7-5 complex quaternion operation

subject

7-5 Four operations of complex number
fraction 25
author Jolin
Company Tsinghua University
This question requires the preparation of procedures , Calculation 2 The sum of two plurals 、 Bad 、 product 、 merchant .

Input format ：
Type in a line and press a1 b1 a2 b2 The format of 2 Plural C1=a1+b1i and C2=a2+b2i The real part and the virtual part of . Title assurance C2 Not for 0.

Output format ：
Respectively in 4 In line with (a1+b1i) Operator (a2+b2i) = The format of the results is output sequentially 2 The sum of two plurals 、 Bad 、 product 、 merchant , The number is accurate to the decimal point 1 position . If the real or imaginary part of the result is 0, No output . If the result is 0, The output 0.0.

sample input 1：
2 3.08 -2.04 5.06
sample output 1：
(2.0+3.1i) + (-2.0+5.1i) = 8.1i
(2.0+3.1i) - (-2.0+5.1i) = 4.0-2.0i
(2.0+3.1i) * (-2.0+5.1i) = -19.7+3.8i
(2.0+3.1i) / (-2.0+5.1i) = 0.4-0.6i
sample input 2：
1 1 -1 -1.01
sample output 2：
(1.0+1.0i) + (-1.0-1.0i) = 0.0
(1.0+1.0i) - (-1.0-1.0i) = 2.0+2.0i
(1.0+1.0i) * (-1.0-1.0i) = -2.0i
(1.0+1.0i) / (-1.0-1.0i) = -1.0

Ideas

Multiplication and division formula of complex number

Code

#include<stdio.h>
double re(double e){
	if(e>=0){
		e=(double)(int)(e*10+0.5)/10;
	}else{
		e=(double)(int)(e*10-0.5)/10;
	}
	return e;
}
void Print(double a,double b,double c,double d,double e,double f,char g)
{
	if(e!=0&&f!=0){
		printf("(%.1lf%+.1lfi) %c (%.1lf%+.1lfi) = %.1lf%+.1lfi\n",a,b,g,c,d,
		e,f);
	}else if(e==0&&f==0){
		printf("(%.1lf%+.1lfi) %c (%.1lf%+.1lfi) = 0.0\n",a,b,g,c,d);
	}else if(e==0){
		printf("(%.1lf%+.1lfi) %c (%.1lf%+.1lfi) = %.1lfi\n",a,b,g,c,d,f);
	}else{
		printf("(%.1lf%+.1lfi) %c (%.1lf%+.1lfi) = %.1lf\n",a,b,g,c,d,e);
	}
}
int main()
{
	double a,b,c,d,e,f;
	scanf("%lf %lf %lf %lf",&a,&b,&c,&d);
	e=a+c;
	f=b+d;
	e=re(e);
	f=re(f);
	Print(a,b,c,d,e,f,'+');
	e=a-c;
	f=b-d;
	e=re(e);
	f=re(f);
	Print(a,b,c,d,e,f,'-');
	e=(a*c-b*d);
	f=(b*c+a*d);
	e=re(e);
	f=re(f);
	Print(a,b,c,d,e,f,'*');
	e=(a*c+b*d)/(c*c+d*d);
	f=(b*c-a*d)/(c*c+d*d);
	e=re(e);
	f=re(f);
	Print(a,b,c,d,e,f,'/');	
 }