第九周实验报告四

#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];       for(int i=0; i<row; ++i)          for(int j=0; j<col; ++j)              Array[i*col+j]=0;    }    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::~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  {      for(int i=0; i<d.row; ++i)          for(int j=0; j<d.col; ++j)              cin>>d.Array[i*d.col+j];//注意地址求法      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 {      for(int i=0; i<d.row; ++i)      {          for(int j=0; j<d.col; ++j)              cout<<d.Array[i*d.col+j]<<"\t";           cout<<endl;      }      cout<<endl;      return output;  }    Douary operator+(const Douary &d1,const Douary &d2)//两个矩阵相加  {      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  {      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  {      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;  }    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;  }   

结果：

输入d1(2,3):
1 2 3 4 5 6
输入d2(2,3):
7 8 9 4 5 6
d1=
1       2       3
4       5       6

d2=
7       8       9
4       5       6

d1+d2=
8       10      12
8       10      12

d1-d2=
-6      -6      -6
0       0       0

d1!=d2
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