According to Euler's Formula -- Faces + Vertices — Edges = 2

#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define ff first
#define ss second
 
void solvr()
{
    char a, b;
    cin >> a >> b;
    ll one, two;
    cin >> one >> two;
    bool f = false;
    bool v = false;
    bool e = false;
    if(a == 'F')
    {
        f = true;
    }
    else if(a == 'V')
    {
        v = true;
    }
    else
    {
        e = true;
    }
    if(b == 'F')
    {
        f = true;
    }
    else if(b == 'V')
    {
        v = true;
    }
    else
    {
        e = true;
    }
    if(f && v)
    {
        cout << "E" << endl;
        cout << one + two - 2 << endl;
    }
    else if(f && e)
    {
        cout << "V" << endl;
        if(a == 'E')
        {
            cout << one + 2 - two << endl;
        }
        else
        {
            cout << two + 2 - one << endl;
        }
    }
    else
    {
        cout << "F" << endl;
        if(a == 'E')
        {
            cout << one + 2 - two << endl;
        }
        else
        {
            cout << two + 2 - one << endl;
        }
 
    }
}
 
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int t = 1;
    //cin >> t;
    while(t--)
    {
        solvr();
    }
}

n ==> (k-2)(180) / k [Interior angle formula for k-sided regular polygon] kn = 180k — 360 360 = k(180-n) 180-n should be divisible by 360, and k should be greater than 2

Be careful, n equal to 180 will give runtime error if not handled as division by zero is invalid.

#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define ff first
#define ss second
 
void solvr()
{
    ll n;
    cin >> n;
    if((180-n) > 0 && 360 % (180-n) == 0 && (360 / (180-n)) > 2)
    {
        cout << "YES" << endl;
        cout << 360 / (180-n) << endl;
    }
    else
    {
        cout << "NO" << endl;
    }
}
 
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int t = 1;
    //cin >> t;
    while(t--)
    {
        solvr();
    }
}