Problem: Example

Input

2
10 10


Output

40.00000000

Solution (in CPP):

Solution:
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <functional>
using namespace std;

#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(int (i)=(a); (i)<=(n); (i)++)
#define repr(i,a,n) for(int (i)=(a); (i)>=(n); (i)--)
#define int long long int

template<typename T> void chmax(T &a, T b) {a = max(a, b);}
template<typename T> void chmin(T &a, T b) {a = min(a, b);}
template<typename T> void chadd(T &a, T b) {a = a + b;}

typedef pair<int, int> pii;
typedef long long ll;

int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
constexpr ll INF = 1001001001001001LL;
constexpr ll MOD = 1000000007LL;

signed main() {
    int N; cin >> N;
    vector<double> a(N), rec(N);
    rep(i,0,N) cin >> a[i];
    double ans = 0.0;
    rep(i,0,N) {
        double temp = a[i];
        rep(j,0,i) {
            chmax(temp, rec[j] + 2 * sqrt(a[i] * a[j]));
        }
        rec[i] = temp;
        chmax(ans, rec[i] + a[i]);
    }
    printf("%.12f\n", ans);
    return 0;
}