<endl;}

    ~Line(){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is erased."<<endl;}
    void SetLine(double a,double b,double c,double d){p1.x=a;p1.y=b;p2.x=c;p2.y=d;}
    void show(){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<")"<<endl;}
};


int main()
{
    char c;
    int num, i;
    double x1, x2, y1, y2;
    Point p(1, -2), q(2, -1), t;
    t.show();
    std::cin>>num;
    Line line[num];
    for(i = 1; i <= num; i++)
    {
        std::cout<<"=========================
";
        std::cin>>x1>>c>>y1>>x2>>c>>y2;
        line[i].SetLine(x1, y1, x2, y2);
        line[i].show();
    }
    std::cout<<"=========================
";
    Line l1(p, q), l2(p, t), l3(q, t), l4(t, q);
    l1.show();
    l2.show();
    l3.show();
    l4.show();
}

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Problem H: 平面上的点和线——Point类、Line类 (V)

Problem H: 平面上的点和线——Point类、Line类 (V)

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 3774  Solved: 1425
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Description

在数学上，平面直角坐标系上的点用X轴和Y轴上的两个坐标值唯一确定，两点确定一条线段。现在我们封装一个“Point类”和“Line类”来实现平面上的点的操作。

根据“append.cc”，完成Point类和Line类的构造方法和show()方法，输出各Line对象和Point对象的构造和析构次序。

接口描述：

Point::show()方法：按格式输出Point对象。

Line::show()方法：按格式输出Line对象。

Line::SetLine(double, double, double, double)方法：设置Line对象起点的x,y坐标（第一个和第二参数）和终点的x,y坐标（第三个和第四个坐标），并返回本对象

Line::SetLine(const Point &, const Point &)方法：设置Line对象的起点（第一个参数）和终点（第二个坐标），并返回本对象

Line::SetLine(const Line&)方法：设置Line对象，复制参数的坐标，并返回本对象

Line::readLine()方法：从标准输入上读入坐标，格式见Sample

Input

输入的第一行为N，表示后面有N行测试样例。

 

每行为两组坐标“x,y”，分别表示线段起点和终点的x坐标和y坐标，两组坐标间用一个空格分开，x和y的值都在double数据范围内。

 

Output

输出为多行，每行为一条线段，起点坐标在前终点坐标在后，每个点的X坐标在前，Y坐标在后，Y坐标前面多输出一个空格，用括号包裹起来。输出格式见sample。

C语言的输入输出被禁用。

Sample Input

4 0,0 1,1 1,1 2,3 2,3 4,5 0,1 1,0

Sample Output

Point : (1, -2) is created. Point : (2, -1) is created. Point : (0, 0) is created. Point : (0, 0) Point : (0, 0) is created. Point : (0, 0) is created. Line : (0, 0) to (0, 0) is created. Point : (0, 0) is created. Point : (0, 0) is created. Line : (0, 0) to (0, 0) is created. Point : (0, 0) is created. Point : (0, 0) is created. Line : (0, 0) to (0, 0) is created. Point : (0, 0) is created. Point : (0, 0) is created. Line : (0, 0) to (0, 0) is created. Line : (0, 0) to (1, 1) Line : (1, 1) to (2, 3) Line : (2, 3) to (4, 5) Line : (0, 1) to (1, 0) Point : (1, -2) is copied. Point : (2, -1) is copied. Line : (1, -2) to (2, -1) is created. Point : (1, -2) is copied. Point : (0, 0) is copied. Line : (1, -2) to (0, 0) is created. Point : (2, -1) is copied. Point : (0, 0) is copied. Line : (2, -1) to (0, 0) is created. Point : (1, -2) is copied. Point : (2, -1) is copied. Line : (1, -2) to (2, -1) is copied. Line : (1, -2) to (2, -1) Line : (1, -2) to (2, -1) Line : (2, -1) to (0, 0) Line : (0, 0) to (2, -1) Line : (0, 0) to (2, -1) is erased. Point : (2, -1) is erased. Point : (0, 0) is erased. Line : (2, -1) to (0, 0) is erased. Point : (0, 0) is erased. Point : (2, -1) is erased. Line : (1, -2) to (2, -1) is erased. Point : (2, -1) is erased. Point : (1, -2) is erased. Line : (1, -2) to (2, -1) is erased. Point : (2, -1) is erased. Point : (1, -2) is erased. Line : (0, 1) to (1, 0) is erased. Point : (1, 0) is erased. Point : (0, 1) is erased. Line : (2, 3) to (4, 5) is erased. Point : (4, 5) is erased. Point : (2, 3) is erased. Line : (1, 1) to (2, 3) is erased. Point : (2, 3) is erased. Point : (1, 1) is erased. Line : (0, 0) to (1, 1) is erased. Point : (1, 1) is erased. Point : (0, 0) is erased. Point : (0, 0) is erased. Point : (2, -1) is erased. Point : (1, -2) is erased.

