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>
#define rep(i, n) for (int i=0; i<n; ++i)
#define rep1(i, n) for (int i=1; i<=n; ++i)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
constexpr ll  MOD = (1e9+7);
constexpr int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
constexpr int lcm(int a, int b) { return a / gcd(a, b) * b; }

template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
ll factorial(ll n, ll m=2) {
    // calculate n!
    m = max(2LL, m);
    ll rtn = 1;
    for (ll i=m; i<=n; i++) {
        rtn = (rtn * i) % MOD;
    }
    return rtn;
}
ll modinv(ll a, ll m) {
    ll b = m, u = 1, v = 0;
    while (b) {
        ll t = a / b;
        a -= t * b;
        swap(a, b);
        u -= t * v;
        swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}
ll modpow(ll a, ll n) {
    ll res = 1;
    while (n > 0) {
        if (n & 1)
            res = res * a % MOD;
        a = a * a % MOD;
        n >>= 1;
    }
    return res;
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int l; cin >> l;
    double ans = (double)pow((double)l/3, (double)3);
    cout << setprecision(10) << ans << endl;
    return 0;
}