这个问题具有类似的设置，以找到适合一组和的数组，尽管其目标完全不同。

考虑一个A长度数组n。该数组仅包含正整数。例如A = (1,1,2,2)。让我们将定义f(A)为的所有非空连续子数组之和的集合A。在这种情况下f(A) = {1,2,3,4,5,6}。产生步骤f(A) 如下：

的子阵列A是(1), (1), (2), (2), (1,1), (1,2), (2,2), (1,1,2), (1,2,2), (1,1,2,2)。它们各自的总和为1,1,2,2,2,3,4,4,5,6。因此，您从此列表中获得的集合是{1,2,3,4,5,6}。

如果没有其他相同长度的数组，则我们将数组称为A 唯一数组，除了反向数组。作为示例，但是没有其他长度数组会产生相同的和集。Bf(A) = f(B)Af((1,2,3)) = f((3,2,1)) = {1,2,3,5,6}3

我们将仅考虑元素为给定整数s或的数组s+1。例如，如果s=1数组仅包含1和2。

任务

任务，对于给定n和s是计算该长度的独特阵列的数量。您可以假设s介于1和之间9。

您不应该算数组的倒数以及数组本身。

例子

s = 1答案永远是n+1。

s = 2，从n = 1上往下数的答案是：

2,3,6,10,20,32,52,86

s = 8，从n = 1上往下数的答案是：

2,3,6,10,20,36,68,130

得分了

对于给定n的代码，您的代码应输出s从1到的所有值的答案9。您的分数是n在一分钟内完成的最高分。

测试中

我将需要在ubuntu机器上运行您的代码，因此请尽可能详细地说明如何编译和运行您的代码。

排行榜

• n = 24，安德斯·卡塞格（Anders Kaseorg）in Rust（34秒）
• n = 16，乌鲁斯在清洁区（36秒）
• JRowan在Common Lisp中的n = 14（49秒）

锈，n≈24

需要每晚使用Rust才能获得便利的reverse_bits功能。rustc -O unique.rs用（例如）编译并运行./unique 24。

#![feature(reverse_bits)]
use std::{collections::HashMap, env, mem, process};

type T = u32;
const BITS: u32 = mem::size_of::<T>() as u32 * 8;

fn main() {
    let args = env::args().collect::<Vec<_>>();
    assert!(args.len() == 2);
    let n: u32 = args[1].parse().unwrap();
    assert!(n > 0);
    assert!(n <= BITS);
    let mut unique = (2..=9).map(|_| HashMap::new()).collect::<Vec<_>>();
    let mut sums = vec![0 as T; n as usize];
    for a in 0 as T..=!0 >> (BITS - n) {
        if a <= a.reverse_bits() >> (BITS - n) {
            for v in &mut sums {
                *v = 0;
            }
            for i in 0..n {
                let mut bit = 1;
                for j in i..n {
                    bit <<= a >> j & 1;
                    sums[(j - i) as usize] |= bit;
                }
            }
            for s in 2..=9 {
                let mut sums_s =
                    vec![0 as T; ((n + (n - 1) * s) / BITS + 1) as usize].into_boxed_slice();
                let mut pos = 0;
                let mut shift = 0;
                let mut lo = 0;
                let mut hi = 0;
                for &v in &sums {
                    lo |= v << shift;
                    if BITS - shift < n {
                        hi |= v >> (BITS - shift);
                    }
                    shift += s;
                    if shift >= BITS {
                        shift -= BITS;
                        sums_s[pos] = lo;
                        pos += 1;
                        lo = hi;
                        hi = 0;
                    }
                }
                if lo != 0 || hi != 0 {
                    sums_s[pos] = lo;
                    pos += 1;
                    if hi != 0 {
                        sums_s[pos] = hi;
                    }
                }
                unique[s as usize - 2]
                    .entry(sums_s)
                    .and_modify(|u| *u = false)
                    .or_insert(true);
            }
        }
    }
    let mut counts = vec![n + 1];
    counts.extend(
        unique
            .iter()
            .map(|m| m.values().map(|&u| u as T).sum::<T>())
            .collect::<Vec<_>>(),
    );
    println!("{:?}", counts);
    process::exit(0); // Avoid running destructors.
}

