题解 | #计算一元二次方程#

好恶心的分类处理，同意的点个赞

#include<iostream>
#include<iomanip>
#include<cmath>

using namespace std;

int main()
{
    float a,b,c,x1,x2;
    while(cin>>a>>b>>c)
    {
        if(((b*b-4*a*c)>0)&&a)
        {
            x1=(-b-sqrt(b*b-4*a*c))/(2*a);
            x2=(-b+sqrt(b*b-4*a*c))/(2*a);
            cout<<fixed<<setprecision(2)<<"x1="<<x1<<";"<<"x2="<<x2<<endl;
        }
        else if((b*b-4*a*c==0)&&a)
        {
            if(0==b&&a>0) x1=x2=b/(2*a);
            else x1=x2=-b/(2*a);
            cout<<fixed<<setprecision(2)<<"x1=x2="<<x1<<endl;
        }
        else if((b*b-4*a*c)<0&&a)
        {
            if(b&&a>0)
            {
            cout<<fixed<<setprecision(2)<<"x1="<<-b/(2*a)<<"-"<<sqrt(-(b*b-4*a*c))/(2*a)<<"i;";
            cout<<"x2="<<-b/(2*a)<<"+"<<sqrt(-(b*b-4*a*c))/(2*a)<<"i"<<endl;               
            }
            else if(b&&a<0)
            {
            cout<<fixed<<setprecision(2)<<"x1="<<-b/(2*a)<<sqrt(-(b*b-4*a*c))/(2*a)<<"i;";
            cout<<"x2="<<-b/(2*a)<<"+"<<-sqrt(-(b*b-4*a*c))/(2*a)<<"i";   
            }
            else if (0==b&&a>0)
            {
            cout<<fixed<<setprecision(2)<<"x1="<<b/(2*a)<<"-"<<sqrt(-(b*b-4*a*c))/(2*a)<<"i;";
            cout<<"x2="<<b/(2*a)<<"+"<<sqrt(-1*(b*b-4*a*c))/(2*a)<<"i"; 
            }
        }
        else if (!a)
        {
            cout<<"Not quadratic equation"<<endl;
        }
    }
    return 0;
}