URI Online Judge | 1036 Bhaskara's Formula (Solution)

April 23, 2015

Problem Link:
https://www.urionlinejudge.com.br/judge/en/problems/view/1036

Read 3 floating-point numbers. After, print the roots of bhaskara’s formula. If was impossible to calculate the roots because a division by zero or a square root of a negative number, presents the message “Impossivel calcular”.

Input

Read 3 floating-point numbers A, B and C.

Output

Print the result with 5 digits after the decimal point or the message if it is impossible to calculate.

Sample Inputs	Sample Outputs
10.0 20.1 5.1	R1 = -0.29788 R2 = -1.71212

0.0 20.0 5.0Impossivel calcular

10.3 203.0 5.0R1 = -0.02466
R2 = -19.68408

10.0 3.0 5.0Impossivel calcular

Solution:

#include<stdio.h>

#include<math.h>

int main()

{

    double a,b,c,r1,r2,del;

    scanf("%lf %lf %lf",&a,&b,&c);

    del=(b*b)-(4*a*c);

    r1=(-b+sqrt(del))/(2*a);

    r2=(-b-sqrt(del))/(2*a);

    if(a!=0 && del>0)

{

        printf("R1 = %.5lf\nR2 = %.5lf\n",r1,r2);

}

    else printf("Impossivel calcular\n");

    return 0;

}

Comments

Sadat hasan shehabAugust 28, 2015 at 1:12 PM
This comment has been removed by the author.
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Morra96August 29, 2016 at 8:45 AM
#include
#include
int main()
{
double a,b,c;

scanf("%lf",&a);
scanf("%lf",&b);
scanf("%lf",&c);

double postive_root,negative_root;
double Z;

Z = sqrt( pow(b,2) - (4 * a * c ) );
postive_root = (-b + Z)/(2*a);
negative_root = (-b - Z)/(2*a);

if( a == 0 || Z<0 )
{
printf("Impossivel calcular\n");
}
else
{
printf("R1 =%.5lf\n",postive_root);
printf("R2 =%.5lf\n",negative_root);
}
return 0;
}
//why doesn't this work
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UnknownJuly 19, 2017 at 1:06 AM
#include
#include
#include
using namespace std;

int main()
{
double a , b , c ,r1 , r2 , sq ;
cin >> a >> b >> c ;
sq = sqrt((b*b) - (4*a*c));
r1 = (-b+sq)/(2*a);
r2 = (-b-sq)/(2*a);
if (a!=0 && sq >0)
{
cout << "R1 = "<<fixed<<setprecision(5)<<r1<<endl;
cout << "R2 = "<<fixed<<setprecision(5)<<r2<<endl;
}
else
cout<< "Impossivel calcular" << endl;

return 0;
}
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UnknownJuly 31, 2017 at 11:28 AM
#include
#include
using namespace std;
int main()
{
double a,b,c,r1,r2,d;
cin>>a>>b>>c;
d=pow(b,2)-4*a*c;
r1=(-b+sqrt(pow(b,2)-4*a*c))/(2*a);
r2=(-b-sqrt(pow(b,2)-4*a*c))/(2*a);
if(d>0&&a!=0)
{

cout<<"R1 = "<<fixed<<setprecision(5)<<r1<<"\n"<<"R2 = "<<fixed<<setprecision(5)<<r2<<"\n";
}
else
{
cout<<"Impossivel calcular/n";
}

return 0;
}
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Mohamed WaelApril 25, 2018 at 6:51 AM
A,B,C=map(float,input().strip().split())
if(A==0):
print("Impossivel calcular")
elif ((B**2)-4*A*C)<0:
print("Impossivel calcular")
else:
formula1=float(((-B+((B**2)-4*A*C)**(1/2))/(2*A)))
formula2=float((-B-((B**2)-4*A*C)**(1/2))/(2*A))
print("R1 = %.5f"%formula1)
print("R2 = %.5f"%formula2)
Here's the solution in Python if you want it in C++ i'll do it ;)
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