第九周任务(三)

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/* (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:定义分数类中<<和>>运算符重载,实现分数的输入输出,改造原程序中对运算结果显示方式.

* 作    者:       刘洋                   

* 完成日期:     2012    年 04  月   16日

* 版 本 号:       1.0   

* 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.h>   #include"stdlib.h"      int gcd(int m, int n);      class CFraction  {  private:      int nume;  // 分子      int deno;  // 分母 char unknown; // '/'的符号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.unknown>>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;     }  


 

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