A.三维坐标点的平移（运算符重载）

题目描述

定义一个三维点Point类，利用友元函数重载"++“和”–"运算符，并区分这两种运算符的前置和后置运算。

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class point
{
	int x;
	int y;
	int z;
public:
	point(int X=0,int Y=0,int Z=0);
	{x=X;y=Y;z=Z;}
	friend point operator ++(point &a);
	friend point operator ++(point &,int);
	friend point operator --(point &);
	friend point operator --(point &,int);
	void show()
	{
		cout<<"x="<<x<<' '<<"y="<<y<<''<<"z="<<z<<endl;
	}
}

要求如下：
1.实现Point类；
2.编写main函数，初始化1个Point对象，将这个对象++或–后赋给另外一个对象，并输出计算后对象的坐标信息。

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#include<iostream>
#include<cstring>
using namespace std;

class point 
{
protected:
	int x;
	int y;
	int z;
public:
	point(int X = 0, int Y = 0, int Z = 0)
	{
		x = X;
		y = Y;
		z = Z;
	}
	void set(int _x, int _y, int _z)
	{
		x = _x;
		y = _y;
		z = _z;
	}
	friend point operator++(point& a)
	{
		a.x++;
		a.y++;
		a.z++;
		return a;
	}
	friend point operator++(point& a, int x)
	{
		point p(a);
		a.x++;
		a.y++;
		a.z++;
		return p;
	}
	friend point operator--(point& a)
	{
		a.x--;
		a.y--;
		a.z--;
		return a;
	}
	friend point operator--(point& a, int x)
	{
		point p(a);
		a.x--;
		a.y--;
		a.z--;
		return p;
	}
	void show()
	{
		cout << "x=" << x << ' ' << "y=" << y << ' ' << "z=" << z << endl;
	}
};

int main()
{
	int x, y, z;
	cin >> x >> y >> z;
	point p1(x, y, z);
	point p2;
	p1++;
	p1.show();
	p1.set(x, y, z);
	p2 = p1;
	p2.show();
	p1.set(x, y, z);
	(++p1).show();
	p1.show();
	p1.set(x, y, z);
	p1--;
	p1.show();
	p1.set(x, y, z);
	p1.show();
	p1.set(x, y, z);
	(--p1).show();
	p1.show();


	return 0;
}

B.货币加减（运算符重载）

题目描述

定义CMoney类，包含元、角、分三个数据成员，友元函数重载‘+’、’-’，实现货币的加减运算（假设a-b中a的金额始终大于等于b的金额），重载输入及输出，实现货币的输入，输出。
读入最初的货币值，对其不断进行加、减操作，输出结果。
可根据需要，为CMoney类添加构造函数或其它成员函数。

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#include<iostream>
#include<cstring>
using namespace std;

class CMoney
{
protected:
	int yuan;
	int corn;
	int coin;
public:
	CMoney(){}
	CMoney(int y,int cr,int ci):yuan(y),corn(cr),coin(ci){}
	friend CMoney operator +(CMoney& a, CMoney& b)
	{
		CMoney c;
		int c1, c2, sum;
		c1 = a.yuan * 100 + a.corn * 10 + a.coin;
		c2 = b.yuan * 100 + b.corn * 10 + b.coin;
		sum = c1 + c2;
		c.yuan = sum / 100;
		c.corn = (sum / 10) % 10;
		c.coin = sum % 10;
		
        
		return c;
	}
	friend CMoney operator-(CMoney& a, CMoney& b)
	{
		CMoney c;
		int c1, c2, sum;
		c1 = a.yuan * 100 + a.corn * 10 + a.coin;
		c2 = b.yuan * 100 + b.corn * 10 + b.coin;
		sum = c1 - c2;
		c.yuan = sum / 100;
		c.corn = (sum / 10) % 10;
		c.coin = sum % 10;
		return c;
	}
	friend ostream& operator <<(ostream& out, CMoney& a)
	{
		out << a.yuan << "元" << a.corn << "角" << a.coin << "分" << endl;
		return out;
	}
	friend istream& operator>>(istream& input, CMoney& a)
	{
		input >> a.yuan >> a.corn >> a.coin;
		return input;
	}
	
