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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <set>
#include <queue>
#include <map>
 
using namespace std;
typedef long long int ll;
typedef pair <int,int> pii;
typedef pair<ll,ll> pll;
 
 
/*bool compare_function(const pair<int, int> a, const pair<int, int> b)
{
    return a.first < b.first;
}*/
// use case
//sort(V.begin(), V.end(), compare_function);
 
/*struct compare
{
    bool operator ()(const pair<int, int> &a, const pair<int, int> &b) {
    if (a.second-a.first==b.second-b.first)
    {
        return a.first>b.first;
    }
    return a.second-a.first < b.second-b.first;}
};
priority_queue <pair<int,int>, vector <pair<int,int>>, compare> Q;*/
 
/*void print1(vector <int> X)
{
    for (int i=0; i<X.size(); i++)
    {
        cout<<X[i]<<" ";
    }
    cout<<endl;
    return;
}
 
void print2(vector <vector <int>> X)
{
    for (int i=0; i<X.size(); i++)
    {
    for (int j=0;j<X[i].size();j++) {
        cout<<X[i][j]<<" ";}
    cout<<endl;}
    return;
}*/
int gcd(int a, int b) {
	if (a==b || b==0) {return a;}
	if (b>a) {return gcd(b,a);}
	return gcd(b,a%b);
	}
 
 
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
	int x;
	cin>>x;
	cout<<360/gcd(360,x)<<endl;
    return 0;
}