URI Online Judge Solution 1036 Bhaskara’s Formula – URI 1036 Solution in C, C++, Java, Python and C#
URI Online Judge Solution 1036 Bhaskara’s Formula | Beginner
URI Problem Link – https://www.urionlinejudge.com.br/judge/en/problems/view/1036
Problem Name: 1036 Bhaskara’s Formula solution
Problem Number : URI – 1036 Bhaskara’s Formula code
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)
URI Solution 1036 Bhaskara’s Formula Code in C / URI 1036 solution in C:
#include<stdio.h>
#include <math.h>
int main()
{
double a, b, c, t;
scanf("%lf %lf %lf", &a, &b, &c);
if(((b * b) - 4 * a * c) < 0 || a == 0){
printf("Impossivel calcularn");
}
else{
t = sqrt((b * b) - 4 * a * c);
printf("R1 = %.5lfnR2 = %.5lfn", ((-b + t) / (2 * a)), ((-b - t) / (2 * a)));
}
return 0;
}
URI Solution 1036 Bhaskara’s Formula Code / URI 1036 solution in CPP:
#include<stdio.h>
#include <math.h>
int main()
{
double a, b, c, t;
scanf("%lf %lf %lf", &a, &b, &c);
if(((b * b) - 4 * a * c) < 0 || a == 0){
printf("Impossivel calcularn");
}
else{
t = sqrt((b * b) - 4 * a * c);
printf("R1 = %.5lfnR2 = %.5lfn", ((-b + t) / (2 * a)), ((-b - t) / (2 * a)));
}
return 0;
}
URI Solution 1036 Bhaskara’s Formula Code / URI 1036 solution in Java:
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
double A, B, C, R1, R2;
Scanner input =new Scanner(System.in);
A = input.nextDouble();
B = input.nextDouble();
C = input.nextDouble();
if ((A == 0) || (((B*B) -(4*A*C)) < 0)) {
System.out.print("Impossivel calcularn");
}else {
R1 = ((-B + Math.sqrt(((B*B) -(4*A*C)))) / (2*A));
R2 = ((-B - Math.sqrt(((B*B) -(4*A*C)))) / (2*A));
System.out.printf("R1 = %.5fn", R1);
System.out.printf("R2 = %.5fn", R2);
}
}
}
URI Solution 1036 Bhaskara’s Formula Code / URI 1036 solution in Python:
URI Solution 1036 Bhaskara’s Formula Code / URI 1036 solution in C# (C Sharp):
Demonstration:
Just implement this in coding. Since having any problem just put a comment below. Thanks
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