Problem:
Score : 300 points

Problem Statement

Given is a positive integer L.
Find the maximum possible volume of a rectangular cuboid whose sum of the dimensions (not necessarily integers) is L.

Constraints

1 ≤ L ≤ 1000

L is an integer.

Input

Input is given from Standard Input in the following format:

L

Output

Print the maximum possible volume of a rectangular cuboid whose sum of the dimensions (not necessarily integers) is L.
Your output is considered correct if its absolute or relative error from our answer is at most 10^{-6}.

Sample Input 1

3

Sample Output 1

1.000000000000

For example, a rectangular cuboid whose dimensions are 0.8, 1, and 1.2 has a volume of 0.96.

On the other hand, if the dimensions are 1, 1, and 1, the volume of the rectangular cuboid is 1, which is greater.

Sample Input 2

999

Sample Output 2

36926037.000000000000

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

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

#define int             long long int
#define ld              long double
#define ff              first
#define ss              second
#define mp              make_pair
#define pb              push_back
#define si              set <int>
#define vi              vector <int>
#define pii             pair <int, int>
#define vpii            vector <pii>
#define vpp             vector <pair<int, pii>>
#define mii             map <int, int>
#define mpi             map <pii, int>
#define msi             map < string, int >
#define loop(i,s,e)     for(int i=s;i<e;i++)
#define rloop(i,e,s)    for(int i=e;i>=s;i--)
#define mset(a,f)       memset(a,f,sizeof(a))
#define spi             set <pii>
#define endl            "\n"
#define M               1000000007
#define sz(x)           ((int) x.size())
#define all(p)          p.begin(), p.end()
#define que_max         priority_queue <int>
#define que_min         priority_queue <int, vi, greater<int>>
#define bug(...)        __f (#__VA_ARGS__, __VA_ARGS__)
#define print(a)        for (auto x : a) cout << x << " "; cout << endl
#define Print(a,x,y)    for (int i=x; i<y; i++) cout<< a[i]<< " "; cout << endl

inline int power(int a, int b)
{
	int x = 1;
	while (b)
	{
		if (b & 1) x *= a;
		a *= a;
		b >>= 1;
	}
	return x;
}

template <typename Arg1>
void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }
template <typename Arg1, typename... Args>
void __f (const char* names, Arg1&& arg1, Args&&... args)
{
	const char* comma = strchr (names + 1, ',');
	cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);
}

const int N = 2000005;




void solve()
{
	int s;
	cin >> s;
	long double x = (double)s / 3;
	long double p = x * x * x;
	cout << setprecision(12) << fixed << p;
	//bug(x);
	return;

}

int32_t main()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);


	//cout << setprecision(9) << fixed;
	clock_t z = clock();
	int t = 1;
	//cin >> t;
	while (t--)
	{
		solve();
	}
	cerr << "Total Time : " << ((double)(clock() - z) / CLOCKS_PER_SEC);

	return 0;
}