### **Bitonic Sequences** * **Time limit:** 200...

Создано: 25 октября 2025 г.

Отвечено с использованием GPT-5 Thinking от Chat01

Вопрос

Bitonic Sequences

  • Time limit: 2000 ms
  • Memory limit: 128 MB

Description

We define a positive bitonic sequence of sum N as a sequence of positive integers S₁, S₂, ..., Sₖ such that the sum of all elements equals N, and there exists an index j satisfying:

  • Sᵢ ≤ Sᵢ₊₁ for all 1 ≤ i < j, and
  • Sᵢ ≥ Sᵢ₊₁ for all j ≤ i < k.

Find the number of distinct positive bitonic sequences with a total sum of N, for all 1 ≤ N ≤ M, modulo 10⁹ + 7.

Two bitonic sequences S₁ and S₂ are considered different if they have different lengths, or if there exists an index j' such that S₁,j' ≠ S₂,j'.


Standard input

The first line of input contains a single integer M.


Standard output

Print the number of distinct positive bitonic sequences for each N (1 ≤ N ≤ M), modulo 10⁹ + 7.


Constraints and notes

  • 1 ≤ M ≤ 3 × 10⁵.

Example

Input:

5

Output:

1 2 4 8 15

#include <bits/stdc++.h>
using namespace std;

const int MOD = 1000000007;

int addmod(int a, int b) { a += b; if (a >= MOD) a -= MOD; return a; }
int mulmod(long long a, long long b) { return int((a * b) % MOD); }

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

text
int M; if (!(cin >> M)) return 0; vector<int> f(M + 1, 0); // Q_p coefficients up to degree T_p where needed. We maintain current Q array up to M. vector<int> Q(M + 1, 0), Qnext(M + 1, 0); Q[0] = 1; // Q_1 = 1, since product over v=1..p-1 empty // For p = 1, B_1(x) = x / (1 - x) = sum_{m >= 1} x^m // Contribution to f[N] from p=1 is A_1(N - m) with A_1(t) = coeff of Q_1(t) = [t == 0] // So f[N] += 1 for all N >= 1. for (int N = 1; N <= M; ++N) f[N] = addmod(f[N], 1); // Now iterate p from 2..M, build Q_p from Q_{p-1} by multiplying (1 - x^{p-1})^{-2} for (int p = 2; p <= M; ++p) { int k = p - 1; // Compute Qnext = Q * (1 - x^k)^{-2} truncated to degree M // For each residue r in [0..k-1], process indices t = r + i*k for (int r = 0; r < k; ++r) { // collect sequence a[i] = Q[r + i*k] for i s.t. idx <= M vector<int> a; for (int idx = r; idx <= M; idx += k) a.push_back(Q[idx]); int L = (int)a.size(); if (L == 0) continue; vector<long long> pref(L, 0), prefi(L, 0); pref[0] = a[0]; prefi[0] = 0LL; // 0 * a[0] for (int i = 1; i < L; ++i) { pref[i] = pref[i-1] + a[i]; if (pref[i] >= MOD) pref[i] -= MOD; prefi[i] = prefi[i-1] + 1LL * i * a[i]; prefi[i] %= MOD; } // b[i] = sum_{c=0..i} (c+1) * a[i-c] = (i+1) * sum_{j=0..i} a[j] - sum_{j=0..i} j * a[j] for (int i = 0; i < L; ++i) { long long sumA = pref[i]; long long sumJA = prefi[i]; long long val = ((long long)(i + 1) * sumA - sumJA) % MOD; if (val < 0) val += MOD; int idx = r + i * k; Qnext[idx] = (int)val; } } Q.swap(Qnext); // clear Qnext for next iteration fill(Qnext.begin(), Qnext.end(), 0); // Add contributions for this p: B_p(x) = sum_{c>=1} x^{pc} // For each N, f[N] += sum over m = p, 2p, ... ≤ N of A_p(N - m), where A_p = Q for (int m = p; m <= M; m += p) { int maxN = M; // for N from m..M: f[N] += Q[N - m] // We can do this as a loop; still O(M/p) per m, totals O(M log M) per p? Actually sum over m is M/p terms, each loop M - m ~ M, gives O(M^2/p) for (int N = m; N <= maxN; ++N) { f[N] = addmod(f[N], Q[N - m]); } } } // Output results for N=1..M for (int N = 1; N <= M; ++N) { if (N > 1) cout << ' '; cout << f[N]; } cout << '\n'; return 0;

