第九周实验报告3.0

/* (程序头部注释开始)   
* 程序的版权和版本声明部分   
* Copyright (c) 2011, 烟台大学计算机学院学生    
* All rights reserved.   
* 文件名称：022                              
* 作    者：任小宁                          
* 完成日期：  2012   年   4   月    16 日   
* 版 本 号：201158504431          
* 对任务及求解方法的描述部分   
* 输入描述：

#include<iostream.h>   #include"stdlib.h"      int gcd(int m, int n);      class CFraction  {  private:      int nume;  // 分子      int deno;  // 分母  public:      //构造函数及运算符重载的函数声明      CFraction(int nu=0,int de=1);   //构造函数，初始化用        void Simplify();                    //化简（使分子分母没有公因子）          void output();           //输出：以8/6为例，style为0时，输出8/6；        bool operator > (CFraction &t);        bool operator < (CFraction &t);        bool operator >= (CFraction &t);        bool operator <= (CFraction &t);        bool operator == (CFraction &t);        bool operator != (CFraction &t);        CFraction operator+(CFraction &c);         CFraction operator-(CFraction &c);      CFraction operator*(CFraction &c);        CFraction operator/(CFraction &c);      CFraction operator-();  friend ostream& operator << (ostream&,CFraction &);friend istream& operator >> (istream&,CFraction &);      };  istream& operator >> (istream& input,CFraction & c){ input>>c.nume>>c.deno;return input;}ostream& operator << (ostream& output,CFraction & c){output<<c.nume<<'/'<<c.deno<<endl; return output;}CFraction::CFraction(int nu,int de)   //构造函数，初始化用     {         if (de!=0)           {               nume=nu;               deno=de;           }           else           {               cerr<<"初始化中发生错误，程序退出\n";                   exit(0);           }    }  void CFraction::Simplify()                    //化简（使分子分母没有公因子）   {       int n;      if(nume < 0)        {            n = gcd(-nume, deno);        }        else        {            n = gcd(nume, deno);        }            nume = nume / n;              deno = deno / n;  }  // 求m，n的最大公约数    int gcd(int m, int n)    {        int r;        if (m<n){r=m;m=n;n=r;}        while(r=m%n)  // 求m，n的最大公约数        {            m=n;            n=r;        }        return n;    }    bool CFraction::operator > (CFraction &t)  {      CFraction c2,c3;      c2.nume =nume*t.deno ;      c3.nume =t.nume *deno;      if(c2.nume >c3.nume )          return true;      else          return false;  }  bool CFraction::operator < (CFraction &t)   {      CFraction c2,c3;      c2.nume =nume*t.deno ;      c3.nume =t.nume *deno;      if(c2.nume <c3.nume )          return true;      else          return false;  }  bool CFraction::operator >= (CFraction &t)  {        CFraction c1;        c1.nume =nume;      c1.deno =deno;      if (c1<t)            return false;        return true;    }      bool CFraction::operator <= (CFraction &t)  {        CFraction c1;        c1.nume =nume;      c1.deno =deno;      if (c1>t)            return false;        return true;    }    bool CFraction::operator == (CFraction &t)   {      CFraction c1;        c1.nume =nume;      c1.deno =deno;      if (c1<t)            return false;        if (c1>t)            return false;        return false;  }    bool CFraction::operator != (CFraction &t)  {      CFraction c1;        c1.nume =nume;      c1.deno =deno;      if (c1==t)            return false;      return true;  }  CFraction CFraction::operator+(CFraction &c)  {      CFraction c2,c3,c4;      c2.nume =nume*c.deno ;      c3.nume =c.nume *deno;      c2.deno =deno*c.deno ;      c3.deno =c.deno *deno;      c4.nume=c2.nume +c3.nume ;      c4.deno =c2.deno ;      c4.Simplify ();      return c4;  }  CFraction CFraction::operator-(CFraction &c)  {      CFraction c2,c3,c4;      c2.nume =nume*c.deno ;      c3.nume =c.nume *deno;      c2.deno =deno*c.deno ;      c3.deno =c.deno *deno;      c4.nume=c2.nume -c3.nume ;      c4.deno =c2.deno ;      c4.Simplify ();      return c4;  }  CFraction CFraction::operator*(CFraction &c)   {      CFraction c2,c3,c4;      c2.nume =nume*c.nume  ;      c2.deno =deno*c.deno ;      c2.Simplify ();      return c2;  }  CFraction CFraction::operator/(CFraction &c)  {      CFraction c2,c3;      c2.nume =c.deno ;      c2.deno =c.nume ;      c3.nume =nume*c2.nume ;      c3.deno =deno*c2.deno ;      c3.Simplify ();      return c3;  }  CFraction CFraction::operator-()  {      CFraction c2;      c2.nume =nume;      c2.deno =deno;      c2.Simplify ();      if(c2.nume<0 || c2.deno<0)      {          if(c2.nume <0)          {              c2.nume =-nume;          }          else          {              c2.deno =-deno;          }      }      else      {          c2.nume =-nume;          c2.deno =deno;      }      return c2;  }    //用于测试的main()函数  void main()    {        CFraction c1,c2,c;cout<<"请您输入一个分数c1：（以a b的形式输入）";cin>>c1;cout<<"请您输入一个分数c1：（以a b的形式输入）";cin>>c2;    cout<<"c1为：";        cout<<c1;        cout<<"c2为：";        cout<<c2;        cout<<"下面比较两个时间大小：\n";        if (c1>c2) cout<<"c1>c2"<<endl;        if (c1<c2) cout<<"c1<c2"<<endl;        if (c1==c2) cout<<"c1＝c2"<<endl;         if (c1!=c2) cout<<"c1≠c2"<<endl;        if (c1>=c2) cout<<"c1≥c2"<<endl;        if (c1<=c2) cout<<"c1≤c2"<<endl;        cout<<endl;        cout<<"c1+c2的数值为：";        c=c1+c2;        cout<<c;      cout<<endl;      cout<<"c1-c2的数值为：";        c=c1-c2;        cout<<c;        cout<<endl;        cout<<"c1*c2的数值为：";        c=c1*c2;        cout<<c;      cout<<endl;      cout<<"c1/c2的数值为：";        c=c1/c2;        cout<<c;       cout<<endl;      cout<<"对c1取反的结果为：";      c=-c1;      cout<<c;      cout<<endl;     }      

 
* 问题描述：  

 

 

前三个题都是让定义《和》的运算重载，从而来实现输入，输出，进而改变程序中对运算结果的现实方式，是程序读起来更加自然。总的来说达到了预期的结果，但是我总是觉的单凭麻烦程度上两种做法差不多的啊？？？？？？