C++复数四则运算的实现
程序主要实现复数的加减乘,数乘,取共轭功能。
将所有函数都定义为了成员函数。
使用库函数atof将字符串转换为浮点型数据。
函数主要难点在于处理输入。由于需要判断输入是选择退出还是继续,所以用字符串来接收输入,判断是否为q或Q后将字符串转换为double型。
由于库函数中定义了complex类,因此,这里的类名改为comple。
类声明
#ifndef COMPLEX_H_
#define COMPLEX_H_ #include<iostream>
using namespace std; class comple
{
private:
double re;
double im;
public:
comple(double r=0.0, double i=0.0);
~comple();
comple operator+(const comple & a) const;
comple operator-(const comple & a) const;
comple operator*(const comple & a) const;
comple operator~() const;
comple operator*(const double x) const;
friend ostream & operator<<(ostream & os, const comple & b);
friend int operator>>(istream & is, comple & b);
friend comple operator*(const double x, const comple & b); }; #endif
方法定义
#include"complex0.h"
#include<iostream>
#include<math.h> //strlen函数头文件
#include"string.h"
#include<cstdlib>//atof函数头文件
using namespace std; comple::comple(double r, double i)
{
re=r;
im=i;
} comple::~comple()
{ } comple comple::operator+(const comple & a) const
{
comple temp;
temp.re=re+a.re;
temp.im=im+a.im;
return temp;
} comple comple::operator-(const comple & a) const
{
comple temp;
temp.re=re-a.re;
temp.im=im-a.im;
return temp;
} comple comple::operator*(const comple & a) const
{
comple temp;
temp.re=re*a.re-im*a.im;
temp.im=re*a.im+im*a.re;
return temp;
} comple comple::operator~() const
{
comple temp;
temp.re=re;
temp.im=-im;
return temp;
} comple comple::operator*(const double x ) const
{
return comple(x*re,x*im);
} ostream & operator<<(ostream & os, const comple & b)
{
os << "(" <<b.re <<"," <<b.im << "i)";
return os;
} int operator>>(istream & is, comple & b)
{
double c1,c2;
char s[],*str;
int c,i;
cout << "real:";
is >> s;
c=strlen(s);
str=s;
for(i=;i<c;i++)
if(s[i]=='q'||s[i]=='Q')
return ; if((s[]>'')&&(s[]<''))
c1=atof(str); cout << "imaginary:";
is >> s;
c=strlen(s);
str=s;
for(i=;i<c;i++)
{
if(s[i]=='q'||s[i]=='Q')
return ;
} if(s[]>''&&s[]<'')
c2=atof(str); b=comple(c1,c2);
return ; } comple operator*(double x, const comple & b)
{
return b*x;
}
测试程序
#include<iostream>
#include"complex0.h" using namespace std; int main()
{
comple a(3.0,4.0);
comple c; cout << "Enter a complex number (q to quit):\n";
while(cin >> c)
{
cout << "c is " << c << endl;
cout << "complex conjugate is " << ~c <<endl;
cout << "a is " << a << endl;
cout << "a+c is " << a+c << endl;
cout << "a-c is " << a-c << endl;
cout << "a*c is " << a*c << endl;
cout << "2*c is " << *c << endl;
cout << "Enter a complex number (q to quit):\n";
}
cout << "Done!\n";
return ;
}
测试结果

atof函数补充
包含的头文件在c中为#include<stdlib.h>,c++中为include<cstdlib>
http://www.cppblog.com/cxiaojia/archive/2012/02/24/166436.html
http://my.oschina.net/Tsybius2014/blog/338234
http://www.cnblogs.com/lidabo/archive/2012/07/10/2584706.html
http://blog.csdn.net/cxh342968816/article/details/6627768
标准c++库函数中复数四则运算程序
#ifndef __MYCOMPLEX__
#define __MYCOMPLEX__ class complex;
complex&
__doapl (complex* ths, const complex& r);
complex&
__doami (complex* ths, const complex& r);
complex&
__doaml (complex* ths, const complex& r); class complex
{
public:
complex (double r = , double i = ): re (r), im (i) { }
complex& operator += (const complex&);
complex& operator -= (const complex&);
complex& operator *= (const complex&);
complex& operator /= (const complex&);
double real () const { return re; }
double imag () const { return im; }
private:
double re, im; friend complex& __doapl (complex *, const complex&);
friend complex& __doami (complex *, const complex&);
friend complex& __doaml (complex *, const complex&);
}; inline complex&
__doapl (complex* ths, const complex& r)
{
ths->re += r.re;
ths->im += r.im;
