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 <algorithm>
#include <string>
#include <stdlib.h>
#include <vector>
#include <climits>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <map>
#include <unordered_map>
#include <deque>
#include <set>
#include <stack>
#include <iomanip>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std; 
#define Endl '\n'
#define ll long long int
/*I can do this*/
int main()
{
    int n;
    cin>>n;
    for (int i = 1; i < 10000; ++i)
    {
        if ((i*n)%360==0)
        {
            cout<<i;
            break;
        }
    }
}