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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
 
int main(void){
  
  int N;
  cin >> N;
  int A[N];
  int num[N];
  long long int ans[N];
  long long int wholeAns=0;
  for(int i=0; i<N; i++){//読み込み初期化
    cin >> A[i];
    num[i]=0;
    ans[i]=0;
  }
  for(int i=0; i<N; i++){//数を数える
    num[A[i]-1]=num[A[i]-1]+1;
  }
  for(int i=0; i<N; i++){//Cを計算
    if(num[i]>1){
      ans[i] = (num[i]*(num[i]-1))/2;
      wholeAns += ans[i];
    }
  }
  
  for(int i=0; i<N; i++){//回答
    if(i!=0) cout << "\n";
    long long int rep=wholeAns;
    int nnum = num[A[i]-1]-1;
    if(nnum>0) rep = rep - num[A[i]-1] + 1;
    /*if(nnum>1) rep -= nnum*(nnum-1)/2;
    nnum--;
    if(nnum>1) rep += nnum*(nnum-1)/2;*/
    cout << rep;
  }
  
  return 0;
}