题解 | #计算一元二次方程#

#include <stdio.h>

#include<math.h>
int main()
{
	float a = 0.0;
	float b = 0.0;
	float c = 0.0;
	float derta = 0.0;
	while(scanf("%f %f %f",&a,&b,&c)!=EOF)
	if (a == 0.0)
	{
		printf("Not quadratic equation");
	}
	else
	{
		derta = b * b - 4*a * c;
		if (derta == 0)
		{
                float r = -b / (2 * a);
				printf("x1=x2=%.2f\n", r==0?0:r);   //这里容易想不到，若不这样写，结果是-0.00，不符合题目                                                       要求
		}
		else if (derta > 0)
		{
			float t1 = (-b - sqrt(derta)) / (2 * a);
			float t2 = (-b + sqrt(derta)) / (2 * a);
			printf("x1=%.2f;x2=%.2f\n", t1, t2);
		}
		else
		{
			float shibu = -b / (2 * a);
			float xubu = sqrt(-derta) / (2 * a);
			printf("x1=%.2f-%.2fi;x2=%.2f+%.2fi\n", shibu, xubu, shibu, xubu);
		}
	}
	return 0;
 }