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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println(Math.pow(sc.nextDouble()/3,3));
    }
    static long factorial(long num){
        if (num<2)return 1;
        else return num*factorial(num-1);
    }
    static int arraySum(int[] array){
        int ans = 0;
        for (int value : array) ans += value;
        return ans;
    }
    static long combination(long a, long b){
        long data = 1;
        for (int i=0;i<b;i++){
            data*=a-i;
        }
        return data/factorial(b);
    }
    static double log(double base,double antilogarithm){
        return Math.log(antilogarithm)/Math.log(base);
    }

    static long gcd(long x,long y){
        if (y==0)return x;
        else return gcd(y,x%y);
    }
    static long lcm(long x,long y){
        return x/gcd(x,y)*y;
    }
    static HashMap<Long,Long> factorization(long num){
        HashMap<Long,Long> hash = new HashMap<>();
        long n = num;
        long count = 2;
        while (n>1){
            while (n%count==0){
                n/=count;
                if (hash.containsKey(count))hash.put(count,hash.get(count)+1);
                else hash.put(count,1L);
            }
            count++;
        }
        return hash;
    }

}