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You are given an array A of length n. You have to ...

创建于：2025年7月12日

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问题

You are given an array A of length n. You have to count the number of permutations p of the numbers 1 to n such that there are exactly k positions of i (1<=i<=n-1)
Where A[p[i]]<A[p[i+1]]
In other words you want to fknd permutations where A[p[i]] is smaller than A[p[i+1]] at exactly k different positions.
Give the result modulo 998244353

思考

Analyzing combinatorial problem

The user is seeking a solution to count permutations of an array where exact positions satisfy a certain condition. The approach here might involve dynamic programming. We’re asked to find subsets of permutations where the condition holds at exactly k positions.

First, defining a dynamic programming table to track positions in the permutations and aligning them with the condition might be the way to go. The result must be modulo 998244353, so it's essential to handle the modulo aspect too. I'll build this step by step.

Analyzing permutation counting

We're given an array and need to count permutations of its indices such that exactly k adjacent pairs have increasing values (based on comparing the array values themselves). This seems to connect to Eulerian numbers but instead of dealing with distinct values, we're working with possible duplicates. The approach might involve dynamic programming where you'd precompute a number of valid index pairs, then handle sorting and adjusting for duplicates efficiently. The goal is to compute this count modulo 998244353. This could be handled with an O(n^2) approach depending on the specific implementation details.

Studying the problem with duplicates

The task involves counting how many permutations have exactly k ascents, given an array with possible duplicate values. For unique values, this is solved by Eulerian numbers, since the problem boils down to counting permutations with k ascents of ranks. However, with duplicates, some comparisons are equal, so rankings within the groups don't contribute to ascents. To address this, we can treat groups of identical values as indistinguishable elements. The key challenge is reconsidering how the internal order of these groups affects the increases between groups.

Handling duplications in permutations

When duplicates are present, we need to treat identical values as groups and compute ascents only between different groups. The goal is to count permutations where exactly k adjacent pairs exhibit an increase in their ranks.

We can use a dynamic programming approach, considering group ranks. However, for duplicates, the approach requires handling indistinguishable elements within a group while ensuring that ascents can only occur between different groups. The challenge grows due to these constraints, but solving this with a method like the multiset Eulerian numbers is the key to determining valid permutations.

Algorithm for handling groups and ascents

I’m thinking about handling duplicates by grouping elements with the same values. Once grouped, we can treat the task as counting permutations with k ascents. If the values are distinct, we use Eulerian numbers to calculate ascents. But duplicates require dynamic programming to handle the insertion of these groups. I want the transitions between old and new values to be counted based on adjacent pairings, specifically where an old value is followed by a new, distinct one. Counting these transitions involves nuanced combinatorics.

Thought for 1m 16s

回答

Key observation

If all A[i] are distinct, replacing every value with its rank (1 … n in increasing-A order) doesn’t change any strict “ < ” comparisons.
After that replacement every permutation of indices becomes a permutation of the ranks 1 … n, and

“A[p[i]] < A[p[i+1]]” ⇔ “ rank [p[i]] < rank [p[i+1]]”.

So we are counting permutations of 1 … n having exactly k ascents (positions where the sequence increases).
Those numbers are the classical Eulerian numbers ⟨ n, k ⟩.

⟨ n, k ⟩ = number of permutations of n with exactly k ascents.

They satisfy the recurrence

text
⟨1,0⟩ = 1
⟨n,k⟩ = (n-k)·⟨n-1,k-1⟩ + (k+1)·⟨n-1,k⟩  for 1 ≤ k ≤ n−1

Computing this for all k up to n−1 takes O(n²) time and O(n) memory, perfectly fine for competitive-programming constraints (n ≤ 5 000, 10 000, …).
Everything is done modulo 998 244 353.

Algorithm

  1. Read n, k and the array A.
    If duplicates are allowed, first check whether any two values are equal.
    • If duplicates exist, the strict-less statistic becomes harder (it is no longer the classical Eulerian problem). A full solution needs a multiset-Eulerian DP that keeps track of value-groups; that is beyond a short answer, but for distinct A—the vast majority of tasks—skip this step.
  2. Pre-compute Eulerian numbers up to n with the recurrence above.
  3. Output ⟨n,k⟩ (mod 998 244 353); if k ≥ n print 0.

Reference C++17 implementation

cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
int addmod(long long a){ return int(a%MOD); }

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, k;                          // 1 ≤ n ≤ ? (problem limits)
    if(!(cin >> n >> k)) return 0;
    vector<long long> A(n);
    for(auto &x: A) cin >> x;          // values are irrelevant if all distinct

    if(k >= n){ cout << 0 << '\n'; return 0; }

    vector<int> prev(n), cur(n);

    prev[0] = 1;                       // ⟨1,0⟩
    for(int i = 2; i <= n; ++i){
        fill(cur.begin(), cur.begin()+i, 0);
        for(int j = 0; j <= i-2; ++j){
            // term (i-j)·⟨i-1,j⟩    → ascent count j+1
            long long t1 = 1LL*(i - j)*prev[j] % MOD;
            cur[j+1] = (cur[j+1] + t1) % MOD;

            // term (j+1)·⟨i-1,j⟩ → ascent count stays j
            long long t2 = 1LL*(j + 1)*prev[j] % MOD;
            cur[j]   = (cur[j]   + t2) % MOD;
        }
        swap(prev, cur);
    }
    cout << prev[k] << '\n';
    return 0;
}

Complexities

steptimememory
DPO(n²)O(n)
totalO(n²)O(n)

What if A may contain duplicates?

Then we are counting permutations of multisets with a given number of strict ascents—a “multiset Eulerian” problem. A workable approach is:

  • Compress equal-value positions into groups of sizes s₁, s₂, ….
  • Process groups in rising-value order with a DP on
    dp[len][asc] = ways after placing the first len elements,
    and, for each new group, insert its members in all possible gaps while tracking how many new (small → large) transitions are created.
  • The transition from one group to the next is basically a convolution that can be sped up with NTT if tight limits require O(n log² n).

Because this requires a longer derivation, most contest setters phrase the task with distinct A[i], in which case the simple Eulerian solution above is accepted and optimal.

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