题解 | 计算一元二次方程

#include <cstdio>
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main()
{
    float a,b,c;
    while(scanf("%f %f %f",&a,&b,&c)!=EOF)
    {
        float d=b*b-4*a*c;
    float f=(-b-sqrt(d))/(2*a);
    float g=(-b+sqrt(d))/(2*a);
    if (a==0)
    cout<<"Not quadratic equation";
    if(-b==-0&&sqrt(d)/(2*a)==0) f=g=0;
    if(a>0)
    {
    if(d==0) cout<<fixed<<setprecision(2)<<"x1=x2="<<f<<endl;
    else if(d>0) cout<<fixed<<setprecision(2)<<"x1="<<f<<';'<<"x2="<<g<<endl;
    else cout<<fixed<<setprecision(2)<<"x1="<<-b/(2*a)<<'-'<<sqrt(-d)/(2*a)<<'i'<<';'<<"x2="<<-b/(2*a)<<'+'<<sqrt(-d)/(2*a)<<'i'<<endl;
    }
    if(a<0)
    {
     if(d==0) cout<<fixed<<setprecision(2)<<"x1=x2="<<f<<endl;
    else if(d>0) cout<<fixed<<setprecision(2)<<"x1="<<g<<';'<<"x2="<<f<<endl;
    else 
    cout<<fixed<<setprecision(2)<<"x1="<<-b/(2*a)<<'+'<<sqrt(-d)/(2*a)<<'i'<<';'<<"x2="<<-b/(2*a)<<'-'<<sqrt(-d)/(2*a)<<endl;   
    }
    }
    return 0;
}