题解 | #计算一元二次方程#

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

int main() {
  float a,b,c,d,x1,x2,m,n;
  while(~scanf("%f%f%f",&a,&b,&c))
  { 
    d=b*b-4*a*c;
    if (a==0)
   {
    printf("Not quadratic equation");
   }
    else if(d==0)
    {
      x1=x2=-b/(2*a);
       if (fabs(x1) < 1e-6) 
            {
                printf("x1=x2=0.00\n");
            } else {
                printf("x1=x2=%.2f\n", x1);
            }
    
    }
     else if(d>0)
     {
      x1=(-b/(2*a))-sqrt(d)/(2*a);
      x2=(-b/(2*a))+sqrt(d)/(2*a);
      printf("x1=%.2f;x2=%.2f",x1,x2);
     }
      else
      {
       m=-b / (2*a),n=sqrt(-d ) / (2*a);
       if (n > 0) {
                printf("x1=%.2f-%.2fi;x2=%.2f+%.2fi\n", m, n, m, n);
            } else if (n < 0) {
                printf("x1=%.2f+%.2fi;x2=%.2f-%.2fi\n", m, -n, m, -n);
            } else {
                if (fabs(m) < 1e-6) {
                    printf("x1=x2=0.00\n");
                } else {
                    printf("x1=x2=%.2f\n", m);
                }
      }
  }}
    return 0;
}

修修改改，终于做出来了