c++实现重载分数的四则运算

[cpp] 
#include<iostream> 
using namespace std; 
 
class Point 
{ 
private: 
    int c_point; 
    int m_point; 
    char code; 
public: 
    Point() 
    { 
        code='/'; 
    } 
    Point(int n1,int n2) 
    { 
        c_point=n1; 
        m_point=n2; 
        code='/'; 
    } 
    void Output() 
    { 
        if(m_point==0)cout<<"error"; 
        else 
            if(m_point==1)cout<<c_point; 
            else 
                cout<<"("<<c_point<<code<<m_point<<")"; 
    } 
    int Big(int n1,int n2)          //最大公约数 
    { 
        int s=1; 
        int n=n1,m=n2; 
        while(s!=0) 
        { 
            s=n1%n2; 
            n1=n2;                
            n2=s; 
        } 
        if(n1!=1&&n%n1==0&&m%n1==0) 
            return n1; 
        else 
            return 0; 
    } 
    //int Small_Multiple() 
    Point operator +(Point b) 
    { 
        Point c; 
        int n1,n2,s=1; 
        n1=m_point; 
        n2=b.m_point; 
        if(n1==n2) 
        { 
            c.c_point=c_point+b.c_point; 
            c.m_point=m_point=b.m_point; 
        } 
        else 
        { 
            while(s!=0) 
            { 
                s=n1%n2; 
                n1=n2;               
                n2=s; 
            } 
            s=m_point*b.m_point/n1;         //最小公倍数 
            c.m_point=s; 
            c.c_point=c_point*(s/m_point)+b.c_point*(s/b.m_point); 
        } 
        int t; 
        t=Big(c.c_point,c.m_point); 
        while(t!=0) 
        { 
            c.c_point=c.c_point/t; 
            c.m_point=c.m_point/t; 
            t=Big(c.c_point,c.m_point); 
        } 
        return c; 
    } 
    Point operator -(Point b) 
    { 
        Point c; 
        int n1,n2,s=1; 
        n1=m_point; 
        n2=b.m_point; 
        if(n1==n2) 
        { 
            c.c_point=c_point-b.c_point; 
            c.m_point=m_point=b.m_point; 
        } 
        else 
        { 
            while(s!=0) 
            { 
                s=n1%n2; 
                n1=n2;               //最大公约数 
                n2=s; 
            } 
            s=m_point*b.m_point/n1;         //最小公倍数 
            c.m_point=s; 
            c.c_point=c_point*(s/m_point)-b.c_point*(s/b.m_point); 
        } 
        int t; 
        t=Big(c.c_point,c.m_point); 
        while(t!=0) 
        { 
            c.c_point=c.c_point/t; 
            c.m_point=c.m_point/t; 
            t=Big(c.c_point

补充：软件开发 , C++ ,