Problem: Example

Input

2
10 10


Output

40.00000000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) for(int i = 0; i < (n); i++)
#define repp(i, l, r) for(int i = (l); i < (r); i++)
#define per(i, n) for(int i = ((n)-1); i >= 0; i--)
#define perr(i, l, r) for(int i = ((r)-1); i >= (l); i--)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define IINF 1000000000
#define LINF 1000000000000000000
#define SP <<" "<<
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define CFS cin.tie(0);ios::sync_with_stdio(false)

typedef long long LL;
typedef long double LD;

int main(){
  int n;
  cin >> n;
  vector<double> r(n),p(n);
  rep(i,n){
    cin >> r[i];
  }
  p[0]=r[0];
  repp(i,1,n){
    double ans=r[i];
    rep(j,i){
      ans=max(ans,2*sqrt(r[i]*r[j])+p[j]);
    }
    p[i]=ans;
  }
  double ans=0;
  rep(i,n){
    ans=max(ans,p[i]+r[i]);
  }
  cout << fixed << setprecision(10) << ans << endl;
  return 0;
}