C语言程序设计教程（第三版）课后习题8.2 （C语言代码）

解题思路:

注意事项:

参考代码:#include<stdio.h>
#include<math.h>
typedef struct OUTPUT
{
double root_1;
double vir_root_1;
double root_2;
double vir_root_2;
}OUTPUT;
OUTPUT get_equation_root_more_zero(double num_a,double num_b,double num_c);
OUTPUT get_equation_root_less_zero(double num_a,double num_b,double num_c);
OUTPUT get_equation_root_equal_zero(double num_a,double num_b,double num_c);
int main()
{
double input_a,input_b,input_c,delta = 0;
scanf("%lf%lf%lf",&input_a,&input_b,&input_c);
OUTPUT output;
delta = (input_b*input_b) - 4*input_a*input_c;
if (delta > 0)
{
output = get_equation_root_more_zero(input_a,input_b,input_c);
printf("x1=%.3lf x2=%.3lf\n",output.root_1,output.root_2);
}
else if (0 == delta)
{
output = get_equation_root_equal_zero(input_a,input_b,input_c);
printf("x1=%.3lf x2=%.3lf\n",output.root_1,output.root_2);
}
else
{
output = get_equation_root_less_zero(input_a,input_b,input_c);
printf("x1=%.3lf+%.3lfi x2=%.3lf+%.3lf\n",output.root_1,output.vir_root_1,output.root_2,output.vir_root_2);
}
return 0;
}
OUTPUT get_equation_root_more_zero(double num_a,double num_b,double num_c)
{
OUTPUT output;
output.root_1 = (-num_b+sqrt(num_b*num_b-4*num_a*num_c))/(2*num_a);
output.root_2 = (-num_b-sqrt(num_b*num_b-4*num_a*num_c))/(2*num_a);
return output;
}
OUTPUT get_equation_root_equal_zero(double num_a,double num_b,double num_c)
{
OUTPUT output;
output.root_1 = (-num_b)/(2*num_a);
output.root_2 = (-num_b)/(2*num_a);
return output;
}
OUTPUT get_equation_root_less_zero(double num_a,double num_b,double num_c)
{
OUTPUT output;
output.root_1 = (-num_b)/(2*num_a);
output.root_2 = (-num_b)/(2*num_a);
output.vir_root_1 = (sqrt(-(num_b*num_b-4*num_a*num_c)))/(2*num_a);
output.vir_root_2 = -(sqrt(-(num_b*num_b-4*num_a*num_c)))/(2*num_a);
return output;
}