};

int main()
{
	int t;
	int y, j, f;
	string a;
	cin >> t;
	while (t--)
	{
		cin >> y >> j >> f;
		CMoney m1(y, j, f);
		CMoney m;
		cin >> a;
		while(a!="stop")
		{
			if (a == "add")
			{
				cin >> y >> j >> f;
				CMoney m2(y, j, f);
				m = m1 + m2;
			}
			else if (a == "minus")
			{
				cin >> y >> j >> f;
				CMoney m2(y, j, f);
				m = m1 - m2;
			}
			m1 = m;
			cin >> a;
		}
		cout << m;
	}
	return 0;
}

C.时钟调整（运算符前后增量）

题目描述

假定一个时钟包含时、分、秒三个属性，取值范围分别为0~11，0~59，0~59，具体要求如下：
1、用一元运算符++，并且是前增量的方法，实现时钟的调快操作。例如要把时钟调快5秒，则执行5次”  ++<对象> “ 的操作
2、用一元运算符–，并且是后增量的方法，实现时钟的调慢操作。例如要把时钟调慢10秒，则执行10次” <对象>-- “的操作
3、用构造函数的方法实现时钟对象的初始化，用输出函数实现时钟信息的输出
clock和time是系统内部函数，所以不要用来做类名或者其他

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#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
class CClock
{
protected:
	int hour;
	int minute;
	int second;
public:
	CClock(int h, int m, int s) :hour(h), minute(m), second(s) {}
	void show()
	{
		cout << hour << ":" << minute << ":" << second << endl;
	}
	 CClock operator++()
	{
		second++;
		if (second > 59)
		{
		     second = 0;
			minute++;
		}
		if (minute > 59)
		{
			minute = 0;
			hour++;
		}
		if (hour > 11)
		{
			hour = 0;
		}
		return *this;
	}
	 CClock operator--(int)
	{
		 CClock temp(*this);
		second--;
		if (second < 0)
		{
			second = 59;
			minute--;
		}
		if (minute < 0)
		{
			minute = 59;
			hour--;
		}
		if (hour < 0)
			hour = 11;
		return *this;
	}
};
int main()
{
	int hour, minute, second;
	int t;
	cin >> hour >> minute >> second;
	CClock c(hour, minute, second);
	cin >> t;
	while (t--)
	{
		int tmp;
		cin >> tmp;
		int num = abs(tmp);
		while (num--)
		{
			if (tmp > 0)
			{
				++c;
			}
			else
			{
				c--;
			}
		}
		c.show();
	}
	return 0;
}

D.X的放大与缩小（运算符重载）

题目描述

X字母可以放大和缩小，变为n行X（n=1,3,5,7,9,…,21）。例如，3行x图案如下：
image.png

现假设一个n行（n>0，奇数）X图案，遥控器可以控制X图案的放大与缩小。遥控器有5个按键，1）show，显示当前X图案；2）show++, 显示当前X图案，再放大图案，n+2；3）++show，先放大图案，n+2，再显示图案；4）show–，显示当前X图案，再缩小图案，n-2；5）–show，先缩小图案，n-2，再显示图案。假设X图案的放大和缩小在1-21之间。n=1时，缩小不起作用，n=21时，放大不起作用。

用类CXGraph表示X图案及其放大、缩小、显示。主函数模拟遥控器，代码如下，不可修改。请补充CXGraph类的定义和实现。

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int main()
{
    int t, n;
    string command;
    cin >> n;
    CXGraph xGraph(n);
    cin >> t;
    while (t--)
    {
        cin >> command;
        if (command == "show++")
        {
            cout << xGraph++ << endl;
        }
        else if(command == "++show")
        {
            cout << ++xGraph << endl;
        }
        else if (command == "show--")
        {
            cout << xGraph-- << endl;
        }
        else if (command == "--show")
        {
            cout << --xGraph << endl;
        }
        else if (command == "show")
        {
            cout << xGraph << endl;
        }
    }
    return 0;
}
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#include<iostream>
#include<cstring>
using namespace std;

