Problem: Example

Input

2
10 10


Output

40.00000000

Solution (in CPP):

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

typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;
typedef pair<LL, LL> PLL;

#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

#define FF first
#define SS second
template<class S, class T>
istream& operator>>(istream& is, pair<S,T>& p){
  return is >> p.FF >> p.SS;
}

const double EPS = 1e-10;
const double PI  = acos(-1.0);
const LL MOD = 1e9+7;

int main(){
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  int N; cin >> N;
  VI rs(N); REP(i,N) cin >> rs[i];
  vector<double> xs(N);
  double ans = 0.;
  REP(i,N){
	xs[i] = rs[i];
	REP(j,i)
	  xs[i] = max(xs[i], xs[j]+2*sqrt(rs[i]*rs[j]));
	ans = max(ans, xs[i]+rs[i]);
  }
  cout << fixed << setprecision(9) << ans << endl;
  
  
  return 0;
}