第八周实验报告任务3 实现分数类中的运算符重载，在分数类中可以完成分数的加减乘除，求反，比较6种关系的运算。

/* (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：实现分数类中的运算符重载，在分数类中可以完成分数的加减乘除（运算后再化简）、求反、比较（6种关系）的运算

* 作    者：         陶锋                       

* 完成日期：     2012    年 04  月    09 日

* 版 本 号：       V1.0   

* 对任务及求解方法的描述部分

* 输入描述： 

* 问题描述： 

* 程序输出：

 * 程序头部的注释结束

*/

 

#include<iostream>  using namespace std;  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-();};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";     system("pause");     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;  }  void CFraction::output()          //输出：以8/6为例，style为0时，输出8/6； {cout<<nume<<'/'<<deno<<endl;}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;elsereturn 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;elsereturn 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(8,20),c2(11,20),c;        cout<<"c1为：";      c1.output();      cout<<"c2为：";      c2.output();      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;      c.output();cout<<endl;    cout<<"c1-c2的数值为：";      c=c1-c2;      c.output();      cout<<endl;      cout<<"c1*c2的数值为：";      c=c1*c2;      c.output();cout<<endl;    cout<<"c1/c2的数值为：";      c=c1/c2;      c.output();     cout<<endl;cout<<"对c1取反的结果为：";c=-c1;c.output ();cout<<endl;    system("PAUSE");  }