#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;
}
第九周项目五 (不要轻信贺老师在BB上贴出的代码……)
最新推荐文章于 2024-10-31 12:20:24 发布