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

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

Time Limit: 1 Sec  Memory Limit:
128 MB
Submit: 2867  Solved: 2021

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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

40,0 1,11,1 2,32,3 4,50,1 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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#include <iostream>
using namespace std;
class  Point
{
private:
double x_,y_;
static int pnum,ptotal;
public:
Point(){x_ = 0; y_ = 0;pnum++;ptotal++;}
Point(double xx,double yy){x_ = xx;y_  = yy;pnum++;ptotal++;}
Point(const Point & p){x_ = p.x_;y_ = p.y_;pnum++;ptotal++;;}
~Point(){ pnum--;}
static void showCounter(){cout << "Current : " << pnum <<" points." << endl;}
static void showSum(){cout << "In total : " << ptotal << " points." <<endl;}
void setXY(double xx,double yy){ x_=  xx;y_ = yy;}
};
class Line
{
private:
Point st,ed;
static int lnum,ltotal;
public:
// Line():p1(),p2(){}
Line():st(0,0),ed(0,0){lnum++;ltotal++;}
Line(const Point &sst, const Point &eed):st(sst),ed(eed)
{
lnum++;ltotal++;
}
Line(const Line & line):st(line.st),ed(line.ed){ lnum++;ltotal++; }
void show() const
{

}
Line &setLine(double x1,double y1,double x2,double y2)
{
st.setXY(x1,y1);
ed.setXY(x2,y2);
return *this;
}
Line & setLine(const Line &pt)
{
st = pt.st;//copy
ed = pt.ed;
return *this;
}
Line & setLine(const Point &sst,const Point &eed)
{
st = sst;//copy
ed = eed;
return *this;
}
void readLine()
{
double x1,y1,x2,y2;
char ch ;
cin >> x1 >>ch >>y1>> x2 >> ch >>y2;
st.setXY(x1,y1);
ed.setXY(x2,y2);
}
const Point  &start() const {return st;}
const Point  &end() const  {return ed;}
void setStart(const Point & pt){ st = pt;}
void setEnd(const Point & pt){ed = pt;}
static void showCounter()  {cout << "Current : " << lnum << " lines." <<endl; }
static void showSum()  {cout << "In total : " << ltotal << " lines." <<endl; }
~Line()
{
lnum--;
}
};
int Point ::pnum = 0;
int Line :: lnum = 0;
int Line :: ltotal = 0;
int Point ::ptotal   = 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();
}