Check if it is possible to create a polygon with a given angle

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Given an angle $a$ where, $1\le a< 180$ . The task is to check whether it is possible to make a regular polygon with all of its interior angle equal to $a$ . If possible then print “YES”, otherwise print “NO” (without quotes).
Examples:

Input: angle = 90
Output: YES
Polygons with sides 4 is
possible with angle 90 degrees.

Input: angle = 30
Output: NO

Approach: The Interior angle is defined as the angle between any two adjacent sides of a regular polygon.
It is given by    $\;Interior\;angle = \frac{180 \times (n-2)}{n}\;$ where, n is the number of sides in the polygon.
This can be written as    $\;a = \frac{180 \times (n-2)}{n}\;$ .
On rearranging terms we get,    $\;n = \frac{360}{180 - a}\;$ .
Thus, if n is an Integer the answer is “YES” otherwise, answer is “NO”.
Below is the implementation of the above approach:

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check whether it is possible
// to make a regular polygon with a given angle.
void makePolygon(float a)
{
    // N denotes the number of sides
    // of polygons possible
    float n = 360 / (180 - a);
    if (n == (int)n)
        cout << "YES";
    else
        cout << "NO";
}
 
// Driver code
int main()
{
    float a = 90;
 
    // function to print the required answer
    makePolygon(a);
 
    return 0;
}

Java

class GFG 
{
// Function to check whether 
// it is possible to make a
// regular polygon with a given angle. 
static void makePolygon(double a) 
{ 
    // N denotes the number of 
    // sides of polygons possible 
    double n = 360 / (180 - a); 
    if (n == (int)n) 
        System.out.println("YES"); 
    else
        System.out.println("NO"); 
} 
 
// Driver code 
public static void main (String[] args) 
{
    double a = 90; 
 
    // function to print
    // the required answer 
    makePolygon(a); 
}
}
 
// This code is contributed by Bilal

Python3

# Python 3 implementation 
# of above approach 
 
# Function to check whether 
# it is possible to make a
# regular polygon with a 
# given angle. 
def makePolygon(a) :
 
    # N denotes the number of sides 
    # of polygons possible
    n = 360 / (180 - a)
 
    if n == int(n) :
        print("YES")
 
    else :
        print("NO")
 
# Driver Code
if __name__ == "__main__" :
    a = 90
 
    # function calling 
    makePolygon(a)
     
# This code is contributed 
# by ANKITRAI1

C#

// C# implementation of 
// above approach
using System;
 
class GFG 
{
// Function to check whether 
// it is possible to make a
// regular polygon with a 
// given angle. 
static void makePolygon(double a) 
{ 
    // N denotes the number of 
    // sides of polygons possible 
    double n = 360 / (180 - a); 
    if (n == (int)n) 
        Console.WriteLine("YES"); 
    else
        Console.WriteLine("NO"); 
} 
 
// Driver code 
static void Main() 
{
    double a = 90; 
 
    // function to print
    // the required answer 
    makePolygon(a); 
}
}
 
// This code is contributed by mits

PHP

<?php 
// PHP implementation of above approach
 
// Function to check whether it 
// is possible to make a regular
// polygon with a given angle.
function makePolygon($a)
{
    // N denotes the number of 
    // sides of polygons possible
    $n = 360 / (180 - $a);
    if ($n == (int)$n)
        echo "YES";
    else
        echo "NO";
}
 
// Driver code
$a = 90;
 
// function to print the
// required answer
makePolygon($a);
 
// This code is contributed 
// by ChitraNayal
?>

Javascript

<script>
 
      // JavaScript implementation of above approach
      // Function to check whether it is possible
      // to make a regular polygon with a given angle.
       
      function makePolygon(a) 
      {
        // N denotes the number of sides
        // of polygons possible
        var n = parseFloat(360 / (180 - a));
        if (n === parseInt(n)) 
        document.write("YES");
        else
        document.write("NO");
      }
 
      // Driver code
      var a = 90;
       
      // function to print the required answer
      makePolygon(a);
       
</script>

Output:

YES

Time Complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.

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