HINT

 

Append Code

append.cc,

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한국어<   中文  فارسی  English  ไทย
All Copyright Reserved 2010-2011 SDUSTOJ TEAM
GPL2.0 2003-2011 HUSTOJ Project TEAM
Anything about the Problems, Please Contact Admin:admin

#include<iostream>
using namespace std;

class Point{
        friend class Line;
private:
    double x,y;
public:
    Point(double a=0,double b=0):x(a),y(b){cout<<"Point : ("<<x<<", "<<y<<") is created."<<endl;}
    Point(const Point&p){x=p.x;y=p.y;cout<<"Point : ("<<x<<", "<<y<<") is copied."<<endl;}
    ~Point(){cout<<"Point : ("<<x<<", "<<y<<") is erased."<<endl;}
    void show(){cout<<"Point : ("<<x<<", "<<y<<")"<<endl;}
};
class Line{
        friend class Point;
private:
    Point p1,p2;
public:
    Line(Point &p,Point &q):p1(p),p2(q){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is created."<<endl;}
    Line(double x1=0,double y1=0,double x2=0,double y2=0):p1(x1,y1),p2(x2,y2){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is created."<<endl;}
    ///Line(const Line&l){p1=l.p1;p2=l.p2;cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is copied."<<endl;}
    Line(const Line&l):p1(l.p1),p2(l.p2){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is copied."<<endl;}
    ~Line(){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is erased."<<endl;}
    Line& setLine(double a,double b,double c,double d){p1.x=a;p1.y=b;p2.x=c;p2.y=d;return *this;}
    Line& setLine(const Point &p, const Point &q){p1=p;p2=q;return *this;}
    Line& setLine(const Line& l){*this=l;return *this;}
    void show(){cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<")"<<endl;}
    void readLine(){char c;cin>>p1.x>>c>>p1.y>>p2.x>>c>>p2.y;}
};

int main()
{
    int num, i;
    Point p(1, -2), q(2, -1), t;
    t.show();
    std::cin>>num;
    Line line[num];
    for(i = 0; i < num; i++)
    {
        line[i].readLine();
        line[i].show();
    }
    Line l1(p, q), l2(p,t), l3(q,t), l4(l1);
    l1.show();
    l2.setLine(l1).show();
    l3.show();
    l4.setLine(t,q).show();
}

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Problem I: 平面上的点和线——Point类、Line类 (VI)

Problem I: 平面上的点和线——Point类、Line类 (VI)

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 3275  Solved: 1182
[Submit][Status][Web Board]

Description

在数学上，平面直角坐标系上的点用X轴和Y轴上的两个坐标值唯一确定，两点确定一条线段。现在我们封装一个“Point类”和“Line类”来实现平面上的点的操作。

根据“append.cc”，完成Point类和Line类的构造方法和show()方法，输出各Line对象和Point对象的构造和析构次序。

接口描述：

Point::show()方法：按格式输出Point对象。

Point::x()方法：取x坐标。

Point::y()方法：取y坐标。

Line::show()方法：按格式输出Line对象。

Line::SetLine(double, double, double, double)方法：设置Line对象起点的x,y坐标（第一个和第二参数）和终点的x,y坐标（第三个和第四个坐标），并返回本对象

Line::SetLine(const Point &, const Point &)方法：设置Line对象的起点（第一个参数）和终点（第二个坐标），并返回本对象

