A.分数的加减乘除（运算符重载）

题目描述

Fraction类的基本形式如下：

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// 定义Fraction类
class Fraction
{
private:
    int fz, fm;
    int commonDivisor(); // 计算最大公约数
    void contracted(); // 分数化简
public:
    Fraction(int = 0, int = 1);
    Fraction(Fraction&);
    Fraction operator+(Fraction);
    Fraction operator-(Fraction);
    Fraction operator*(Fraction);
    Fraction operator/(Fraction);
    void set(int = 0, int = 1);
    void display();
};

要求如下：

1.实现Fraction类；common_divisor()和contracted()函数体可为空，不实现具体功能。
2.编写main函数，初始化两个Fraction对象的，计算它们之间的加减乘除。

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

class Fraction
{
private:
    int fz, fm;
    int commonDivisor() {};
    void contracted() {};
public:
    Fraction(int a = 0, int b = 1)
    {
        fz = a;
        fm = b;
    }
    Fraction(Fraction& a)
    {
        fz = a.fz;
        fm = a.fm;
    }
    Fraction operator+(Fraction& a)
    {
        Fraction A;
        A.fz = fz * a.fm + a.fz * fm;
        A.fm = fm * a.fm;
        return A;
    }
    Fraction operator-(Fraction& a)
    {
        Fraction A;
        A.fz = fz * a.fm - a.fz * fm;
        A.fm = fm * a.fm;
        return A;
    }
    Fraction operator*(Fraction& a)
    {
        Fraction A;
        A.fz = fz * a.fz;
        A.fm = fm * a.fm;
        return A;
    }
    Fraction operator/(Fraction& a)
    {
        Fraction A;
        A.fz = fz * a.fm;
        A.fm = fm * a.fz;
        return A;
    }
    void set(int a = 0, int b = 1)
    {
        fz = a;
        fm = b;
    }
    void display()
    {
        cout << "fraction=" << fz << "/" << fm << endl;
    }
};

int main()
{
    int a, b;
    cin >> a >> b;
    Fraction x(a, b);
    cin >> a >> b;
    Fraction y(a, b);
    Fraction z;
    z = x + y;
    z.display();
    z = x - y;
    z.display();
    z = x * y;
    z.display();
    z = x / y;
    z.display();
        

    return 0;
}

B.复数的加减乘运算（运算符重载）

题目描述

定义一个复数类，通过重载运算符：+、-、*，实现两个复数之间的各种运算。

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class Complex
{
private:
    float real, image;
public:
    Complex(float x = 0, float y = 0);
    friend Complex operator+(Complex&, Complex&);
    friend Complex operator-(Complex&, Complex&);
    friend Complex operator*(Complex&, Complex&);
    void show();
};

要求如下：

1.实现Complex类；
2.编写main函数，初始化两个Complex对象，计算它们之间的加减乘，并输出结果。
复数相乘的运算规则
设z1=a+bi，z2=c+di(a、b、c、d∈R)是任意两个复数，那么它们的积(a+bi)(c+di)=(ac-bd)+(bc+ad)i.

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#include<iostream>
using namespace std;

class Complex
{
private:
	float real, image;
public:
	Complex(){}
	Complex(float x , float y)
	{
		real = x;
		image = y;
	}
	friend Complex operator+(Complex& a, Complex& b)
	{
		Complex c;
		c.real = a.real + b.real;
		c.image = a.image + b.image;
		return c;
	}
	friend Complex operator-(Complex& a, Complex& b)
	{
		Complex c;
		c.real = a.real - b.real;
		c.image = a.image - b.image;
		return c;
	}
	friend Complex operator*(Complex& a, Complex& b)
	{
		Complex c;
		c.real = a.real * b.real - a.image * b.image;
		c.image = a.image * b.real + a.real * b.image;
		return c;
	}
	void show()
	{
		cout << "Real=" << real << " Image=" << image << endl;
	}
};

int main()
{
	int a, b, c, d;
	cin >> a >> b >> c >> d;
	Complex c1(a, b);
	Complex c2(c, d);
	Complex c3;
	c3 = c1 + c2;
	c3.show();
	c3 = c1 - c2;
	c3.show();
	c3 = c1 * c2;
	c3.show();

	return 0;
}

C.向量的加减（运算符重载）

题目描述

设向量X=(x1,x2,…,xn)和Y=(y1,y2…,yn),它们之间的加、减分别定义为：

X+Y=(x1+y1,x2+y2,…,xn+yn)
  X-Y=(x1-y1,x2-y2,…,xn-yn)

编程序定义向量类Vector ,重载运算符“+”、“-”,实现向量之间的加、减运算;并编写print函数作为向量的输出操作。

Vector类的基本形式如下：

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class Vector
{
private:
    int vec[5];
public:
    Vector(int v[]);
    Vector();
    Vector(const Vector& obj);
    Vector operator +(const Vector& obj);
    Vector operator -(const Vector& obj);
    void print();
};

要求如下：
1.实现Vector类；
2.编写main函数，初始化两个Vector对象的，计算它们之间的加减，并输出结果。

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#include<iostream>
using namespace std;

class Vector
{
private:
	int vec[5];
public:
	Vector(int v[5])
	{
		for (int i = 0; i < 5; i++)
		{
			vec[i] = v[i];
		}
	}
	Vector(){}
	Vector(const Vector& obj)
	{
		for (int i = 0; i < 5; i++)
		{
			vec[i] = obj.vec[i];
		}
	}
	Vector operator+(const Vector& obj)
	{
		Vector tmp;
		for (int i = 0; i < 5; i++)
		{
			tmp.vec[i] = vec[i] + obj.vec[i];
		}
		return tmp;
	}
	Vector operator-(const Vector& obj)
	{
		Vector tmp;
		for (int i = 0; i < 5; i++)
		{
			tmp.vec[i] = vec[i] - obj.vec[i];
		}
		return tmp;
	}
	void print()
	{
		for (int i = 0; i < 5; i++)
		{
			cout << vec[i] << " ";
		}
		cout << endl;
	}
	
