Problem:
Score : 500 points

Problem Statement

There are N children standing in a line from left to right. The activeness of the i-th child from the left is A_i.

You can rearrange these children just one time in any order you like.

When a child who originally occupies the x-th position from the left in the line moves to the y-th position from the left, that child earns A_x \times |x-y| happiness points.

Find the maximum total happiness points the children can earn.

Constraints

2 \leq N \leq 2000

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum total happiness points the children can earn.

Sample Input 1

4
1 3 4 2

Sample Output 1

20

If we move the 1-st child from the left to the 3-rd position from the left, the 2-nd child to the 4-th position, the 3-rd child to the 1-st position, and the 4-th child to the 2-nd position, the children earns 1 \times |1-3|+3 \times |2-4|+4 \times |3-1|+2 \times |4-2|=20 happiness points in total.

Sample Input 2

6
5 5 6 1 1 1

Sample Output 2

58

Sample Input 3

6
8 6 9 1 2 1

Sample Output 3

85

Solution in Rust (1.42.0) (judge verdict: Time Limit Exceeded)

extern crate core;

use std::collections::HashMap;
use std::io::{self, Read};

#[macro_export]
macro_rules! input {
  ($s:expr=>$($t:tt)*) => {
    let mut lines=$s.split("\n");
    $(
        line_parse!(lines,$t);
    )*
  };
}

macro_rules! line_parse {
  ($lines:expr,($($name:ident:$t:tt)*)) => {
    let mut line=$lines.next().unwrap().split_whitespace();
    $(value_def!(line,$name,$t);)*
  };

  //複数行
  ($lines:expr,{$n:expr;$name:ident:$t:tt}) => {
    values_def!($lines,$n,$name,$t);
  };
}

macro_rules! value_def {
  ($line:expr, $name:ident, $t:tt) => {
    let $name = value!($line, $t);
  };
}

macro_rules! values_def {
  ($lines:expr, $n:expr, $name:ident, $t:tt) => {
    let $name = {
      let mut vec = Vec::new();
      for i in 0..$n {
        let mut next = $lines.next().unwrap().split_whitespace();
        vec.push(value!(next, $t));
      }
      vec
    };
  };
}

macro_rules! value {
  //配列
  ($line:expr,[$t:tt]) => {
    $line.map(|x|{
      let mut iter=::std::iter::once(x);
      value!(iter,$t)
    }).collect::<Vec<_>>()
  };
  //タプル
  ($line:expr,($($t:tt),*)) => {
    ($(value!($line,$t),)*)
  };
  //文字列
  ($line:expr,#) => {
    $line.next().unwrap()
  };
  //インデックス(-1)
  ($line:expr,@) => {
    $line.next().unwrap().parse::<usize>().unwrap()-1
  };
  //単一値
  ($line:expr,$t:ty) => {
    $line.next().unwrap().parse::<$t>().unwrap()
  };
}

fn main() {
  let mut input = String::new();
  io::stdin().read_to_string(&mut input).unwrap();
  let output = solve(input.trim().to_string());
  println!("{}", output);
}

fn solve(input: String) -> String {
  input!(input=>(n:usize)(list:[i64]));
  f(&list, &mut HashMap::new(), 0, {
    let mut v = Vec::with_capacity(n);
    v.resize(n, false);
    v
  })
  .to_string()
}

// リスト、次並べる場所、各幼児が利用済みか、嬉しさの合計
// i以降の席に並べるときの嬉しさの最大値
fn f(
  list: &Vec<i64>,
  memo: &mut HashMap<(usize, Vec<bool>), i64>,
  i: usize,
  used: Vec<bool>,
) -> i64 {
  if let Some(res) = memo.get(&(i, used.clone())) {
    return *res;
  }

  let res = used
    .clone()
    .into_iter()
    .enumerate()
    .filter(|(_, u)| !u)
    .map(|(x, _)| x)
    .map(|p| {
      // pをiに移動する時
      let mut used = used.clone();
      used[p] = true;
      list[p] * (i as i64 - p as i64).abs() + f(list, memo, i + 1, used)
    })
    .max()
    .unwrap_or(0);

  memo.insert((i, used), res);
  res
}

macro_rules! tests {
    ($($name:ident: $input:expr=>$output:expr,)*) => {
        mod tests {
            $(
                #[test]
                fn $name() {
                    assert_eq!($output.trim().to_string(),super::solve($input.trim().to_string()));
                }
            )*
        }
    }
}
tests! {
  test1: "4\n1 3 4 2\n" => "20\n",
  test2: "6\n5 5 6 1 1 1\n" => "58\n",
  test3: "6\n8 6 9 1 2 1\n" => "85\n",
}