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;
typedef long long ll;
typedef long double ld;
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define endl '\n'
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define rep(i, a, b) for (ll i = a; i < (b); ++i)
#define reb(i, a, b) for (ll i = a; i >= (b); --i)
#define trav(a, x) for (auto &a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (ll)(x).size()
typedef pair<ll, ll> pii;
typedef vector<ll> vi;
int main()
{
   fast;
  int k;
  cin>>k;
  rep(i,1,360+1)
  {
    if((k*i)%360==0)
    {
      cout<<i;
      return 0;
    }
  }
    
}