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<bits/stdc++.h>
using namespace std;
#define ll long long int
#define endl "\n"
#define F first
#define S second
#define def(x) cerr << #x << " is " << x << endl;
#define mod 1000000007
#define pb push_back
#define FOR(i,a,n) for(ll i=a;i<n;i++)
#define ios ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define all(a) a.begin(),a.end()
const int NUM = 2e5 + 5;
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }

void err(istream_iterator<string> it) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
    cerr << *it << " = " << a << endl;
    err(++it, args...);
}
int main(){
    ios
    int num_tests=1; 
    //cin>>num_tests;
    while(num_tests-->0){
        ll ang; cin>>ang;
        cout<<360/__gcd((ll)360,ang);
    }   
}