Problem: Example

Input

4
5
8
58
85


Output

2970.000000000

Solution (in CPP):

Solution:
#define _USE_MATH_DEFINES
#include <algorithm>
#include <cstdio>
#include <functional>
#include <iostream>
#include <cfloat>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <time.h>
#include <vector>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> i_i;
typedef pair<ll, int> ll_i;
typedef pair<double, int> d_i;
typedef pair<ll, ll> ll_ll;
typedef pair<double, double> d_d;
struct edge { int u, v; ll w; };

ll MOD = 1000000007;
ll _MOD = 1000000009;
double EPS = 1e-9;

int main() {
	int N; cin >> N;
	vector<int> r(N);
	for (int i = 0; i < N; i++)
		cin >> r[i];
	sort(r.begin(), r.end());
	double ma = 0;
	for (int n = 3; n <= N; n++) {
		vector<int> a(r.end() - n, r.end());
		do {
			vector<double> x(n), y(n);
			for (int i = 0; i < n; i++) {
				x[i] = a[i] * cos(M_PI*2 / n * i);
				y[i] = a[i] * sin(M_PI*2 / n * i);
			}
			for (int t = 0; t < 10000; t++) {
				int i = rand() % n;
				int j = (i + n - 1) % n, k = (i + 1) % n;
				double dx = x[k] - x[j], dy = y[k] - y[j];
				double _dx = dy, _dy = -dx;
				double r = sqrt(_dx * _dx + _dy * _dy);
				x[i] = _dx / r * a[i];
				y[i] = _dy / r * a[i];
			}
			double sum = 0;
			for (int i = 0; i < n; i++) {
				int j = (i + 1) % n;
				sum += x[i] * y[j] - y[i] * x[j];
			}
			ma = max(ma, sum);
		} while (next_permutation(a.begin() + 1, a.end()));
	}
	printf("%.15f\n", ma / 2);
}