40pts code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main() 
{
    ll n,a,k,b;
    cin>>n>>a>>k>>b;
    if(n%2!=0)
	{
        cout<<-1<<endl;
        return 0;
    }
    ll sum1=((n/2)/k)*b+((n/2)%k)*a,sum2=(n/2)*a;
    ll ans=min(sum1,sum2);
    cout<<ans*2<<endl;
    return 0;
}

求大佬一调，违规紫珊