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

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++ 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;
}

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

# 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# 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 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
?>

<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>

YES

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