PAT : 结构-06. 复数四则运算(15)

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+b1*i和C2=a2+b2*i的实部和虚部。题目保证C2不为0。

输出格式：

分别在4行中按照“(a1+b1i) 运算符 (a2+b2i) = 结果”的格式顺序输出2个复数的和、差、积、商，数字精确到小数点后1位。如果结果的实部或者虚部为0，则不输出。如果结果为0，则输出0.0。
输入样例1：

2 3.08 -2.04 5.06

输出样例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

输入样例2：

1 1 -1 -1.01

输出样例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

题目说输出的结果若实部或虚部为0则不输出，全为零则输出0.0，这暗示如果输出的不是结果，不管实部虚部是不是0，都得输出，我未看清题意错了好几次

其次就是-0.0的问题了，这反倒不难，下面的clean（）函数就是做这个的，因为精确到小数点后1位，故当数>-0.05或<0.05均为0

#include <stdio.h>
#include <math.h>

#define EPSILON 0.05

struct comlexnum
{
double real;
double imaginary;
};

double clean (double x)
{
if(fabs(x) < EPSILON)
return 0;
return x;
}

void printcomplex(comlexnum a)
{
if(a.real == 0 && a.imaginary == 0)
printf("0.0");
else if(a.real == 0)
printf("%.1fi", a.imaginary);
else if(a.imaginary == 0)
printf("%.1f", a.real);
else
printf("%.1f%+.1fi", a.real, a.imaginary);
}

void printexpression(comlexnum a, comlexnum b, comlexnum res, char op)
{
printf("(%.1f%+.1fi) %c (%.1f%+.1fi) = ", a.real, a.imaginary, op, b.real, b.imaginary);
printcomplex(res);
printf("\n");
}

void add (comlexnum a, comlexnum b)
{
comlexnum res;
res.real = clean(a.real + b.real);
res.imaginary = clean(a.imaginary + b.imaginary);

printexpression(a, b, res, '+');
}

void sub (comlexnum a, comlexnum b)
{
comlexnum res;
res.real = clean(a.real - b.real);
res.imaginary = clean(a.imaginary - b.imaginary);

printexpression(a, b, res, '-');
}

void mul (comlexnum a, comlexnum b)
{
comlexnum res;
res.real = clean(a.real*b.real - a.imaginary*b.imaginary);
res.imaginary = clean(a.real*b.imaginary + a.imaginary*b.real);

printexpression(a, b, res, '*');
}

void div (comlexnum a, comlexnum b)
{
comlexnum res;
int c;

c = b.real*b.real + b.imaginary*b.imaginary;
res.real = clean((a.real*b.real + a.imaginary*b.imaginary)/c);
res.imaginary = clean((a.imaginary*b.real - a.real*b.imaginary)/c);

printexpression(a, b, res, '/');

}

int main()
{
comlexnum a,b;
scanf("%lf %lf %lf %lf", &a.real, &a.imaginary, &b.real, &b.imaginary);

add(a, b);
sub(a, b);
mul(a, b);
div(a, b);

return 0;
}