实验3
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Problem A: 平面上的点——Point类 (IV)
Time Limit: 1 Sec Memory Limit: 4 MBSubmit: 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
Sample Output
HINT
对象计数通过静态成员来实现
Append Code
#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)
Time Limit: 1 Sec Memory Limit: 4 MBSubmit: 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
Sample Output
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
#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)
Time Limit: 1 Sec Memory Limit: 4 MBSubmit: 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
Sample Output
HINT
给函数正确的返回值。注意常量对象调用的函数。
Append Code
#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)
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 2814 Solved: 1374
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Description
Input
输入的第一行为N,表示后面有N行测试样例。
Output
Sample Input
Sample Output
HINT
Append Code
#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)
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 2641 Solved: 1357
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Description
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
Sample Input
Sample Output
HINT
Append Code
#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)
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 3854 Solved: 1367
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Description
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
Sample Input
Sample Output
HINT
Append Code
#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<<"=========================\n"; std::cin>>x1>>c>>y1>>x2>>c>>y2; Line line(x1, y1, x2, y2); line.show(); } std::cout<<"=========================\n"; 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)
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 2619 Solved: 1277
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Description
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
Sample Input
Sample Output
HINT
Append Code
#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<<"=========================\n"; std::cin>>x1>>c>>y1>>x2>>c>>y2; line[i].SetLine(x1, y1, x2, y2); line[i].show(); } std::cout<<"=========================\n"; 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)
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 3774 Solved: 1425
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Description
Input
Output
Sample Input
Sample Output
HINT
Append Code
#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)
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 3275 Solved: 1182
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Description
Input
Output
Sample Input
Sample Output
HINT
Append Code
#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)
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 2151 Solved: 1171
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Description
Input
Output
Sample Input
Sample Output
HINT
Append Code
#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(); }