a006. 一元二次方程式 | From: [140.117.147.41] | 發表日期 : 2014-07-22 16:55

此解法 考量到  a=0,b=0,c=0的可能

 

 

 

#include <iostream>

#include <cmath>

using namespace std;

int main()

{

  double a,b,c,out1,out2;

  while(cin>>a>>b>>c)

  {  

    if(c==0){out1=0;out2=((b==0)?0:((a==0)?0:-b/a));}

    else {out1=((-b+sqrt(abs(b*b-4*a*c)))/2/a );out2=((-b-sqrt(abs(b*b-4*a*c)))/2/a );}

    if(abs(out1)==0)out1=0;    if(abs(out2)==0) out2=0;

    

    if(b*b-4*a*c<0||(a==0&&b==0&&c!=0))cout<<"No real root"<<endl;

    else if(a==0&&b==0&&c==0) cout<<"Infinitely solutions"<<endl;

    else {if(out1==out2) cout<<"Two same roots x="<<out1<<endl;

          else cout<<"Two different roots x1="<<((a>0)?out1:out2)<<" , x2="<<(a>0?out2:out1)<<endl;

          }

    

          

  }

}