第九周项目五 （不要轻信贺老师在BB上贴出的代码……）

#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0
    Douary(const Douary &d); //因为有指针数据成员，复制构造函数需要定义
    ~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
    friend ostream &operator<<(ostream &output, Douary &d);//重载运算符“<<”以m行n列形式输出
    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，构造函数中分配空间用Array = new int[row*col];
    int row; //row 为二维数组的行数。
    int col; //col 为二维数组的列数。
};
int main()
{
    Douary d1(2,3),d2(2,3),d3(0,0);
    cout<<"输入d1:"<<endl; cin>>d1;
    cout<<"输入d2:"<<endl; cin>>d2;
    cout<<"d1="<<endl; cout<<d1;
    cout<<"d2="<<endl; cout<<d2;
    cout<<"d1+d2="<<endl;
    d3=(d1+d2);
    cout<<d3;
    cout<<"d1-d2="<<endl;
    d3=d1-d2;
    cout<<d3;
    cout<<"d1"<<((d1==d2)?"==":"!=")<<"d2"<<endl;
    return 0;
}
Douary::Douary(int m, int n)
{
    row=m;
    col=n;
    Array = new int[m*n];
        for(int i=0; i<row; ++i)
       for(int j=0; j<col; ++j)
          Array[i*col+j]=0;

}
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)
{
    cout<<"该矩阵行为"<<d.row<<"列为"<<d.col<<endl;
    int i=0,j=0;
    for(i;i<d.row;i++)
    {
        for(j;j<d.col;j++)
        {
            cin>>d.Array[i*d.col+j];
        }
    }
    return input;
}
 ostream &operator<<(ostream &output, Douary &d)
{
    cout<<"该矩阵行为"<<d.row<<"列为"<<d.col<<endl;
    int a,b;
    for(a=0;a<d.row;a++)
        {
            for(b=0;b<d.col;b++)
            {
                cout<<d.Array[a*d.col+b]<<"    ";
            }cout<<endl;
        }
        return output;
}
Douary::Douary(const Douary &d)
{
    row=d.row;
    col=d.col;
    Array = new int[row*col];
    for(int i=0; i<row; ++i)
        for(int j=0; j<col; ++j)
            Array[i*col+j]=d.Array[i*col+j];

}
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
{
    //在此可以先判断d1和d2的行列是否相同，如果不相同可以报错退出，不做运算。本参考解答忽略了这一前提
   Douary d(d1.row,d1.col);
    for(int i=0; i<d1.row; ++i)
   {
       for(int j=0; j<d1.col; ++j)
            d.Array[i*d1.col+j]=d1.Array[i*d1.col+j]+d2.Array[i*d1.col+j];
    }
  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
{
    //在此可以先判断d1和d2的行列是否相同，如果不相同可以报错退出，不做运算。本参考解答忽略了这一前提
    Douary d(d1.row,d1.col);
    for(int i=0; i<d1.row; ++i)
    {
        for(int j=0; j<d1.col; ++j)
            d.Array[i*d1.col+j]=d1.Array[i*d1.col+j]-d2.Array[i*d1.col+j];
    }
    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; ++i)
    {
        for(int j=0; j<col; ++j)
           if (Array[i*col+j]!=d.Array[i*col+j])
        {
                eq=false;
            break;
            }
            if (!eq) break;
}
   return eq;
}