平面上的点和线——Point类、Line类 (VII)

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

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 2497 Solved: 1744
[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::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

#include <iostream>using namespace std;class Point{private:    double x,y;    friend class Line;   static int num;   static int t;public:      Point(double x1=0,double y1=0)        {            x=x1;            y=y1;            num++;            t++;        }//        Point()//        {//            x=0;y=0;//            num++;//            t++;////   //     }        ~Point()        {           num--;        }        Point(const Point &p):x(p.x),y(p.y)        {            num++;            t++;        }       static void showCounter()        {            cout<<"Current : "<<num<<" points."<<endl;        }       static void showSum()        {            cout<<"In total : "<<t<<" points."<<endl;        }};class Line{private:    Point a,b;    static int num1;    static int t1;    friend class Point;public:    Line(Point &a1,Point &b1):a(a1),b(b1)    {        num1++;        t1++;    }    Line():a(0,0),b(0,0)    {        num1++;        t1++;    }     Line(double a11,double b11,double c11,double d11)    {        a.x=a11;        a.y=b11;        b.x=c11;        b.y=d11;       num1++;       t1++;    }    ~Line()    {        num1--;;    }    Line(const Line &p):a(p.a),b(p.b)    {        num1++;        t1++;    }      void readLine()    {        char c;        int i;       cin>>a.x>>c>>a.y>>b.x>>c>>b.y;    }    static void showCounter()    {        cout<<"Current : "<<num1<<" lines."<<endl;    }    static void showSum()    {        cout<<"In total : "<<t1<<" lines."<<endl;    }};int Point::num=0;int Point::t=0;int Line::num1=0;int Line::t1=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();}

误删一个大括号，报错。