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2012年 上半年 第八周 C++程序设计 (二十五)

2012-04-16 17:57 323 查看
【任务】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。

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

};

//重载函数的实现及用于测试的main()函数

 

代码

 

 

#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;
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(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");
}


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