普通Lisp SBCL，N = 14

呼叫功能（goahead ns）

    (defun sub-lists(l m &optional(x 0)(y 0))
  (cond; ((and(= y (length l))(= x (length l)))nil)
        ((= y (length l))m)
        ((= x (length l))(sub-lists l m 0(1+ y)))
    (t (sub-lists l (cons(loop for a from x to (+ x y)

             when (and(nth (+ x y)l)(nth a l)(< (+ x y)(length l)))
                ;   while (nth a l)
             ;while(and(< (+ x y)(length l))(nth a l))
                    collect (nth a l))m) (1+ x)y))
    ))
(defun permutations(size elements)
  (if (zerop size)'(())
 (mapcan (lambda (p)
                    (map 'list (lambda (e)
                           (cons e p))
                         elements))
     (permutations (1- size) elements))))
(defun remove-reverse(l m)
  (cond ((endp l)m)
    ((member (reverse (first l))(rest l) :test #'equal)(remove-reverse (rest l)m))
    (t (remove-reverse (rest l)(cons (first l)m)))))
(defun main(n s)
  (let((l (remove-reverse (permutations n `(,s ,(1+ s)))nil)))

  (loop for x in l
     for j = (remove 'nil (sub-lists x nil))
       collect(sort (make-set(loop for y in j
        collect (apply '+ y))nil)#'<)
     )
  ))
(defun remove-dups(l m n)
  (cond ((endp l)n)
        ((member (first l) (rest l) :test #'equal)(remove-dups(rest l)(cons (first l) m) n))
    ((member (first l) m :test #'equal)(remove-dups(rest l)m n))
    (t(remove-dups (rest l) m (cons (first l) n))))

  )
(defun goahead(n s)
  (loop for a from 1 to s
  collect(length (remove-dups(main n a)nil nil))))
(defun make-set (L m)
  "Returns a set from a list. Duplicate elements are removed."
  (cond ((endp L) m)
    ((member (first L) (rest L)) (make-set (rest L)m))
    ( t (make-set (rest L)(cons (first l)m)))))

这是运行时间

CL-USER> (time (goahead 14 9))
Evaluation took:
  34.342 seconds of real time
  34.295000 seconds of total run time (34.103012 user, 0.191988 system)
  [ Run times consist of 0.263 seconds GC time, and 34.032 seconds non-GC time. ]
  99.86% CPU
  103,024,254,028 processor cycles
  1,473,099,744 bytes consed

(15 1047 4893 6864 7270 7324 7328 7328 7328)
CL-USER> (time (goahead 15 9))
Evaluation took:
  138.639 seconds of real time
  138.511089 seconds of total run time (137.923824 user, 0.587265 system)
  [ Run times consist of 0.630 seconds GC time, and 137.882 seconds non-GC time. ]
  99.91% CPU
  415,915,271,830 processor cycles
  3,453,394,576 bytes consed

(16 1502 8848 13336 14418 14578 14594 14594 14594)

清洁

当然，这不是最有效的方法，但是我很想知道一个简单的按值过滤器的性能如何。

也就是说，使用此方法仍需改进。

module main
import StdEnv, Data.List, System.CommandLine

f l = sort (nub [sum t \\ i <- inits l, t <- tails i])

Start w
	# ([_:args], w) = getCommandLine w
	= case map toInt args of
		[n] = map (flip countUniques n) [1..9]
		_ = abort "Wrong number of arguments!"

countUniques 1 n = inc n
countUniques s n = length uniques
where
	lists = [[s + ((i >> p) bitand 1) \\ p <- [0..dec n]] \\ i <- [0..2^n-1]]
	pairs = sortBy (\(a,_) (b,_) = a < b) (zip (map f lists, lists))
	groups = map (snd o unzip) (groupBy (\(a,_) (b,_) = a == b) pairs)
	uniques = filter (\section = case section of [a, b] = a == reverse b; [_] = True; _ = False) groups

放置在名为的文件中main.icl，或将顶行更改为module <your_file_name_here>。

用编译clm -h 1500m -s 50m -fusion -t -IL Dynamics -IL StdEnv -IL Platform main。

您可以从标题中的链接获得TIO（和我自己）使用的版本，或者从此处获得更新的版本。