Given an angle 
where, 
. The task is to check whether it is possible to make a regular polygon with all of its interior angle equal to . 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 
where, n is the number of sides in the polygon.
This can be written as 
.
On rearranging terms we get, 
.
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 codeint 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 Codeif __name__ == "__main__" : a = 90 # function calling makePolygon(a) # This code is contributed # by ANKITRAI1 |
C#
// C# implementation of // above approachusing 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 answermakePolygon($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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