return *ths;
} inline complex&
complex::operator += (const complex& r)
{
return __doapl (this, r);
} inline complex&
__doami (complex* ths, const complex& r)
{
ths->re -= r.re;
ths->im -= r.im;
return *ths;
} inline complex&
complex::operator -= (const complex& r)
{
return __doami (this, r);
} inline complex&
__doaml (complex* ths, const complex& r)
{
double f = ths->re * r.re - ths->im * r.im;
ths->im = ths->re * r.im + ths->im * r.re;
ths->re = f;
return *ths;
} inline complex&
complex::operator *= (const complex& r)
{
return __doaml (this, r);
} inline double
imag (const complex& x)
{
return x.imag ();
} inline double
real (const complex& x)
{
return x.real ();
} inline complex
operator + (const complex& x, const complex& y)
{
return complex (real (x) + real (y), imag (x) + imag (y));
} inline complex
operator + (const complex& x, double y)
{
return complex (real (x) + y, imag (x));
} inline complex
operator + (double x, const complex& y)
{
return complex (x + real (y), imag (y));
} inline complex
operator - (const complex& x, const complex& y)
{
return complex (real (x) - real (y), imag (x) - imag (y));
} inline complex
operator - (const complex& x, double y)
{
return complex (real (x) - y, imag (x));
} inline complex
operator - (double x, const complex& y)
{
return complex (x - real (y), - imag (y));
} inline complex
operator * (const complex& x, const complex& y)
{
return complex (real (x) * real (y) - imag (x) * imag (y),
real (x) * imag (y) + imag (x) * real (y));
} inline complex
operator * (const complex& x, double y)
{
return complex (real (x) * y, imag (x) * y);
} inline complex
operator * (double x, const complex& y)
{
return complex (x * real (y), x * imag (y));
} complex
operator / (const complex& x, double y)
{
return complex (real (x) / y, imag (x) / y);
} inline complex
operator + (const complex& x)
{
return x;
} inline complex
operator - (const complex& x)
{
return complex (-real (x), -imag (x));
} inline bool
operator == (const complex& x, const complex& y)
{
return real (x) == real (y) && imag (x) == imag (y);
} inline bool
operator == (const complex& x, double y)
{
return real (x) == y && imag (x) == ;
} inline bool
operator == (double x, const complex& y)
{
return x == real (y) && imag (y) == ;
} inline bool
operator != (const complex& x, const complex& y)
{
return real (x) != real (y) || imag (x) != imag (y);
} inline bool
operator != (const complex& x, double y)
{
return real (x) != y || imag (x) != ;
} inline bool
operator != (double x, const complex& y)
{
return x != real (y) || imag (y) != ;
} #include <cmath> inline complex
polar (double r, double t)
{
return complex (r * cos (t), r * sin (t));
} inline complex
conj (const complex& x)
{
return complex (real (x), -imag (x));
} inline double
norm (const complex& x)
{
return real (x) * real (x) + imag (x) * imag (x);
} #endif //__MYCOMPLEX__
测试程序
#include <iostream>
#include "complex.h" using namespace std; ostream&
operator << (ostream& os, const complex& x)
{
return os << '(' << real (x) << ',' << imag (x) << ')';
} int main()
{
complex c1(, );
complex c2(, ); cout << c1 << endl;
cout << c2 << endl; cout << c1+c2 << endl;
cout << c1-c2 << endl;
cout << c1*c2 << endl;
cout << c1 / << endl; cout << conj(c1) << endl;
cout << norm(c1) << endl;
cout << polar(,) << endl; cout << (c1 += c2) << endl; cout << (c1 == c2) << endl;
cout << (c1 != c2) << endl;
cout << +c2 << endl;
cout << -c2 << endl; cout << (c2 - ) << endl;
cout << ( + c2) << endl; return ;
}
C++复数四则运算的实现的更多相关文章
- C语言 · 复数四则运算
算法提高 6-17复数四则运算 时间限制:1.0s 内存限制:512.0MB 设计复数库,实现基本的复数加减乘除运算. 输入时只需分别键入实部和虚部,以空格分割,两个复数之间用运算符 ...
