Description

从键盘任意输入a，b，c的值，编程计算并输出一元二次方程ax2+bx+c=0的根，当a=0时，输出“Sorry!”，当a≠0时，分b2-4ac=0、b2-4ac>0、b2-4ac<0三种情况，计算并输出方程的根。

Input

一行输入三个实数，以空格隔开

Output

输出“Sorry!”，或者一个实数，或者两个以空格隔开的实数，或者a+bi a-bi，其中a为实数，b为正实数（为保证两个虚根输出顺序的唯一性），所有实数小数点后保留两位有效数字。

Sample Input


2 3 2

Sample Output


-0.75+0.66i -0.75-0.66i

Hint

比较一个实数a是否等于0，要写成if(fabs(a)<=1e-6),其中1e-6可以使用宏定义 #define EPS 1e-6
#include<stdio.h>
#include<math.h>
#define ESP 1e-6
int main()
{
    double a,b,c,d,x1,x2;
    scanf("%lf %lf %lf",&a,&b,&c);
   
    if(fabs(a)<=ESP)
            printf("Soory\n");
    else
    {
   
       d=b*b-4*a*c;
        
    if(d>0)
    {    x1=(-b+sqrt(d))/(2*a);
        x2=(-b+sqrt(d))/(2*a);
        printf("%.2f %.2f\n",x1,x2);
    }
    if(d==0)
    {
        x1=(-b)/(2*a);
        printf("%.2f\n",x1);
    }
    if(d<0)
    {
        x1=(-b)/(2*a);
        x2=(sqrt(-d))/(2*a);
         printf("%.2lf+%.2lfi %.2lf-%.2lfi\n",x1,x2,x1,x2);
        
    }
}
    return 0;
}

应该是printf("sorry！\n");吧

|俺也弱 弱的回一个，
# include <stdio.h>
# include <math.h>
int main(void)
{
     double a, b, c;
     double delta;
     double x1, x2;
     char ch;
 
     do
     {
          printf ("请输入一元二次方程的三个系数:\n");
          printf ("a = ");
          scanf ("%lf", &a);
  
          printf ("b = ");
          scanf ("%lf", &b);
  
          printf ("c = ");
          scanf ("%lf", &c);
  
          delta = b*b - 4*a*c;
            if (delta > 0)
          {   
               x1 = (-b + sqrt(delta)) / (2*a);
               x2 = (-b - sqrt(delta)) / (2*a);
               printf ("方程有两个解: x1 =%lf, x2 =%lf\n", x1, x2 );
           }        
          else if (0 == delta)
           {   
               x1 = x2 = (-b) / (2*a);
               printf ("方程有唯一解：x1 = x2 = %lf\n",x1, x2);
           }
            else
            {        
                   printf ("方程无解\n");
           }   
           printf ("您想继续吗 (Y/N):");
           scanf (" %c", &ch);                  //%c前面必须加一个空格
         }while('y' == ch || 'Y' == ch);
 
     return 0;
}

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

#define ESP 1e-6

int main()
{
    double a  = 0.0;
    double b  = 0.0；
    double c  = 0.0；
    double d  = 0.0；
    double x1 = 0.0；
    double x2 = 0.0；
    scanf("%lf %lf %lf",&a,&b,&c);
   
    if(fabs(a) <= ESP){
         printf("Soory\n");
    }else{
            d = b*b-4*a*c;
        
            if(d > 0){   
                x1 = (-b+sqrt(d))/(2*a);
                x2 = (-b-sqrt(d))/(2*a);
                printf("%.2f %.2f\n",x1,x2);
            }
            if(d == 0){
                x1=(-b)/(2*a);
                printf("%.2f\n",x1);
             }
            if(d < 0){
                x1 = (-b)/(2*a);
                x2 = (sqrt(-d))/(2*a);
                printf("%.2lf+%.2lfi %.2lf-%.2lfi\n",x1,x2,x1,x2);
            }
        }
    return 0;
}
--------------
运行了一下没错误啊
可能你写代码的习惯不好
给你大体改了一下 你看下
有个小错误 ，给你改好了
x1 = (-b+sqrt(d))/(2*a);
x2 = (-b-sqrt(d))/(2*a);

#include<stdio.h>
#include<math.h>
#define ESP 1e-6
int main()
{
    double a,b,c,d,x1,x2;
    scanf("%lf %lf %lf",&a,&b,&c); 

    if(fabs(a)<=ESP)
        printf("Sorry!\n");
    else
    {       

        d=b*b-4*a*c;   

        if(d>0)
        {   

            x1=(-b+sqrt(d))/(2*a);
            x2=(-b-sqrt(d))/(2*a);//这边太大意啦
            printf("%.2f %.2f\n",x1,x2);
        }
        else if(d==0)
        {
            x1=(-b)/(2*a);
            printf("%.2f\n",x1);
        }
        else if(d<0)
        {
            x1=(-b)/(2*a);
            x2=(sqrt(-d))/(2*a);
            printf("%.2lf+%.2lfi %.2lf-%.2lfi\n",x1,x2,x1,x2);           

        }
    }
    return 0;
}