第九周任务（5）

#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(const Douary &d);//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;  
    Array = new int[m*n]; //Array只能指向一维数组，将m*n个元素的一维数组当作m行n列的数组看待了  
    for(int i=0; i<row*col; ++i)
		Array[i]=0;    
}  
  
Douary::Douary(const Douary &d)  
{  
    row=d.row;  
    col=d.col;  
    Array = new int[row*col];  
    for(int i=0; i<row*col; ++i)    
            Array[i]=d.Array[i];   
}  
  
Douary::~Douary() //析构函数：用于释放动态数组所占用的存储空间  
{  
    delete [] Array;  
}  
  
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  
{  
	for(int i=0; i<d.row*d.col; ++i)    
            cin>>d.Array[i];
    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  
{  
	int a=1;
    for(int i=0; i<d.row*d.col; ++i)  
	{
        cout<<d.Array[i]<<" ";
		if(a%d.col==0)
		{
			cout<<endl;
		}
		a=a+1;
    }  
    cout<<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  
{  
	/*while(1)
	{
      if(d1.row!=d2.row||d1.col!=d2.col) 
		 break;
	}*/
    Douary d(d1.row,d1.col);  
    for(int i=0; i<d1.row*d1.col; ++i)  
    {    
       d.Array[i]=d1.Array[i]+d2.Array[i];  
    }  
    return d;  
}  
  
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  
{  
	/*while(1)
	{
	if(d1.row!=d2.row||d1.col!=d2.col) 
	{
		break;
	}
	}*/
    Douary d(d1.row,d1.col);  
    for(int i=0; i<d1.row*d1.col; ++i)  
    { 
         d.Array[i]=d1.Array[i]-d2.Array[i];
    }
    return d;  
}  
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  
{  
    if(row!=d.row||col!=d.col) 
        return false;  
    bool eq = true;  
    for(int i=0; i<row*col; ++i)  
    {  
            if (Array[i]!=d.Array[i])   
            {  
                eq=false;  
                break;  
            }  
            if (!eq) break;  
    }  
	
    return eq;  
}  
  
int main()  
{  
    Douary d1(2,3),d2(2,3);  
    cout<<"输入d1(2,3):"<<endl;  
    cin>>d1;  
    cout<<"输入d2(2,3):"<<endl;  
    cin>>d2;  
    cout<<"d1="<<endl;  
    cout<<d1;  
    cout<<"d2="<<endl;  
    cout<<d2;  
    cout<<"d1+d2="<<endl;  
    Douary d3=(d1+d2);  
    cout<<d3;  

    cout<<"d1-d2="<<endl;  
    cout<<(d1-d2);  
    cout<<"d1"<<((d1==d2)?"==":"!=")<<"d2"<<endl;  
    system("pause");  
    return 0;  
}