#include<bits/stdc++.h> using namespace std; int main() { double a,b,c,d,x,y; cin>>a>>b>>c; d=sqrt(bb-4ac); x=(-b+d)/(2.00a); y=(-b-d)/(2.00*a); if(x>y) { cout<<fixed<<setprecision(2)<<y<<" "<<x; } else { cout<<fixed<<setprecision(2)<<x<<" "<<y; } return 0; }

思路

求根公式：

$$x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

建议不要定义 x1 之类的全局变量，极有可能出错！！！

code

#include<bits/stdc++.h>
using namespace std;
double a,b,c;
int main(){
	cin>>a>>b>>c;
	double x1=(-b+sqrt(b*b-4*a*c))/2.0/a,x2=(-b-sqrt(b*b-4*a*c))/2.0/a;
	printf("%.2lf %.2lf",min(x1,x2),max(x1,x2));
	return 0;
}

#include<bits/stdc++.h> using namespace std; int main() { double a,b,c,d,e,f; cin>>a>>b>>c; d=sqrt(bb-4ac); e=(-b+d)/(2a); f=(-b-d)/(2*a); cout<<fixed<<setprecision(2)<<min(e,f)<<" "<<max(e,f); return 0; }

#include<bits/stdc++.h> using namespace std; int main() { double a,b,c,x1,x2,d; cin>>a>>b>>c; d=sqrt(bb-4ac); x1=(-b+d)/(2.00a); x2=(-b-d)/(2.00*a); if(x1<x2) { cout<<fixed<<setprecision(2)<<x1<<" "<<x2; } else { cout<<fixed<<setprecision(2)<<x2<<" "<<x1; } return 0; }

#include<bits/stdc++.h>
using namespace std;
int main()
{
	double a,b,c,d,x1,x2;
	cin>>a>>b>>c;
	d=sqrt(b*b-4*a*c);
	x1=(-b+d)/(2*a);
	x2=(-b-d)/(2*a);
	if(x1>x2)
	{
		cout<<fixed<<setprecision(2)<<x2<<" "<<x1<<endl;
	}
	else
	{
	    	cout<<fixed<<setprecision(2)<<x1<<" "<<x2<<endl;
	}
	return 0;
}

记得用double

#include<bits/stdc++.h> using namespace std; int main() { double a,b,c,d,x,y; cin>>a>>b>>c; d=sqrt(bb-4ac), x=(0-b+d)/(2a), y=(0-b-d)/(2*a); if(x<y){cout<<fixed<<setprecision(2)<<x<<" "<<y<<endl;} else{cout<<fixed<<setprecision(2)<<y<<" "<<x<<endl;} return 0; }

#include<bits/stdc++.h> using namespace std; int main() { double a,b,c,d,e,f; cin>>a>>b>>c; d=sqrt(bb-4ac); if((bb-4ac)>=0){ e=(-b+d)/(2a); f=(-b-d)/(2a); if(e>=f) {cout<<fixed<<setprecision(2)<<f<<" "<<e; } if(e<f) {cout<<fixed<<setprecision(2)<<e<<" "<<f; } } }

#include<bits/stdc++.h> using namespace std; int main() { double a,b,c; double x1,x2,d; cin>>a>>b>>c; d=sqrt(bb-4.00ac); x1=(-b+d)/(2.00a); x2=(-b-d)/(2.00*a); cout<<fixed<<setprecision(2)<<x1<<' '<<x2<<' '<<endl; return 0; }

#include<bits/stdc++.h> using namespace std; int main() { double a,b,c; double x1,x2,d; cin>>a>>b>>c; d=bb-4ac; x1=(-b+d)/(2a); x2=(-b-d)/(2*a); cout<<fixed<<setprecision(2)<<x1<<" "<<x2<<endl; return 0; }

aasds

1. #include<bits/stdc++.h> using namespace std; int main() { double a,b,c; double x1,x2,d; cin>>a>>b>>c; d=bb-4.00ac; x1=(-b+d)/(2.00a); x2=(-b-d)/(2.00*a); cout<<fixed<<setprecision(2)<<x1<<" "<<x2<<endl; return 0; }