Algorithm
Problem Name: beecrowd | 2313
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/2313
Which Triangle
By Alexandre A. Melo, IFSC 
Brazil
Timelimit: 1
Given three values, find out if they form a triangle. If so, check if the triangle is scalene, isoceles or equilateral and if it is a triangle rectangle or not.
Input
Input is given by three integers A,B e C (0 < A,B,C < 105).
Output
The output must be the one single line containing the string "Invalido" if the input values do not represent a triangle.
If the values can be the sides of a triangle the output must be "Valido-Equilatero" if such triangle is equilateral, "Valido-Escaleno" if it is scalene or "Valido-Isoceles" if it is isoceles. The next line of output must read "Retangulo: S" if the triangle is rectangle or "Retangulo: N" otherwise, as shown in the examples.
| Input Samples | Output Samples |
|
4 6 2 |
Invalido |
|
4 3 3 |
Valido-Isoceles |
|
3 4 5 |
Valido-Escaleno |
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
int main(void)
{
int i, j, a[3], temp;
for (i = 0; i < 3; ++i)
scanf("%i", &a[i]);
for (i = 0; i < 3-1; ++i)
{
for (j = i+1; j < 3; ++j)
{
if (a[j] < a[i])
{
temp = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
if (a[2] >= a[0]+a[1])
printf("Invalido\n");
else
{
if (a[0] == a[1] && a[1] == a[2])
printf("Valido-Equilatero\n");
else if (a[0] != a[1] && a[0] != a[2] && a[1] != a[2])
printf("Valido-Escaleno\n");
else
printf("Valido-Isoceles\n");
if (a[2]*a[2] == (a[0]*a[0]+a[1]*a[1]))
printf("Retangulo: S\n");
else
printf("Retangulo: N\n");
}
return 0;
}
Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
#include<bits/stdc++.h>
using namespace std;
int main()
{
int a,b,c;
string ans = "" ;
cin >> a >> b >> c ;
if(a+c > b){
if((a != b and b == c) or ( a == c and a != b) or ( a == b and c != b))
ans = "Valido-Isoceles";
else if(a == b and a == c)
ans = "Valido-Equilatero" ;
else if(a != b and b != c and a != c)
ans = "Valido-Escaleno";
}
else{
ans = "Invalido";
}
if(ans != "Invalido"){
if(pow(a,2) == pow(b,2) + pow(c,2) or pow(b,2) == pow(a,2) + pow(c,2) or pow(c,2) == pow(a,2) + pow(b,2))
cout << ans << endl << "Retangulo: S" << endl ;
else
cout << ans << endl << "Retangulo: N" << endl;
}
else
cout << ans << endl;
return 0;
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
var input = require('fs').readFileSync('/dev/stdin', 'utf8');
var lines = input.split('\n');
const numbers = lines.shift().split(" ");
numbers.sort((a,b) => a - b).reverse();
const [a, b, c] = numbers
const A = parseFloat(a);
const B = parseFloat(b);
const C = parseFloat(c);
if(A >= B + C){
console.log("Invalido");
return;
}
if(A == B && B == C){
console.log("Valido-Equilatero");
}
else if(A == B || A == C || B == C){
console.log("Valido-Isoceles");
}
else if(A != B && B != C && C != A){
console.log("Valido-Escaleno");
}
if(Math.pow(A, 2) == (Math.pow(B,2) + Math.pow(C,2))){
console.log("Retangulo: S");
}
else{
console.log("Retangulo: N");
}
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
A, B, C = input().split()
A = int(A)
B = int(B)
C = int(C)
if(A < B and A < C and B < C):
a = C
b = B
c = A
elif(A < B and A < C and C < B):
a = B
b = C
c = A
elif(B < A and B < C and A < C):
a = C
b = A
c = B
elif(B < A and B < C and C < A):
a = A
b = C
c = B
elif(C < A and C < B and A < B):
a = B
b = A
c = C
elif(C < A and C < B and B < A):
a = A
b = B
c = C
elif(A == B and A < C):
a = C
b = A
c = A
elif(A == B and C < A):
a = A
b = A
c = C
elif(A == C and A < B):
a = B
b = A
c = A
elif(A == C and B < A):
a = A
b = B
c = A
elif(B == C and B < A):
a = A
b = B
c = B
elif(B == C and A < B>:
a = B
b = A
c = B
else:
a = A
b = B
c = C
if(a >= b + c):
print("Invalido")
exit()
elif(a == b and b == c):
print("Valido-Equilatero")
elif(a == b or a == c or b == c):
print("Valido-Isoceles")
else:
print("Valido-Escaleno")
if(((a * a) == b * b + c * c) == True):
print("Retangulo: S")
else:
print("Retangulo: N")
Input
Output