class CXGraph
{
private:
	int n;
public:
	CXGraph(int _n):n(_n){}
	friend CXGraph operator ++(CXGraph& a)
	{
		if(a.n!=21)
		{
			a.n+=2;
		}
		return a;
	}
	friend CXGraph operator ++(CXGraph& a, int)
	{
		CXGraph b=a;
		if (a.n != 21)
		{
			a.n+=2;
		}
		return b;
	}
	friend CXGraph operator --(CXGraph& a)
	{
		if (a.n != 1)
		{
			a.n-=2;	
		}
		return a;
	}
	friend CXGraph operator --(CXGraph& a, int)
	{
		CXGraph b = a;
		if (a.n != 1)
		{
			a.n-=2;
		}
		return b;
	}
	friend ostream& operator<<(ostream& out, CXGraph const& a)
	{
		int i, j, k;
		k = a.n;
		for(i=0;i<(a.n+1)/2;i++)
		{
			for (j = 0; j < (a.n - k) / 2; j++)
			{
				out << " ";
			}
			for (j = k; j > 0; j--)
			{
				out << "X";
			}
			out << endl;
			k -= 2;
		}
		k = 3;
		for (i = (a.n + 1) / 2; i < a.n; i++)
		{
			for (j = (a.n - k) / 2; j > 0; j--)
			{
				out << " ";
			}
			for (j = 0; j < k; j++)
			{
				out << "X";
			}
			out << endl;
			k += 2;
		}
		return out;
	}
};

int main()
{
	int t, n;
	string command;
	cin >> n;
	CXGraph xGraph(n);
	cin >> t;
	while (t--)
	{
		cin >> command;
		if (command == "show++")
		{
			cout << xGraph++ << endl;
		}
		else if (command == "++show")
		{
			cout << ++xGraph << endl;
		}
		else if (command == "show--")
		{
			cout << xGraph-- << endl;
		}
		else if (command == "--show")
		{
			cout << --xGraph << endl;
		}
		else if (command == "show")
		{
			cout << xGraph << endl;
		}
	}
	return 0;
}

E.矩形关系（运算符重载）

题目描述

假设坐标采用二维平面坐标。

定义点类CPoint，包含属性x,y（整型）。方法有：带参构造函数，getX，getY分别返回点的x坐标，y坐标。
定义矩形类CRectangle，包含属性：矩形的左上角坐标leftPoint，右下角坐标rightPoint。类中方法有：
1）带参构造函数，初始化矩形的左上角、右下角
2）重载>运算符，参数为CPoint点对象，假设为p，若p在矩形内，返回true,否则返回false。
3）重载>运算符，第一个矩形若包含第二个矩形（部分边界可以相等），返回true，否则返回false。（要求该函数调用2）实现）
4）重载==运算符，判断两个矩形是否一致，返回true或false。
5）重载*运算符，判断两个矩形是否有重叠部分，返回true或false。
6）重载类型转换运算符，计算矩形的面积并返回，面积是整型。
7）重载<<运算符，输出矩形的两个角坐标，具体格式见样例。
输入2个矩形，计算面积，判断矩形的关系。主函数如下，不可修改。

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int main()
{
    int t, x1, x2, y1, y2;
    cin >> t;
    while (t--)
    {
        // 矩形1的左上角、右下角
        cin >> x1 >> y1 >> x2 >> y2;
        CRectangle rect1(x1, y1, x2, y2);
        // 矩形2的左上角、右下角
        cin >> x1 >> y1 >> x2 >> y2;
        CRectangle rect2(x1, y1, x2, y2);
        // 输出矩形1的坐标及面积
        cout << "矩形1:" << rect1 << " " << (int)rect1 << endl;
        // 输出矩形2的坐标及面积
        cout << "矩形2:" << rect2 << " " << (int)rect2 << endl;
        if (rect1 == rect2)
        {
            cout << "矩形1和矩形2相等" << endl;
        }
        else if (rect2 > rect1)
        {
            cout << "矩形2包含矩形1" << endl;
        }
        else if (rect1 > rect2)
        {
            cout << "矩形1包含矩形2" << endl;
        }
        else if (rect1 * rect2)
        {
            cout << "矩形1和矩形2相交" << endl;
        }
        else
        {
            cout << "矩形1和矩形2不相交" << endl;
        }
        cout << endl;
    }
    return 0;
}

