7-3 复数四则运算 (15分)

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：
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

**虽然没有全通过 **，但认为这道题还是比较直观的，难点应该是在于小数点精度的判断上。这道题还启示我写码时得安静，脑子不能乱（笑~

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

#define MARK 0.05

int demo_1(double a, double b){
    //一部分    
    if(abs(a) < MARK && abs(b) < MARK) printf("(0.0+0.0i)");
        
    else if(abs(a) < MARK && abs(b) >= MARK) printf("(0.0+%0.1lfi)", b);
        
    else {
            
        if(abs(b) < MARK) printf("(%0.1lf+0.0i)", a);        
        else if(abs(b) >= MARK && b < 0) printf("(%0.1lf-%0.1lfi)", a, -b);        
        else printf("(%0.1lf+%0.1lfi)", a, b);   
     }    
    return 0;
}

int path_0(double i, double j){
    // 固定过程 //    
    if (abs(i) < MARK && abs(j) < MARK)        printf("0.0");    
    else if (abs(i) < MARK && abs(j) >= MARK)        printf("%0.1lfi", j);     
    else    {
     //abs(i) >= 0.05        
        if (abs(j) < MARK)            printf("%0.1lf", i);        
        else if (abs(j) >= MARK && j < 0)            printf("%0.1lf-%0.1lfi", i, -j);        
        else            printf("%0.1lf+%0.1lfi", i, j);    }   
 return 0；
}
 
double rounding(double num){
        //四舍五入
if (num > 0)        
num = (int)(num * 10 + 0.5) / 10.0;    
else        num = (int)(num * 10 - 0.5) / 10.0;    
return num;
}

//start to solve where is down by this. 

int CL_add(double a, double b, double c, double d){
       //加法
 double i = a + c, j = b + d;    
 path_0(rounding(i), rounding(j));    puts("");    
 return 0;}
 
int CL_sub(double a, double b, double c, double d){
        //减法
double i = a - c, j = b - d;    
path_0(rounding(i), rounding(j));    puts("");    
return 0;
}

int CL_mul(double a, double b, double c, double d){
        //乘法
double i = a*c - b*d, j = a*d + b*c;    path_0(rounding(i),rounding(j));     puts("");    
return 0;
}

int CL_div(double a, double b, double c, double d){
        //除法
//double i = (a*c + b*d)/(c*c + d*d), j = (b*c - a*d)/(c*c + d*d);    
path_0(rounding((a * c + b * d) / (c * c + d * d)), rounding((b * c - a * d) / (c * c + d * d)));    puts("");   
 return 0;
 }
 
//divide into four parts
int main(){
       
double a1, a2, b1, b2;
   scanf("%lf%lf%lf%lf", &a1, &b1, &a2, &b2);
      
   demo_1(a1,b1);    printf(" + ");    demo_1(a2,b2);    printf(" = ");   CL_add(a1,b1,a2,b2);
   
   demo_1(a1, b1);   printf(" - ");   demo_1(a2, b2);   printf(" = ");   CL_sub(a1,b1,a2,b2);
   
   demo_1(a1, b1);   printf(" * ");   demo_1(a2, b2);   printf(" = ");   CL_mul(a1,b1,a2,b2);
   
   demo_1(a1, b1);   printf(" / ");   demo_1(a2, b2);   printf(" = ");   CL_div(a1,b1,a2,b2);
   system("pause");  
    return 0;
    }

**过了三个测试点，第一个总是过不了，还得再想想。 **

注意fabs（）和abs（）的区别！！！