- Java实现 蓝桥杯 算法提高 复数四则运算
算法提高 6-17复数四则运算 时间限制:1.0s 内存限制:512.0MB 提交此题 设计复数库,实现基本的复数加减乘除运算. 输入时只需分别键入实部和虚部,以空格分割,两个复数之间用运算符分隔:输 ...
- PTA 复数四则运算
本题要求编写程序,计算2个复数的和.差.积.商. 输入格式: 输入在一行中按照a1 b1 a2 b2的格式给出2个复数C1=a1+b1i和C2=a2+b2i的实部和虚部.题目保证C2不为0. 输出格式 ...
- 算法笔记_156:算法提高 6-17复数四则运算(Java)
目录 1 问题描述 2 解决方案 1 问题描述 设计复数库,实现基本的复数加减乘除运算. 输入时只需分别键入实部和虚部,以空格分割,两个复数之间用运算符分隔:输出时按a+bi的格式在屏幕上打印结果 ...
- [C++][重载]
运算符重载 C++中预定义的运算符的操作对象只能是基本数据类型,实际上,对于很多用户自定义类型,也需要有类似的运算操作.例如: class complex { public: complex(d ...
- 基于C++任意点数的FFT/IFFT(时域和频域)实现
函数说明:更改主函数体中的N和length(=log2(N))既可以实现任意点数(2的幂次)的FFT/ IFFT的实现,fft函数中flag标志位控制是正变换还是逆变换. 1.复数操作类 定 ...
- C++运算符重载讲解与经典实例
最近在学C++,找到一篇详细讲解运算符重载的文章,贴在这里分享和收藏. C++中预定义的运算符的操作对象只能是基本数据类型,实际上,对于很多用户自定义类型,也需要有类似的运算操作.例如: class ...
- [Swift]复数的表示和四则运算
我们把形如z=a+bi(a,b均为实数)的数称为复数,其中a称为实部,b称为虚部,i称为虚数单位. 当虚部等于零时,这个复数可以视为实数:当z的虚部不等于零时,实部等于零时,常称z为纯虚数. 复数域是 ...
- C复数的四则运算
#include<stdio.h> void judge(int True,int Fake) { if (True == 0) { if (Fake == ...
随机推荐
- jsp+oracle实现数据库内容以表格形式在前台显示(包含分页)
jsp+oracle实现数据库内容以表格形式在前台显示(包含分页) 在文件夹内新增list_emp.jsp 代码如下: <%@ page contentType="text/html& ...
- poj2528 线段树+离散化
由于坐标可能很大,此时需要离散化,将值转化为对应的坐标. #include<stdio.h> #include<algorithm> using namespace std; ...
- 【ZOJ 3844】Easy Task
题意 每次把序列中最大的数a的一个和最小的数b的一个变成a-b.求最后是否能使序列里的数全部相同,能则输出这个相同的数. 分析 一定是有解的,不断减少最大数的个数,最大数减少为0个时,就是减少了不同数 ...
- Unix/Linux 命令技巧
锁定一个文件夹 为了我的数据隐私,我想要锁定我文件服务器下的/downloads文件夹.因此我运行了: chmod 0000 /downloads root用户仍旧可以访问,而ls和cd命令则不工作. ...
- 【bzoj1179】 Apio2009—Atm
www.lydsy.com/JudgeOnline/problem.php?id=1179 (题目链接) 题意 给出一张有向图,每个节点有点权.标记一些点,找出一条路径,可以重复经过一条边,使得总点权 ...
- POJ3784 Running Median
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 1670 Accepted: 823 Description For th ...
- Latex 笔记
A source file is made up of text, formulas, and instructions (commands) to $\LaTeX.$ Commands start ...
- Using a Comparison Function for the Key Type
(这是C++系列随笔的第二篇,这一系列是我练习C++而查的资料) C++ Primer 5th. Ed. pp. 425 ---------------------- Using a Comparis ...
- IOS学习笔记—苹果推送机制APNs
转自:唐韧_Ryan http://blog.csdn.net/ryantang03/article/details/8482259 推送是解决轮询所造成的流量消耗和 电量消耗的一个比较好的解决方案, ...
- linux vsftpd搭建
1.yum install vsftpd; 2.关闭SELinux的方法:修改/etc/selinux/config文件中的SELINUX="" 为 disabled ,然后重启 ...