/* (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？？？？？？