URI Online Judge Solution 1045 Triangle Types Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1045 Triangle Types Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1045 Triangle Types | Beginner
URI Problem Link – https://www.urionlinejudge.com.br/judge/en/problems/view/1045

Problem Name: 1045 Triangle Types
Problem Number : URI – 1045 Triangle Types
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)

URI 1045 Triangle Types Code in C / URI 1045 solution in C:

#include <stdio.h>

int main()

{

    double a, b, c, temp;

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

    if (a < b)

    {
        temp = a;
        a = b;
        b = temp;
    }

    if (b < c)

    {
        temp = b;
        b = c;
        c = temp;
    }

    if (a < b)
    {
        temp = a;
        a = b;
        b = temp;
    }

    if (a >= b + c)

    {
        printf("NAO FORMA TRIANGULOn");
    }

    else if (a * a == b * b + c * c)

    {
        printf("TRIANGULO RETANGULOn");
    }

    else if (a * a > b * b + c * c)

    {
        printf("TRIANGULO OBTUSANGULOn");
    }

    else if (a * a < b * b + c * c)

    {
        printf("TRIANGULO ACUTANGULOn");
    }

     

    if (a == b && b == c)

    {
        printf("TRIANGULO EQUILATEROn");
    }

    else if (a == b || b == c)

    {
        printf("TRIANGULO ISOSCELESn");
    }

    return 0;

}

URI 1045 Triangle Types Code in C++ / URI 1045 solution in CPP:

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main()
{
 float x;
 vector<float> v;

 for (int i = 0; i < 3; ++i)
 {
  cin >> x;
  v.push_back(x);
 }

 sort(v.begin(), v.begin() + 3);
 float a, b, c;

 c = v[0];
 b = v[1];
 a = v[2];

 if(a >= (b + c)){
  cout << "NAO FORMA TRIANGULO" << endl;
 }else{

  if(a*a == (b*b + c*c)){
   cout << "TRIANGULO RETANGULO" << endl;
  }else if(a*a > (b*b + c*c)){
   cout << "TRIANGULO OBTUSANGULO" << endl;
  }else{
   cout << "TRIANGULO ACUTANGULO" << endl;
  }

  if(a == b && b == c){
   cout << "TRIANGULO EQUILATERO" << endl;
  }

  if((a == b && a != c) || (b == c && b != a) || (c == a && c != b)){
   cout << "TRIANGULO ISOSCELES" << endl;
  }
 }

 return 0;
}

import java.io.IOException;
import java.util.Scanner;

public class Main {
 
    public static void main(String[] args) throws IOException {
 
        double A, B, C;
  Scanner input =new Scanner(System.in);
  A = input.nextDouble();
  B = input.nextDouble();
  C = input.nextDouble();
  double tempA = Math.max(A, Math.max(B, C));
  double tempB = 0;
  double tempC = 0;
  
  if (tempA == A) {
   tempB =Math.max(B, C);
   tempC =Math.min(B, C);
  }
  if (tempA == B) {
   tempB =Math.max(A, C);
   tempC =Math.min(A, C);
  }
  if (tempA == C) {
   tempB =Math.max(B, A);
   tempC =Math.min(B, A);
  }
  //------------------------------
  if (tempA >= (tempB + tempC)) {
   System.out.print("NAO FORMA TRIANGULOn");
   
  }else if (tempA*tempA > ((tempB*tempB)+(tempC*tempC))) {
   System.out.print("TRIANGULO OBTUSANGULOn");
  }
  if (tempA*tempA == ((tempB*tempB)+(tempC*tempC))) {
   System.out.print("TRIANGULO RETANGULOn");
  }

  if (tempA*tempA < ((tempB*tempB)+(tempC*tempC))) {
   System.out.print("TRIANGULO ACUTANGULOn");
  }
  if ((tempA == tempB) &&(tempA == tempC)) {
   System.out.print("TRIANGULO EQUILATEROn");
  }
  if (((tempA == tempB) &&(tempA != tempC)) || ((tempA == tempC) &&(tempA != tempB)) || ((tempB == tempC) &&(tempB != tempA)) ) {
   System.out.print("TRIANGULO ISOSCELESn");
  }
 
    }
 
}

URI Solution 1045 Triangle Types Code / URI 1045 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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