URI Online Judge Solution 1012 Area - URI 1012 Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1012 Area - URI 1012 Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1012 Area | Beginner
URI Problem Link - https://www.urionlinejudge.com.br/judge/en/problems/view/1012

Problem Name: 1012 Area code
Problem Number : URI - 1012 Area solution
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)

URI Solution 1012 Area Code in C/ URI 1012 solution in C:

URI 1012 in simple way:

#include <stdio.h>

int main()
{
 double a, b, c;

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

 printf("TRIANGULO: %.3lf\n", (a * c) / 2);
 printf("CIRCULO: %.3lf\n", c * c * 3.14159);
 printf("TRAPEZIO: %.3lf\n", ((a + b) / 2) * c);
 printf("QUADRADO: %.3lf\n", b * b);
 printf("RETANGULO: %.3lf\n", a * b);

 return 0;
}

URI 1012 in advanced way (Using while loop):

#include <stdio.h>
#define pi 3.14159
int main ()
{
    float a,b,c;
    while (scanf("%f %f %f",&a,&b,&c) != EOF)
    {
 
        printf ("TRIANGULO: %.3f\n",.5*(a*c));
        printf ("CIRCULO: %.3f\n",pi*(c*c));
        printf ("TRAPEZIO: %.3f\n",.5*(a+b)*c);
        printf ("QUADRADO: %.3f\n",b*b);
        printf ("RETANGULO: %.3f\n",a*b);
 
    }
    return 0;
 
}

URI Solution 1012 Area Code / URI 1012 Area solution in CPP:

#include <cstdio>

int main()
{
 double a, b, c;

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

 printf("TRIANGULO: %.3lf\n", (a * c) / 2);
 printf("CIRCULO: %.3lf\n", c * c * 3.14159);
 printf("TRAPEZIO: %.3lf\n", ((a + b) / 2) * c);
 printf("QUADRADO: %.3lf\n", b * b);
 printf("RETANGULO: %.3lf\n", a * b);

 return 0;
}

URI Solution 1012 Area Code / URI 1012 Area solution in Java:

import java.util.Scanner;

public class Main {

 public static void main(String[] args) {
  double a, b, c;

  Scanner sc = new Scanner(System.in);
  a = sc.nextDouble();
  b = sc.nextDouble();
  c = sc.nextDouble();

  System.out.printf("TRIANGULO: %.3f\n", (a * c) / 2);
  System.out.printf("CIRCULO: %.3f\n", c * c * 3.14159);
  System.out.printf("TRAPEZIO: %.3f\n", ((a + b) / 2) * c);
  System.out.printf("QUADRADO: %.3f\n", b * b);
  System.out.printf("RETANGULO: %.3f\n", a * b);

 }

}

URI Solution 1012 Area Code / URI 1012 Area solution in  Python:

valor = input().split(" ")

a, b, c = valor
pi = 3.14159

triangulo = (float(a) * float(c))/2
circulo = pi * (float(c)* float(c))
trapezio = float(c) *(float(a) + float(b)) / 2
quadrado = float(b) * float(b)
retangulo = float(a) * float(b)


print("TRIANGULO: %0.3f\nCIRCULO: %0.3f\nTRAPEZIO: %0.3f\nQUADRADO: %0.3f\nRETANGULO: %0.3f" % (triangulo, circulo, trapezio, quadrado, retangulo))

URI Solution 1012 Area Code / URI 1012 Area solution in  C# (C Sharp):

Demonstration:

By the rules of all of the rectangles, just do this..
Just implement this in coding. Since having any problem just put a comment below. Thanks



Tags: URI Online Judge Solution, URI OJ Solution list, URI Problems Solution, URI solver, URI all problem solution list, URI area code in C, URI 1012 Area code in C++, URI Area solution in C, URI solution, URI 1012 Area solution in C,URI 1012 Area solution in C++-CPP,URI 1012 Area solution in C# (C sharp),URI 1012 Area solution in Java,URI 1012 Area solution in Python,