Cut a rod for the most money

def cut_rod(prices, n):
    best = [0] * (n + 1)                  # best[t]: most a rod of length t earns
    for t in range(1, n + 1):
        for L in range(1, min(t, len(prices)) + 1):
            best[t] = max(best[t], prices[L - 1] + best[t - L])   # first piece of length L
    return best[n]

#include <algorithm>
#include <vector>

long long solve(const std::vector<int>& prices, int n) {
    std::vector<long long> best(n + 1, 0);
    for (int t = 1; t <= n; ++t)
        for (int L = 1; L <= std::min(t, (int)prices.size()); ++L)
            best[t] = std::max(best[t], prices[L - 1] + best[t - L]);   // first piece of length L
    return best[n];
}

class Solution {
    public long cutRod(int[] prices, int n) {
        long[] best = new long[n + 1];
        for (int t = 1; t <= n; t++)
            for (int L = 1; L <= Math.min(t, prices.length); L++)
                best[t] = Math.max(best[t], prices[L - 1] + best[t - L]);   // first piece of length L
        return best[n];
    }
}

function cutRod(prices, n) {
  const best = new Array(n + 1).fill(0);
  for (let t = 1; t <= n; t++)
    for (let L = 1; L <= Math.min(t, prices.length); L++) best[t] = Math.max(best[t], prices[L - 1] + best[t - L]);
  return best[n];
}

fn solve(prices: &[i64], n: i64) -> i64 {
    let n = n as usize;
    let mut best = vec![0i64; n + 1];
    for t in 1..=n {
        for l in 1..=t.min(prices.len()) {
            best[t] = best[t].max(prices[l - 1] + best[t - l]);   // first piece of length l
        }
    }
    best[n]
}

import Data.Array

solve :: [Int] -> Int -> Int
solve prices n = best ! n
  where
    p = listArray (1, length prices) prices
    best = listArray (0, n) (0 : [maximum [p ! l + best ! (t - l) | l <- [1 .. min t (length prices)]] | t <- [1 .. n]]) :: Array Int Int