可根据需要，添加构造函数和析构函数。

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#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;

class CPoint
{
private:
	int x;
	int y;
public:
	CPoint() {}
	CPoint(int _x, int _y) :x(_x), y(_y) {}
	int getX()
	{
		return x;
	}
	int getY()
	{
		return y;
	}
};

class CRectangele
{
private:
	CPoint leftPoint;
	CPoint rightPoint;
public:
	CRectangele() {}
	CRectangele(int x1, int y1, int x2, int y2) :leftPoint(x1, y1), rightPoint(x2, y2) {}
	friend  ostream& operator<<(ostream& out, CRectangele& a)
	{
		out << a.leftPoint.getX() << " " << a.leftPoint.getY() << " " << a.rightPoint.getX() << " " << a.rightPoint.getY();
		return out;
	}
	bool  operator>(CPoint& p)
	{
		if (p.getX() >= leftPoint.getX() && p.getX() <= rightPoint.getX() && p.getY() >= rightPoint.getY() && p.getY() <= leftPoint.getY())
			return true;
		else
			return false;
	}
	bool operator>(CRectangele& r)
	{
		if (r.leftPoint.getX() >= leftPoint.getX() && r.leftPoint.getX() <= rightPoint.getX() && r.leftPoint.getY() >= rightPoint.getY() && r.leftPoint.getY() <= leftPoint.getY() && r.rightPoint.getX() >= leftPoint.getX() && r.rightPoint.getX() <= rightPoint.getX() && r.rightPoint.getY() >= rightPoint.getY() && r.rightPoint.getY() <= leftPoint.getY())
		{
			return true;
		}
		else
		{
			return false;
		}
	}
	bool operator==(CRectangele& r)
	{
		if (leftPoint.getX() == r.leftPoint.getX() && leftPoint.getY() == r.leftPoint.getY() && rightPoint.getX() == r.rightPoint.getX() && rightPoint.getY() == r.rightPoint.getY())
		{
			return true;
		}
		else
		{
			return false;
		}
	}
	bool operator*(CRectangele& r)
	{
		if (abs(((leftPoint.getX() + rightPoint.getX()) / 2) - ((r.leftPoint.getX() + r.rightPoint.getX()) / 2)) <= (((rightPoint.getX() - leftPoint.getX()) + (r.rightPoint.getX() - r.leftPoint.getX())) / 2) &&
			abs(((leftPoint.getY() + rightPoint.getY()) / 2) - ((r.leftPoint.getY() + r.rightPoint.getY()) / 2)) <= (((leftPoint.getY() - rightPoint.getY()) + (r.leftPoint.getY() - r.rightPoint.getY())) / 2))
			return true;
		else
		    return false;

	}
	operator int()
	{
		int sum;
		sum = (rightPoint.getX() - leftPoint.getX()) * (leftPoint.getY() - rightPoint.getY());
		return sum;
	}
};



int main()
{
	int t, x1, x2, y1, y2;
	cin >> t;
	while (t--)
	{
		cin >> x1 >> y1 >> x2 >> y2;
		CRectangele rect1(x1, y1, x2, y2);
		cin >> x1 >> y1 >> x2 >> y2;
		CRectangele rect2(x1, y1, x2, y2);
		cout << "矩形1:" << rect1 << " " << (int)rect1 << endl;
		cout << "矩形2:" << rect2 << " " << (int)rect2 << endl;
		if (rect1 == rect2)
		{
			cout << "矩形1和矩形2相等" << endl;
		}
		else if (rect2 > rect1)
		{
			cout << "矩形2包含矩形1" << endl;
		}
		else if (rect1 > rect2)
		{
			cout << "矩形1包含矩形2" << endl;
		}
		else if (rect1 * rect2)
		{
			cout << "矩形1和矩形2相交" << endl;
		}
		else
		{
			cout << "矩形1和矩形2不相交" << endl;
		}
		cout << endl;
	}
	return 0;
}