C++面向对象编程解决三阶矩阵相加减

/*此处用面向对象编程*/

#include<iostream>
#include<string.h>
using namespace std;
class Matrices
{
private:
    int mat[3][3];
public:
    Matrices();
    void input()
    {
        for(int i=0; i<3; i++)
        {
            for(int j=0; j<3; j++)
            {
                cin>>mat[i][j];
            }
        }
    }
    friend Matrices operator+(Matrices &,Matrices &);
    friend Matrices operator-(Matrices &a,Matrices &b);
    friend ostream &operator <<(ostream &output,Matrices &);
    friend istream &operator >>(istream &input,Matrices &);
};

Matrices ::Matrices()
{
    memset(mat,0,sizeof(mat));
}

istream &operator >>(istream &input,Matrices &a)
{
    for(int i=0; i<3; i++)
    {
        for(int j=0; j<3; j++)
        {
            input>>a.mat[i][j];
        }
    }
    return input;
}

ostream &operator <<(ostream &output,Matrices &a)
{
    for(int i=0; i<3; i++)
    {
        for(int j=0; j<3; j++)
        {
            output<<a.mat[i][j]<<"   ";
        }
        cout<<endl;
    }
    return output;
}

Matrices operator+(Matrices &a,Matrices &b)
{
    Matrices c;
    for(int i=0; i<3; i++)
    {
        for(int j=0; j<3; j++)
        {
            c.mat[i][j]=a.mat[i][j]+b.mat[i][j];
        }
    }
    return c;
}

Matrices operator-(Matrices &a,Matrices &b)
{
    Matrices c;
    for(int i=0; i<3; i++)
    {
        for(int j=0; j<3; j++)
        {
            c.mat[i][j]=a.mat[i][j]-b.mat[i][j];
        }
    }
    return c;
}

int main()
{
    cout<<"请输入需要进行的三阶加法运算(+)或减法运算(-)"<<endl;
    char c;
    Matrices A,B,C;
    while(cin>>c)
    {
        if(c=='+'||c=='-')
        {
            if(c=='+')
            {
                cout<<"转换成功!即将进行加法运算"<<endl;
                cout<<"请依次输入矩阵1,以回车换行(确认)"<<endl;
                cin>>A;
                cout<<"请依次输入矩阵2,以回车换行(确认)"<<endl;
                cin>>B;
                C=A+B;
                cout<<"运算成功!两矩阵的和为:"<<endl;
            }
            else
            {
                cout<<"转换成功!即将进行减法运算"<<endl;
                cout<<"请依次输入矩阵1,以回车换行(确认)"<<endl;

                cin>>A;
                cout<<"请依次输入矩阵2,以回车换行(确认)"<<endl;
                cin>>B;
                C=A-B;

                cout<<"运算成功!两矩阵的差为:"<<endl;
            }
            cout<<endl;
            cout<<C;
            cout<<"可输入'+'或'-'继续进行运算"<<endl;
        }
        else cout<<"数据格式错误!请重新输入:"<<endl;
    }
}


/*

测试数据

1 1 1
2 2 2
3 3 3

1 1 1
2 2 2
3 3 3
*/