Problem: Example

Input

4
5
8
58
85


Output

2970.000000000

Solution (in CPP):

Solution:
#include<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<list>
#include<bitset>
#include<stack>
#include<memory>
#include<numeric>
using namespace std;
using namespace std::chrono;
typedef long long int llint;
typedef long double lldo;
#define mp make_pair
#define mt make_tuple
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define fir first
#define sec second
#define res resize
#define ins insert
#define era erase
/*cout<<fixed<<setprecision(20);cin.tie(0);ios::sync_with_stdio(false);*/
const llint mod=1000000007 ;
const llint big=2.19e15+1;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-10;
const long double epsA=1e-4;
template <class T,class U>bool mineq(T& a,U b){if(a>b){a=b;return true;}return false;}
template <class T,class U>bool maxeq(T& a,U b){if(a<b){a=b;return true;}return false;}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint b){if(a==0){return b;}return a/gcd(a,b)*b;}
template<class T> void SO(T& ve){sort(ve.begin(),ve.end());}
template<class T> void REV(T& ve){reverse(ve.begin(),ve.end());}
template<class T>llint LBI(const vector<T>&ar,T in){return lower_bound(ar.begin(),ar.end(),in)-ar.begin();}
template<class T>llint UBI(const vector<T>&ar,T in){return upper_bound(ar.begin(),ar.end(),in)-ar.begin();}
double solve(vector<double>r){
	//53124
	int i,j,n=r.size();
	vector<double>s(n);
	for(i=1;i+1<n;i++){s[i]=r[i-1]*r[i+1];}
	s[0]=r[0]*r[1];s[n-1]=r[n-2]*r[n-1];
	if(n==4){return s[0]+s[1]+s[2]+s[3];}
	double bmin=0,bmax=s[0]*s[1]*1.01;
	//角度合計を2pai にする ただし、n=3のみpaiにする。
	double moku=pai*2;
	if(n==3){moku=pai;}
	for(int h=0;h<40;h++){
		double bgen=(bmin+bmax)/2;
		double kak=0;
		for(i=0;i<n;i++){
			double sor=bgen/s[i];
			if(sor<1){kak+=acos(sor);}
		}
		if(kak>moku){bmin=bgen;}
		else{bmax=bgen;}
	}
	//bmaxを採用
	double ans=0;
	for(i=0;i<n;i++){
		double sor=bmax/s[i];
		//cerr<<sor<<endl;
		if(sor<1){ans+=sqrt(1-sor*sor)*s[i];}
	}
	//cerr<<"n="<<n<<"ans="<<ans<<endl;
	return ans;
}
int main(void){
	cout<<fixed<<setprecision(20);cin.tie(0);ios::sync_with_stdio(false);
	int i,n;cin>>n;
	double ans=0;
	vector<double>r(n);
	for(i=0;i<n;i++){cin>>r[i];}
	SO(r);REV(r);
	while(r.size()>=3){
		maxeq(ans,solve(r));
		r.pob();
	}
	cout<<ans/2<<endl;
	return 0;
}