这个挑战是从入学考试到封闭数字网络安全课程。无论如何，它与网络安全无关，只是为了测试学生的逻辑和编码技能。

任务

编写一个程序，该程序从数组中删除条目，以便以严格的降序对其余值进行排序，并且在所有其他可能的降序中将其总和最大化。

输入输出

输入将是整数值的数组严格地大于0和所有彼此不同。您可以自由选择是从文件，命令行还是标准输入中读取输入。

输出将是输入之一的降序子数组，其总和大于任何其他可能的降序子数组。

注意： [5, 4, 3, 2]是的子数组[5, 4, 1, 3, 2]，即使4和3不相邻。只是因为1弹出了。

暴力解决方案

当然，最简单的解决方案是在给定数组的所有可能组合中进行迭代，并搜索具有最大和的排序后的数组，即在Python中：

import itertools

def best_sum_desc_subarray(ary):
    best_sum_so_far = 0
    best_subarray_so_far = []
    for k in range(1, len(ary)):
        for comb in itertools.combinations(ary, k):
            if sum(comb) > best_sum_so_far and all(comb[j] > comb[j+1] for j in range(len(comb)-1)):
                best_subarray_so_far = list(comb)
                best_sum_so_far = sum(comb)
    return best_subarray_so_far

不幸的是，由于检查数组是否已排序并计算其元素的总和为，并且由于此操作的完成时间为从到，因此渐近时间复杂度为

挑战

您的目标是实现比上述蛮力更好的时间复杂性。渐近时间复杂度最小的解决方案是挑战的胜者。如果两个解具有相同的渐近时间复杂度，则获胜者将是最小的渐近空间复杂度。

注意： 即使大量，您也可以考虑读取，写入和比较原子。

注意： 如果有两个或多个解决方案，请返回其中一个。

测试用例

Input:  [200, 100, 400]
Output: [400]

Input:  [4, 3, 2, 1, 5]
Output: [4, 3, 2, 1]

Input:  [50, 40, 30, 20, 10]
Output: [50, 40, 30, 20, 10]

Input:  [389, 207, 155, 300, 299, 170, 158, 65]
Output: [389, 300, 299, 170, 158, 65]

Input:  [19, 20, 2, 18, 13, 14, 8, 9, 4, 6, 16, 1, 15, 12, 3, 7, 17, 5, 10, 11]
Output: [20, 18, 16, 15, 12, 7, 5]

Input:  [14, 12, 24, 21, 6, 10, 19, 1, 5, 8, 17, 7, 9, 15, 23, 20, 25, 11, 13, 4, 3, 22, 18, 2, 16]
Output: [24, 21, 19, 17, 15, 13, 4, 3, 2]

Input:  [25, 15, 3, 6, 24, 30, 23, 7, 1, 10, 16, 29, 12, 13, 22, 8, 17, 14, 20, 11, 9, 18, 28, 21, 26, 27, 4, 2, 19, 5]
Output: [25, 24, 23, 22, 17, 14, 11, 9, 4, 2]

佩尔

时间应为O（n ^ 2），空间应为O（n）

将数字用空格隔开，在一行上给STDIN

#!/usr/bin/perl -a
use strict;
use warnings;

# use Data::Dumper;
use constant {
    INFINITY => 9**9**9,
    DEBUG    => 0,
};

# Recover sequence from the 'how' linked list
sub how {
    my @z;
    for (my $h = shift->{how}; $h; $h = $h->[1]) {
        push @z, $h->[0];
    }
    pop @z;
    return join " ", reverse @z;
}

use constant MINIMUM => {
    how  => [-INFINITY, [INFINITY]],
    sum  => -INFINITY,
    next => undef,
};

# Candidates is a linked list of subsequences under consideration
# A given final element will only appear once in the list of candidates
# in combination with the best sum that can be achieved with that final element
# The list of candidates is reverse sorted by final element
my $candidates = {
    # 'how' will represent the sequence that adds up to the given sum as a
    # reversed lisp style list.
    # so e.g. "1, 5, 8" will be represented as [8, [5, [1, INFINITY]]]
    # So the final element will be at the front of 'how'
    how  => [INFINITY],
    # The highest sum that can be reached with any subsequence with the same
    # final element
    sum  => 0,
    # 'next' points to the next candidate
    next => MINIMUM,   # Dummy terminator to simplify program logic
};

