[面向对象课]复数的输入输出

大概是前几天面向对象期中考的时候有这么一个题…需要读入和输出复数（还都是浮点数），当时写了好久才过，虽然确实不算难，但当时也卡了好久…

#include<bits/stdc++.h>
using namespace std;
double read(string str){
//	cout<<"str="<<str<<endl;
	double ans=0;
	int len=str.length(),pos=-1,f=1;
	if(str[len-1]=='i'){
		if(len-2>=0){
			if(str[len-2]=='-')return -1;
			if(str[len-2]=='+')return 1;
		}else return 1;
		len--;
	}
	for(int i=0;i<len;i++){
		if(str[i]=='+')continue;
		if(str[i]=='-'){
			f=-1;
			continue;
		}
		if(str[i]!='.')ans=ans*10+str[i]-'0';
		else pos=i;
	}
	if(pos!=-1){
		pos=len-1-pos;
		while(pos){
			pos--;
			ans/=10;
		}
	}
	return ans*f;
}
int main(){
// 	freopen("input.txt","r",stdin);
	string str;cin>>str;
	cout<<"complex"<<' '<<str<<endl;
	int len=str.length();
	
	double x,y;
	if(str[len-1]!='i')x=read(str),y=0;
	else{
		int p=len-1;
		while(p&&str[p]!='+'&&str[p]!='-')p--;
		y=read(str.substr(p,len-p));
		p--;
		if(p<0)x=0;
		else x=read(str.substr(0,p+1));
	}
	cout<<"the real part is "<<x<<endl;
	cout<<"and the imaginary part is "<<y;
	return 0;
}

以及还有一个封装过的复数的输入输出：

#include<bits/stdc++.h>
using namespace std;
int read(string str);
class Complex{
private:
	int real,image;
public:
	Complex(int _r=0,int _i=0):real(_r),image(_i){}
	Complex operator *(const Complex &rhs){
		return Complex(real*rhs.real-image*rhs.image,
						real*rhs.image+image*rhs.real);
	}
	friend istream& operator >>(istream &in,Complex &t){
		string str;in>>str;
		int len=str.length();
		if(str[len-1]!='i')t.real=read(str),t.image=0;
		else{
			int p=len-1;
			while(p&&str[p]!='+'&&str[p]!='-')p--;
			t.image=read(str.substr(p,len-p));
			p--;
			if(p<0)t.real=0;
			else t.real=read(str.substr(0,p+1));
		}
		return in;
	}
	friend ostream& operator<<(ostream &out,const Complex &t){
		if(!t.real&&!t.image){out<<0;return out;}
		if(t.real)out<<t.real;
		if(t.image){
			if(t.real||t.image<0){
				if(t.image<0)out<<'-';
				else out<<'+';
			}
			if(abs(t.image)!=1)out<<abs(t.image);
			out<<'i';
		}
		return out;
	}
};
int read(string str){
	int ans=0,len=str.length(),f=1;
	if(str[len-1]=='i'){
		if(len-2>=0){
			if(str[len-2]=='-')return -1;
			if(str[len-2]=='+')return 1;
		}else return 1;
		len--;
	}
	for(int i=0;i<len;i++){
		if(str[i]=='+')continue;
		if(str[i]=='-'){f=-1;continue;}
		ans=ans*10+str[i]-'0';
	}
	return ans*f;
}
int main(){
	Complex a,b;
	cin>>a>>b;
	cout<<a*b;
	return 0;
}