第九周任务四

﹏ヽ暗。殇╰゛Y 2022-06-13 12:40 191阅读 0赞

/* 
    实验内容:建立一个二维数组类Douary，使该类中有以下数据成员、成员函数及友员函数，完成矩阵的输入、输出、加、减、相等判断等操作。
    * 程序的版权和版本声明部分  
    * Copyright (c) 2011, 烟台大学计算机学院学生  
    * All rights reserved.  
    * 文件名称：矩阵的输入、输出、加、减、相等判断                            
    * 作    者：  薛广晨                             
    * 完成日期：  2012       年  4    月  15      日  
    * 版 本号：  x1.0           
    */
    //【任务4】建立一个二维数组类Douary，使该类中有以下数据成员、成员函数及友员函数，完成矩阵的输入、输出、加、减、相等判断等操作。
    #include <iostream>
    
    using namespace std;
    
    class Douary
    {
    public:
    	Douary(int m, int n);//构造函数：用于建立动态数组存放m行n列的二维数组（矩阵）元素，并将该数组元素初始化为
    	~Douary(); //析构函数：用于释放动态数组所占用的存储空间。
    	friend istream &operator>>(istream &input, Douary &d);//重载运算符“>>”输入二维数组，其中d为Dousry类对象；
    	friend ostream &operator<<(ostream &output, Douary &d);//重载运算符“<<”以m行n列矩阵的形式输出二维数组，其中d为Douary类对象。
    	friend Douary &operator+(const Douary &d1,const Douary &d2);//两个矩阵相加，规则：对应位置上的元素相加
    	friend Douary &operator-(const Douary &d1,const Douary &d2);//两个矩阵相减，规则：对应位置上的元素相减
    	bool operator==(const Douary &d);//判断两个矩阵是否相等，即对应位置上的所有元素是否相等
    private:
    	int ** Array;      //Array 为动态数组指针。
    	int row;          //row  为二维数组的行数。
    	int col;          //col   为二维数组的列数。
    };
    
    Douary :: Douary(int m, int n) //构造函数：用于建立动态数组存放m行n列的二维数组（矩阵）元素，并将该数组元素初始化为；
    {
        row = m;
        col = n;
        if(m != 0 && n!= 0)
        {
            Array = new int *[row];
            for (int i = 0; i < row; i++)
            {
                Array[i] = new int[col];
            }
        }
        else
            Array = NULL;
    }
    
    Douary :: ~Douary()
    {
        if (Array != NULL)
        {
            for (int i = 0; i < row; i++)
            {
                delete Array[i];
            }
            delete[] Array;
            Array = NULL;
        }
    }
    
    istream &operator>>(istream &input, Douary &d)//重载运算符“>>”输入二维数组，其中d为Douary类对象；
    {
        //input >> d.row >> d.col;
        if(d.row != 0 && d.col != 0)
        {
            int i,j;
            d.Array = new int *[d.row];
            for (i = 0; i < d.row; i++)
            {
                d.Array[i] = new int[d.col];
            }
            for (i = 0; i < d.row; ++i)
                for(j = 0; j < d.col; ++j)
                {
                    input >> d.Array[i][j];
                }
                
        }
        else
            d.Array = NULL;
        return input;
    }
    
    ostream &operator<<(ostream &output, Douary &d)//重载运算符“<<”以m行n列矩阵的形式输出二维数组，其中d为Douary类对象。
    {
        output << d.row << '/' << d.col << endl; 
        if (d.row != 0 && d.col != 0)
        {
            int i,j;
            for(i = 0; i < d.row; ++i)
            {
                for (j = 0; j < d.col; ++j)
                    output << d.Array[i][j] << " ";
                output << endl;
            }
        }
        return output;
    }
    
    Douary &operator +(const Douary &d1, const Douary &d2)//两个矩阵相加，规则：对应位置上的元素相加
    {
        Douary *p = new Douary(0,0);
        if (d1.row == d2.row && d1.col == d2.col)
        {
            int i, j;
            p -> row = d2.row;
            p -> col = d2.col;
            p -> Array = new int *[d1.row];
            for (i = 0; i < d1.row; i++)
            {
                p -> Array[i] = new int[d1.col];
            }
            for (i = 0; i < d2.row; ++i)
                for(j = 0; j < d2.col; ++j)
                {
                    p -> Array[i][j] = d1.Array[i][j] + d2.Array[i][j];
                }
        }
        else
        {
            cout << "不能构成矩阵!" << endl;
        }
        return *p;
    }
    
    Douary &operator-(const Douary &d1, const Douary &d2)//两个矩阵相减，规则：对应位置上的元素相减
    {
        Douary *p = new Douary(0, 0);
        if (d1.row == d2.row && d1.col == d2.col)
        {
            int i, j;
            p -> row = d2.row;
            p -> col = d2.col;
            p -> Array = new int *[d1.row];
            for (i = 0; i < d1.row; i++)
            {
                p -> Array[i] = new int[d1.col];
            }
            for (i = 0; i < d2.row; ++i)
                for(j = 0; j < d2.col; ++j)
                {
                    p -> Array[i][j] = d1.Array[i][j] - d2.Array[i][j];
                }
        }
        else
        {
            cout << "不能构成矩阵!" << endl;
        }
        return *p;
    }
    
    bool Douary::operator==(const Douary &d)//判断两个矩阵是否相等，即对应位置上的所有元素是否相等
    {
        bool b = false;
        if (row == d.row && col == d.col)
        {
            int i,j;
            for(i = 0; i < row;++i)
                for(j = 0; j < col; ++j)
                    if(d.Array[i][j] != Array[i][j])
                        return false;
            b = true;
        }
        return b;
    }
    
    int main()
    {
    	Douary d1(2, 3), d2(2, 3);
    
    	cout << "输入d1:" << endl;
    	cin >> d1;
    	cout << "输入d2:"<<endl;
    	cin >> d2;
    
    	cout << "d1=" << endl << d1;
    	cout << "d2=" << endl << d2;
    	cout << "d1+d2=" << endl << (d1 + d2);
    	cout << "d1-d2=" << endl << (d1 - d2);
    	cout << "d1" << ((d1 == d2) ? "==" : "!=") << "d2" << endl;
    
    	system("pause");
    	return 0;
    }

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上机感言：这个题有难度，有挑战性，就是数组的地方比较难

[1334492970_5433.jpg]: https://img-my.csdn.net/uploads/201204/15/1334492970_5433.jpg