Line::SetLine(const Line&)方法：设置Line对象，复制参数的坐标，并返回本对象

Line::readLine()方法：从标准输入上读入坐标，格式见Sample

Line::start()方法：取Line的起点

Line::end()方法：取Line的终点

Line::setStart()方法：设置Line的起点

Line::setEnd()方法：设置Line的终点

 

以下三个函数用于输出Line对象，格式同sample

showLineCoordinate(const Line&)

showLinePoint(const Line&)

showLine(const Line&)

Input

输入的第一行为N，表示后面有N行测试样例。

 

每行为两组坐标“x,y”，分别表示线段起点和终点的x坐标和y坐标，两组坐标间用一个空格分开，x和y的值都在double数据范围内。

 

Output

输出为多行，每行为一条线段，起点坐标在前终点坐标在后，每个点的X坐标在前，Y坐标在后，Y坐标前面多输出一个空格，用括号包裹起来。输出格式见sample。

C语言的输入输出被禁用。

Sample Input

4 0,0 1,1 1,1 2,3 2,3 4,5 0,1 1,0

Sample Output

Point : (1, -2) is created. Point : (2, -1) is created. Point : (0, 0) is created. Point : (0, 0) Point : (0, 0) is created. Point : (0, 0) is created. Line : (0, 0) to (0, 0) is created. Point : (0, 0) is created. Point : (0, 0) is created. Line : (0, 0) to (0, 0) is created. Point : (0, 0) is created. Point : (0, 0) is created. Line : (0, 0) to (0, 0) is created. Point : (0, 0) is created. Point : (0, 0) is created. Line : (0, 0) to (0, 0) is created. Point : (0, 0) is created. Point : (0, 0) is created. Line : (0, 0) to (0, 0) is created. Line : (0, 0) to (1, 1) Line : (1, 1) to (2, 3) Line : (2, 3) to (4, 5) Line : (0, 1) to (1, 0) Point : (1, -2) is copied. Point : (2, -1) is copied. Line : (1, -2) to (2, -1) is created. Point : (1, -2) is copied. Point : (0, 0) is copied. Line : (1, -2) to (0, 0) is created. Point : (2, -1) is copied. Point : (0, 0) is copied. Line : (2, -1) to (0, 0) is created. Point : (1, -2) is copied. Point : (2, -1) is copied. Line : (1, -2) to (2, -1) is copied. Line : (1, -2) to (2, -1) Line : Point : (1, -2) to Point : (0, 0) Line : Point : (1, -2) to Point : (2, -1) Line : (0, 0) to (2, -1) Line : (0, 0) to (2, -1) is erased. Point : (2, -1) is erased. Point : (0, 0) is erased. Line : (1, -2) to (2, -1) is erased. Point : (2, -1) is erased. Point : (1, -2) is erased. Line : (1, -2) to (0, 0) is erased. Point : (0, 0) is erased. Point : (1, -2) is erased. Line : (1, -2) to (2, -1) is erased. Point : (2, -1) is erased. Point : (1, -2) is erased. Line : (0, 1) to (1, 0) is erased. Point : (1, 0) is erased. Point : (0, 1) is erased. Line : (2, 3) to (4, 5) is erased. Point : (4, 5) is erased. Point : (2, 3) is erased. Line : (1, 1) to (2, 3) is erased. Point : (2, 3) is erased. Point : (1, 1) is erased. Line : (0, 0) to (1, 1) is erased. Point : (1, 1) is erased. Point : (0, 0) is erased. Line : (0, 0) to (2, -1) is erased. Point : (2, -1) is erased. Point : (0, 0) is erased. Point : (0, 0) is erased. Point : (2, -1) is erased. Point : (1, -2) is erased.

HINT

 

Append Code

append.cc,

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All Copyright Reserved 2010-2011 SDUSTOJ TEAM
GPL2.0 2003-2011 HUSTOJ Project TEAM
Anything about the Problems, Please Contact Admin:admin

