C 解法, 执行用时: 2ms, 内存消耗: 300KB, 提交时间: 2022-05-18

#include <stdio.h>
#include <math.h>
int main()
{
    double a = 0.0;
    double b = 0.0;
    double c = 0.0;
    while(scanf("%lf %lf %lf",&a,&b,&c) != EOF)
    {
        if(a == 0)
            printf("Not quadratic equation\n");
        else
        {
            double drct = (b*b)-(4*a*c);
           
            if(drct == 0)
            {
                if(b == 0)
                {
                    printf("x1=x2=%.2lf\n",0);
                }
                else
                {
                    printf("x1=x2=%.2lf\n",(-b)/(2*a));
                }
            }
            else if(drct > 0)
                printf("x1=%.2lf;x2=%.2lf\n",(-b-sqrt(drct))/(2*a),(-b+sqrt(drct))/(2*a));
            else
            {
               double real = (-b)/(2*a);
               double dash = sqrt(-drct)/(2*a);
                if(dash < 0)
                    dash = -dash;
                printf("x1=%.2lf-%.2lfi;x2=%.2lf+%.2lfi\n", real, dash, real, dash);
                
            }
        }
    }
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 312KB, 提交时间: 2022-05-07

#include<stdio.h>
#include<math.h>
int main(){
    float a,b,c,o;
    while(~scanf("%f %f %f",&a,&b,&c)){
        o=b*b-4*a*c;
        if(a==0){
            printf("Not quadratic equation");
        }else{
            if(o==0){
                    if(-b/(2.0*a)==-0.00){
                        printf("x1=x2=0.00");
                    }else{printf("x1=x2=%.2f",-b/(2.0*a));}
                    
             }else if(o>0){
                     printf("x1=%.2f;x2=%.2f",(-b-sqrt(o))/(2.0*a),(-b+sqrt(o))/(2.0*a));
             }else{
                    printf("x1=%.2f-%.2fi;x2=%.2f+%.2fi",-b/(2.0*a),fabs(sqrt(-o)/(2.0*a)),-b/(2.0*a),fabs(sqrt(-o)/(2.0*a)));
             }
            printf("\n");
        }
    }
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 316KB, 提交时间: 2022-07-19

#include <stdio.h>
#include <math.h>
int main()
{
    float a,b,c;
    float d,e;
    while(scanf("%f %f %f",&a,&b,&c)!=EOF)
    {
        if(a==0)
        {
            printf("Not quadratic equation");
        }
        else
        {
            d=(b*b)-(4.0*a*c);
            if(d==0)
            {
                if(b==0)
                {
                    if(a>0)
                    {
                        printf("x1=x2=%.2f\n",(b)/(2.0*(a)));
                    }
                    else
                    {
                        printf("x1=x2=%.2f\n",(b)/(2.0*(-a)));
                    }
                }
                else
                {
                    printf("x1=x2=%.2f\n",(-b)/(2.0*a));
                } 
            }
            else if(d>0)
            {
                printf("x1=%.2f;x2=%.2f\n",((-b)-sqrt(d))/(2.0*a),((-b)+sqrt(d))/(2.0*a));
            }
            else
            {
                e=(sqrt(-d)/(2*a));
                if(e>0)
                {
                    printf("x1=%.2f-%.2fi;x2=%.2f+%.2fi",
                       -(b)/(2*a),(sqrt(-d)/(2*a)),
                       -(b)/(2*a),(sqrt(-d)/(2*a)));
                }
                else
                {
                    printf("x1=%.2f%.2fi;x2=%.2f+%.2fi",
                       -(b)/(2*a),(sqrt(-d)/(2*a)),
                       -(b)/(2*a),-(sqrt(-d)/(2*a)));
                }
            }
        }
    }
    
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 320KB, 提交时间: 2022-07-30

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

C 解法, 执行用时: 2ms, 内存消耗: 324KB, 提交时间: 2022-07-16

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

int main()
{
    float a,b,c,o;
    while(~scanf("%f %f %f",&a,&b,&c)){
        o=b*b-4*a*c;
        if(a==0){
            printf("Not quadratic equation");
        }else{
            if(o==0){
                    if(-b/(2.0*a)==-0.00){
                        printf("x1=x2=0.00");
                    }else{printf("x1=x2=%.2f",-b/(2.0*a));}
                     
             }else if(o>0){
                     printf("x1=%.2f;x2=%.2f",(-b-sqrt(o))/(2.0*a),(-b+sqrt(o))/(2.0*a));
             }else{
                    printf("x1=%.2f-%.2fi;x2=%.2f+%.2fi",-b/(2.0*a),fabs(sqrt(-o)/(2.0*a)),-b/(2.0*a),fabs(sqrt(-o)/(2.0*a)));
             }
            printf("\n");
        }
    }
    return 0;
}