};

int main()
{
	int x[5];
	for (int i = 0; i < 5; i++)
	{
		cin >> x[i];
	}
	Vector a(x);
	for (int i = 0; i < 5; i++)
	{
		cin >> x[i];
	}
	Vector b(x);
	Vector c;
	c = a + b;
	c.print();
	c = a - b;
	c.print();

	return 0;

}

D.矩阵相乘（运算符重载）

题目描述

定义一个矩阵类MyMatrix，并且在类中进行运算符重定义，用*实现矩阵相乘。要求必须对运算符进行重载，如果用函数如multiply（matrix，matrix）去实现矩阵之间的运算一律记0分。

必须包含以下析构函数，其中n是矩阵的阶数，data是存放矩阵数据的二维动态数组：

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MyMatrix::~MyMatrix() 
{ // 释放空间 
	for (int i = 0; i < n; i++) 
	{ 
		delete[] data[i]; 
	} 
	delete[] data; 
}
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#include<iostream>
using namespace std;
class MyMatrix
{
private:
	int a[10][10];
	int n;
public:
	MyMatrix(int _n):n(_n){}
	friend MyMatrix operator*(MyMatrix& m1, MyMatrix& m2)
	{
		int n = m1.n;
		MyMatrix x(n);
		int sum;
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)
			{
				sum = 0;
				for (int k = 0; k < n; k++)
				{
					sum += m1.a[i][k] * m2.a[k][j];
				}
				x.a[i][j] = sum;
			}
		}
		return x;
	}
	void set()
	{
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)
			{
				cin >> a[i][j];
			}
		}
	}
	void show()
	{
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)
			{
				if (j != n - 1)
				{
					cout << a[i][j] << " ";
				}
				else
				{
					cout << a[i][j] << endl;
				}
			}
			
		}
	}
};

int main()
{
	int c, n;
	cin >> c >> n;
	MyMatrix m1(n);
	m1.set();
	for (int i = 0; i < c - 1; i++)
	{
		MyMatrix m2(n);
		m2.set();
		m1 = m1 * m2;
	}
	m1.show();

	return 0;
}

E.最小生日差值计算（运算符重载）

题目描述

定义一个学生类Student，包含该学生的姓名、出生年、月、日 ，重定义 “-”号实现两个学生之间相差多少天的比较。并利用重载的“-”运算符，求所有学生中年龄相差最小的两个人的名字以及相差天数。

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


class Student
{
private:
	string name;
	int year;
	int month;
	int day;
public:
	Student(){}
	Student(string n,int y,int m,int d):name(n),year(y),month(m),day(d){}
	void set(string n, int y, int m, int d)
	{
		year = y;
		month = m;
		day = d;
		name = n;
	}
	string getname()
	{
		return name;
	}
	int operator-(Student& s)
	{
		int tmp = 0;
		int emp = 0;
		int m[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
		int n[12] = { 31,29,31,30,31,30,31,31,30,31,30,31 };
		if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0))
		{
			for (int i = 0; i < month - 1; i++)
			{
				tmp += n[i];
			}
		}
		else
		{
			for (int i = 0; i < month - 1; i++)
			{
				tmp += m[i];
			}
		}
		if ((s.year % 4 == 0 && s.year % 100 != 0) || (s.year % 400 == 0))
		{
			for (int i = 0; i < s.month - 1; i++)
			{
				emp += n[i];
			}
		}
		else
		{
			for (int i = 0; i < s.month - 1; i++)
			{
				emp += m[i];
			}
		}
		tmp += day;
		emp += s.day;
		if (year == s.year)
		{
			if (tmp - emp < 0)
			{
				return emp - tmp;
			}
			else
			{
				return tmp - emp;
			}
		}
		else if (year > s.year)
		{
			int max = year;
			int min = s.year;
			while (max - min)
			{
				if ((min % 4 == 0 && min % 100 != 0) || (min % 400 == 0))
				{
					tmp += 366;
				}
				else
				{
					tmp += 365;
				}
				min++;
			}
			return tmp - emp;
		}
		else
		{
			int min = year;
			int max = s.year;
			while (max - min)
			{
				if ((min % 4 == 0 && min % 100 != 0) || (min % 400 == 0))
				{
					emp += 366;
				}
				else
				{
					emp += 365;
				}
				min++;
			}
			return emp - tmp;
		}
	}
};

int main()
{
	int t, d;
	int min = 10000;
	int s1 = 0;
	int s2 = 0;
	int year, month, day;
	string name;
	Student* s;
	cin >> t;
	s = new Student[t];
	for (int i = 0; i < t; i++)
	{
		cin >> name >> year >> month >> day;
		s[i].set(name, year, month, day);
	}
	for (int i = 0; i < t - 1; i ++ )
	{
		for (int j = 1; j < t; j++)
		{
			d = s[j] - s[i];
			if (d < min && s[i].getname() != s[j].getname())
			{
				min = d;
				s1 = i;
				s2 = j;
			}
		}
	}
	cout << s[s1].getname() << "和" << s[s2].getname() << "年龄相差最小，为" << min << "天。" << endl;
	
	return 0;

}