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

URI Online Judge Solution URI Solution -Beginner

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

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;
}

URI 1045 Triangle Types Code in java/ URI 1045 solution in Java:

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

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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