for my $num (@F) {
    # Among the candidates on which an extension with $num is valid
    # find the highest sum
    my $max_sum = MINIMUM;
    my $c = \$candidates;
    while ($num < $$c->{how}[0]) {
        if ($$c->{sum} > $max_sum->{sum}) {
            $max_sum = $$c;
            $c = \$$c->{next};
        } else {
            # Remove pointless candidate
            $$c = $$c->{next};
        }
    }

    my $new_sum = $max_sum->{sum} + $num;
    if ($$c->{how}[0] != $num) {
        # Insert a new candidate with a never before seen end element
        # Due to the unique element rule this branch will always be taken
        $$c = { next => $$c };
    } elsif ($new_sum <= $$c->{sum}) {
        # An already known end element but the sum is no improvement
        next;
    }
    $$c->{sum} = $new_sum;
    $$c->{how} = [$num, $max_sum->{how}];
    # print(Dumper($candidates));
    if (DEBUG) {
        print "Adding $num\n";
        for (my $c = $candidates; $c; $c = $c->{next}) {
            printf "sum(%s) = %s\n", how($c), $c->{sum};
        }
        print "------\n";
    }
}

# Find the sequence with the highest sum among the candidates
my $max_sum = MINIMUM;
for (my $c = $candidates; $c; $c = $c->{next}) {
    $max_sum = $c if $c->{sum} > $max_sum->{sum};
}

# And finally print the result
print how($max_sum), "\n";

Haskell，时间，空间O （n $O(n \log n)$$O(n)$

{-# LANGUAGE MultiParamTypeClasses #-}

import qualified Data.FingerTree as F

data S = S
  { sSum :: Int
  , sArr :: [Int]
  } deriving (Show)

instance Monoid S where
  mempty = S 0 []
  mappend _ s = s

instance F.Measured S S where
  measure = id

bestSubarrays :: [Int] -> F.FingerTree S S
bestSubarrays [] = F.empty
bestSubarrays (x:xs) = left F.>< sNew F.<| right'
  where
    (left, right) = F.split (\s -> sArr s > [x]) (bestSubarrays xs)
    sLeft = F.measure left
    sNew = S (x + sSum sLeft) (x : sArr sLeft)
    right' = F.dropUntil (\s -> sSum s > sSum sNew) right

bestSubarray :: [Int] -> [Int]
bestSubarray = sArr . F.measure . bestSubarrays

怎么运行的

bestSubarrays xs是xs{最大和，最小第一个元素}的有效边界上的子数组序列，按从左到右的顺序通过增加和和增加第一个元素来排序。

从去bestSubarrays xs到bestSubarrays (x:xs)，我们

1. 将序列分为左侧，第一个元素小于x，右侧和第一个元素大于x，
2. 通过x在左侧最右边的子数组之前找到一个新的子数组，
3. 从右侧删除子数组的前缀，其总和比新子数组小，
4. 连接左侧，新的子数组和右侧的其余部分。

一个手指树支持所有这些操作时间。$O(\log n)$

此答案是Ton Hospel的答案的扩展。

该问题可以通过使用递归的动态编程来解决

其中是输入序列，而是任何以索引结尾的递减子序列的最大可实现总和。然后可以使用追溯实际的解决方案，如下面的rust代码所示。T （i ）i $(a_i)$$T(i)$$i$$T$

fn solve(arr: &[usize]) -> Vec<usize> {
    let mut tbl = Vec::new();
    // Compute table with maximum sums of any valid sequence ending
    // with a given index i.
    for i in 0..arr.len() {
        let max = (0..i)
            .filter(|&j| arr[j] >= arr[i])
            .map(|j| tbl[j])
            .max()
            .unwrap_or(0);
        tbl.push(max + arr[i]);
    }
    // Reconstruct an optimal sequence.
    let mut sum = tbl.iter().max().unwrap_or(&0).clone();
    let mut limit = 0;
    let mut result = Vec::new();

    for i in (0..arr.len()).rev() {
        if tbl[i] == sum && arr[i] >= limit {
            limit = arr[i];
            sum -= arr[i];
            result.push(arr[i]);
        }
    }
    assert_eq!(sum, 0);
    result.reverse();
    result
}

fn read_input() -> Vec<usize> {
    use std::io::{Read, stdin};
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    s.split(|c: char| !c.is_numeric())
        .filter(|&s| !s.is_empty())
        .map(|s| s.parse().unwrap())
        .collect()
}

fn main() {
    println!("{:?}", solve(&read_input()));
}

在线尝试！