#include<iostream>
using namespace std;

class Point{
        friend class Line;
private:
    double m,n;
public:
    ///Point():m(0),n(0){cout<<"Point : ("<<m<<", "<<n<<") is created."<<endl;}
    Point(double a):m(a),n(a){cout<<"Point : ("<<m<<", "<<n<<") is created."<<endl;}
    Point(double a=0,double b=0):m(a),n(b){cout<<"Point : ("<<m<<", "<<n<<") is created."<<endl;}
    Point(const Point&p){m=p.m;n=p.n;cout<<"Point : ("<<m<<", "<<n<<") is copied."<<endl;}
    ~Point(){cout<<"Point : ("<<m<<", "<<n<<") is erased."<<endl;}
    void show(){cout<<"Point : ("<<m<<", "<<n<<")"<<endl;}
    double x()const {return m;}
    double y()const {return n;}
    ///Point &setPoint(double x,double y){m=x;n=y;return *this;}
    void showNoEndOfLine()const{cout<<"Point : ("<<m<<", "<<n<<")";}

};
class Line{
        friend class Point;
private:
    Point p1,p2;
public:
    Line(Point &p,Point &q):p1(p),p2(q){cout<<"Line : ("<<p1.m<<", "<<p1.n<<") to ("<<p2.m<<", "<<p2.n<<") is created."<<endl;}
    Line(double x1=0,double y1=0,double x2=0,double y2=0):p1(x1,y1),p2(x2,y2){cout<<"Line : ("<<p1.m<<", "<<p1.n<<") to ("<<p2.m<<", "<<p2.n<<") is created."<<endl;}
    Line(const Line&l):p1(l.p1),p2(l.p2){cout<<"Line : ("<<p1.m<<", "<<p1.n<<") to ("<<p2.m<<", "<<p2.n<<") is copied."<<endl;}
    ~Line(){cout<<"Line : ("<<p1.m<<", "<<p1.n<<") to ("<<p2.m<<", "<<p2.n<<") is erased."<<endl;}
    Line& setLine(double a,double b,double c,double d){p1.m=a;p1.n=b;p2.m=c;p2.n=d;return *this;}
    Line& setLine(const Point &p, const Point &q){p1=p;p2=q;return *this;}
    Line& setLine(const Line& l){*this=l;return *this;}
    void show()const{cout<<"Line : ("<<p1.m<<", "<<p1.n<<") to ("<<p2.m<<", "<<p2.n<<")"<<endl;}
    void readLine(){char c;cin>>p1.m>>c>>p1.n>>p2.m>>c>>p2.n;}
    const Point &start()const{return p1;}
    const Point &end()const{return p2;}
    void setStart(Point &p){p1=p;}
    void setEnd(Point &p){p2=p;}

};
void showLineCoordinate(const Line& line)
{
    std::cout<<"Line : ";
    std::cout<<"("<<line.start().x()<<", "<<line.start().y()<<")";
    std::cout<<" to ";
    std::cout<<"("<<line.end().x()<<", "<<line.end().y()<<")";
    std::cout<<std::endl;
}

void showLinePoint(const Line& line)
{
    std::cout<<"Line : ";
    line.start().showNoEndOfLine();
    std::cout<<" to ";
    line.end().showNoEndOfLine();
    std::cout<<std::endl;
}

void showLine(const Line& line)
{
    line.show();
}

int main()
{
    int num, i;
    Point p(1, -2), q(2, -1), t;
    t.show();
    std::cin>>num;
    Line line[num + 1];
    for(i = 1; i <= num; i++)
    {
        line[i].readLine();
        showLine(line[i]);
    }
    Line l1(p, q), l2(p,t), l3(q,t), l4(l1);
    showLineCoordinate(l1);
    showLinePoint(l2);
    showLinePoint(l3.setLine(l1));
    showLineCoordinate(l4.setLine(t,q));
    line[0].setStart(t);
    line[0].setEnd(q);
}

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Problem J: 平面上的点和线——Point类、Line类 (VII)

Problem J: 平面上的点和线——Point类、Line类 (VII)

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 2151  Solved: 1171
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Description

在数学上，平面直角坐标系上的点用X轴和Y轴上的两个坐标值唯一确定，两点确定一条线段。现在我们封装一个“Point类”和“Line类”来实现平面上的点的操作。

根据“append.cc”，完成Point类和Line类的构造方法和show()方法，输出各Line对象和Point对象的构造和析构次序。

接口描述：

Point::showCounter()方法：按格式输出当前程序中Point对象的计数。

Point::showSum()方法：按格式输出程序运行至当前存在过的Point对象总数。

Line::showCounter()方法：按格式输出当前程序中Line对象的计数。

Line::showSum()方法：按格式输出程序运行至当前存在过的Line对象总数。

Input

输入的第一行为N，表示后面有N行测试样例。

 

