/* (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 rightsreserved.
* 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:
晁阳
* 完成日期: 2012 年 04 月 16日
* 版本号: T1.0
* 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; 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"); }