Problem:
Score : 200 points

Problem Statement

Takahashi is standing on a two-dimensional plane, facing north. Find the minimum positive integer K such that Takahashi will be at the starting position again after he does the following action K times:

Go one meter in the direction he is facing. Then, turn X degrees counter-clockwise.

Constraints

1 \leq X \leq 179

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

Print the number of times Takahashi will do the action before he is at the starting position again.

Sample Input 1

90

Sample Output 1

4

Takahashi's path is a square.

Sample Input 2

1

Sample Output 2

360

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
class Main{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
  int x=sc.nextInt();
  int ans=0;

  System.out.println(lcm(360,x)/x);
}

 static int gcd(int a, int b) 
    { 
    if (a == 0) 
        return b;  
    return gcd(b % a, a);  
    } 
      
    // method to return LCM of two numbers 
    static int lcm(int a, int b) 
    { 
        return (a*b)/gcd(a, b); 
    } 
}