实例要求
' w: _6 X4 F* I( u 实现⼀个复数类 Complex 。 Complex 类包括两个 double 类型的成员 real 和 image ,分别表示复数的实部和虚部。对 Complex 类,重载其流提取、流插⼊运算符,以及加减乘除四则运算运算符。
( n6 F9 F& B/ H( o3 I 重载流提取运算符 >> ,使之可以读⼊以下格式的输⼊(两个数值之间使⽤空⽩分隔),将第⼀个数值存为复数的实部,将第⼆个数值存为复数的虚部:; p& ^- F6 M Y% o- ?) Y
<p>1</p><p>2</p><p>-1.1 2.0</p><p>+0 -4.5</p>
重载流插⼊运算符 << ,使之可以将复数输出为如下的格式⸺实部如果是⾮负数,则不输出符号位;输出时要包含半⻆左右⼩括号:
& p9 \' f& c1 v <p>1</p><p>2</p><p>(-1.1+2.0i)</p><p> (0-4.5i)</p>
每次输⼊两个复数,每个复数均包括由空格分隔的两个浮点数,输⼊第⼀个复数后,键⼊回⻋,然后继续输⼊第⼆个复数。
$ g2 `. u) c5 y) A* q$ T 输出两个复数,每个复数占⼀⾏;复数是由⼩括号包围的形如 (a+bi) 的格式。注意不能输出全⻆括号。0 t% B) h5 L% Y' h+ o7 s
样例输⼊
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- \$ A8 U0 H) n1 O& K( V9 J -1.1 2.0' L+ a0 V6 m2 F- h& t/ N+ |* r! C
0 -4.5
2 B2 F* X) i6 E5 A' ^( U 样例输出
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3
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(-1.1+2i) (0-4.5i)
|* B, a' s/ u (-1.1-2.5i)
: S5 J$ A6 k, r1 H (-1.1+6.5i)
+ F/ i3 a# @4 h5 q* _ (9+4.95i): O1 D7 ?" `0 j# b
(-0.444444-0.244444i)
6 g6 o0 {# R' z; s* ]& d+ b 提示
% M1 {/ b2 z+ w( y5 d8 p 需要注意,复数的四则运算定义如下所示:0 W1 R- t( I! u- x& n d6 p
加法法则: ( a + b i ) + ( c + d i ) = ( a + c ) + ( b + d ) i (a + bi) + (c + di) = (a + c) + (b + d)i (a+bi)+(c+di)=(a+c)+(b+d)i 减法法则: ( a + b i ) − ( c + d i ) = ( a − c ) + ( b − d ) i (a + bi) − (c + di) = (a − c) + (b − d)i (a+bi)−(c+di)=(a−c)+(b−d)i 乘法法则: ( a + b i ) × ( c + d i ) = ( a c − b d ) + ( b c + a d ) i (a + bi) × (c + di) = (ac − bd) + (bc + ad)i (a+bi)×(c+di)=(ac−bd)+(bc+ad)i 除法法则: ( a + b i ) ÷ ( c + d i ) = [ ( a c + b d ) / ( c 2 + d 2 ) ] + [ ( b c − a d ) / ( c 2 + d 2 ) ] i (a + bi) ÷ (c + di) = [(ac + bd)/(c^2 + d^2 )] + [(bc − ad)/(c^2 + d^2)]i (a+bi)÷(c+di)=[(ac+bd)/(c2+d2)]+[(bc−ad)/(c2+d2)]i
. S) K9 g) M- R. V4 b! e$ {5 o 两个流操作运算符必须重载为 Complex 类的友元函数,此外,在输出的时候,你需要判断复数的虚部是否⾮负⸺例如输⼊ 3 1.0 ,那么输出应该为 3+1.0i 。这⾥向⼤家提供⼀种可能的处理⽅法:使⽤ ostream 提供的 setf() 函数 ⸺它可以设置数值输出的时候是否携带标志位。例如,对于以下代码:+ s3 `5 K5 C3 i! @6 F) b. u0 {9 i
ostream os;
os.setf(std::ios::showpos);
os << 12;
输出内容会是 +12 。3 o: n5 G& h+ Y- \9 w
⽽如果想要取消前⾯的正号输出的话,你可以再执⾏:
6 j9 ^& D) ^3 ~" a1 [7 ^" n os.unsetf(std::ios::showpos);
即可恢复默认的设置(不输出额外的正号)7 J9 S- k) E2 Z7 }, L
代码实现 % W$ B H1 U* p( J5 h4 M8 Q
#include <iostream>
using namespace std;
const double EPISON = 1e-7;
class Complex
{
private:
double real;
double image;
public:
Complex(const Complex& complex) :real{ complex.real }, image{ complex.image } {
}
Complex(double Real=0, double Image=0) :real{ Real }, image{ Image } {
}
//TODO
Complex operator+(const Complex c) {
return Complex(this->real + c.real, this->image + c.image);
}
Complex operator-(const Complex c) {
return Complex(this->real - c.real, this->image - c.image);
}
Complex operator*(const Complex c) {
double _real = this->real * c.real - this->image * c.image;
double _image = this->image * c.real + this->real * c.image;
return Complex(_real, _image);
}
Complex operator/(const Complex c) {
double _real = (this->real * c.real + this->image * c.image) / (c.real * c.real + c.image * c.image);
double _image = (this->image * c.real - this->real * c.image) / (c.real * c.real + c.image * c.image);
return Complex(_real, _image);
}
friend istream &operator>>(istream &in, Complex &c);
friend ostream &operator<<(ostream &out, const Complex &c);
};
//重载>>
istream &operator>>(istream &in, Complex &c) {
in >> c.real >> c.image;
return in;
}
//重载<<
ostream &operator<<(ostream &out, const Complex &c) {
out << "(";
//判断实部是否为正数或0
if (c.real >= EPISON || (c.real < EPISON && c.real > -EPISON)) out.unsetf(std::ios::showpos);
out << c.real;
out.setf(std::ios::showpos);
out << c.image;
out << "i)";
return out;
}
int main() {
Complex z1, z2;
cin >> z1;
cin >> z2;
cout << z1 << " " << z2 << endl;
cout << z1 + z2 << endl;
cout << z1 - z2 << endl;
cout << z1*z2 << endl;
cout << z1 / z2 << endl;
return 0;
}
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