每行为两组坐标“x,y”，分别表示线段起点和终点的x坐标和y坐标，两组坐标间用一个空格分开，x和y的值都在double数据范围内。

 

Output

输出格式见sample。

C语言的输入输出被禁用。

Sample Input

4 0,0 1,1 1,1 2,3 2,3 4,5 0,1 1,0

Sample Output

Current : 3 points. In total : 3 points. Current : 6 lines. In total : 6 lines. Current : 17 points. In total : 17 points. Current : 6 lines. In total : 7 lines. Current : 15 points. In total : 17 points. Current : 6 lines. In total : 8 lines. Current : 17 points. In total : 21 points. Current : 6 lines. In total : 9 lines. Current : 15 points. In total : 21 points. Current : 6 lines. In total : 10 lines. Current : 17 points. In total : 25 points. Current : 6 lines. In total : 11 lines. Current : 15 points. In total : 25 points. Current : 6 lines. In total : 12 lines. Current : 17 points. In total : 29 points. Current : 6 lines. In total : 13 lines. Current : 15 points. In total : 29 points. Current : 9 lines. In total : 17 lines. Current : 21 points. In total : 37 points. Current : 13 lines. In total : 21 lines. Current : 21 points. In total : 45 points.

HINT

 

Append Code

append.cc,

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한국어<   中文  فارسی  English  ไทย
All Copyright Reserved 2010-2011 SDUSTOJ TEAM
GPL2.0 2003-2011 HUSTOJ Project TEAM
Anything about the Problems, Please Contact Admin:admin

#include<iostream>
using namespace std;

class Point{
        friend class Line;
private:
    double m,n;
    static int psum,pnum;
public:
    Point(double a=0,double b=0):m(a),n(b){psum++;pnum++;}
    Point(const Point&p){m=p.m;n=p.n;psum++;pnum++;}
    ~Point(){pnum--;}
    static void showCounter(){cout<<"Current : "<<pnum<<" points."<<endl;}
    static void showSum(){cout<<"In total : "<<psum<<" points."<<endl;}

};
class Line{
        friend class Point;
private:
    Point p1,p2;
    static int lsum,lnum;
public:
    Line(Point &p,Point &q):p1(p),p2(q){lsum++;lnum++;}
    Line(double x1=0,double y1=0,double x2=0,double y2=0):p1(x1,y1),p2(x2,y2){lsum++;lnum++;}
    ///Line(const Line&l){p1=l.p1;p2=l.p2;cout<<"Line : ("<<p1.x<<", "<<p1.y<<") to ("<<p2.x<<", "<<p2.y<<") is copied."<<endl;}
    Line(const Line&l):p1(l.p1),p2(l.p2){lsum++;lnum++;}
    ~Line(){lnum--;}
    void readLine(){char c;cin>>p1.m>>c>>p1.n>>p2.m>>c>>p2.n;}
    static void showCounter(){cout<<"Current : "<<lnum<<" lines."<<endl;}
    static void showSum(){cout<<"In total : "<<lsum<<" lines."<<endl;}
};
int Point::psum=0;
int Point::pnum=0;
int Line::lsum=0;
int Line::lnum=0;


int main()
{
    int num, i;
    Point p(1, -2), q(2, -1), t;
    t.showCounter();
    t.showSum();
    std::cin>>num;
    Line line[num + 1];
    for(i = 1; i <= num; i++)
    {
        Line *l1, l2;
        l1->showCounter();
        l1->showSum();
        l1 = new Line(p, q);
        line[i].readLine();
        p.showCounter();
        p.showSum();
        delete l1;
        l2.showCounter();
        l2.showSum();
        q.showCounter();
        q.showSum();
    }
    Line l1(p, q), l2(p,t), l3(q,t), l4(l1);
    Line::showCounter();
    Line::showSum();
    Point::showCounter();
    Point::showSum();
    Line *l = new Line[num];
    l4.showCounter();
    l4.showSum();
    delete[] l;
    t.showCounter();
    t.showSum();
}