}

I only passed a few use cases, and all the others timed out. How can I optimize this?

Python

python
def compositions(n): res=[] # generate compositions by splitting between positions of n ones def rec(rem, cur): if rem==0: res.append(cur.copy()) return for x in range(1, rem+1): cur.append(x) rec(rem-x, cur) cur.pop() rec(n, []) return res def is_unimodal(seq): k=len(seq) # j index 1..k for j in range(1, k+1): ok=True for i in range(1, j): if seq[i-1] > seq[i]: ok=False; break if not ok: continue for i in range(j, k): if seq[i-1] < seq[i]: ok=False; break if ok: return True return False for n in range(1,11): cnt=sum(1 for s in compositions(n) if is_unimodal(s)) print(n,cnt)

Python

python
seq=[1,2,4,8,15,27,47,79,130,209,336,538,857][:12] # find recurrence of small order using linear algebra mod integers def find_rec(seq, order=4): import numpy as np import fractions n=len(seq) best=None for k in range(1,7): # solve for coeffs c1..ck: s[n]=sum c_i s[n-i] A=[] b=[] for i in range(k, n): A.append(seq[i-1:i-k-1:-1]) b.append(seq[i]) import numpy.linalg as la try: x=la.lstsq(np.array(A,dtype=float), np.array(b,dtype=float), rcond=None)[0] # Round to integers maybe cand=[round(v) for v in x] ok=True for i in range(k, n): pred=sum(cand[j]*seq[i-1-j] for j in range(k)) if

Python

python
# compute partitions into odd parts (equals partitions into distinct parts) from functools import lru_cache def partitions_odd(n): @lru_cache(None) def dp(i, rem): # i odd upto rem if rem==0: return 1 if i>rem: return 0 return dp(i, rem-i) + dp(i+2, rem) return dp(1, n) def partitions(n): @lru_cache(None) def dp(maxp, rem): if rem==0: return 1 if maxp==0: return 0 if maxp>rem: return dp(rem, rem) res=0 for k in range(rem//maxp+1): res+=dp(maxp-1, rem-k*maxp) return res return dp(n,n) vals=[(n, partitions_odd(n), partitions(n)) for n in range(1,11)] vals [(1, 1, 1), (2, 1, 2), (3, 2, 3), (4, 2, 5), (5, 3, 7), (6, 4, 11), (7, 5, 15), (8, 6, 22), (9, 8, 30), (10, 10, 42)]

Мысль

Considering dynamic programming approach

I need counts up to 100 to test the recurrence, but enumerating compositions grows exponentially—up to 25 is already challenging. Testing up to 10 was fine earlier. A DP method could help, focusing on the state of remainder, last value, mode, and peak status. I’ll work through defining sequences based on peaks and avoid double-counting central values.

Considering sequence structure

I’m realizing that a plateau means having multiple p’s in the center, so the left and right sequences include a central length, which adds complexity. I recall a known identity about unimodal compositions involving pyramidal partitions. But optimizing involves reducing the double loop in their code, which is currently O(M^2).

Optimizing convolution method

To optimize, I can use the convolution trick with an arithmetic progression of ones. Instead of looping through each m, I’ll convolve with a vector G_p for each p. However, doing this with NTT individually is expensive. We need to apply the “convolution with periodic indicator” trick using residue classes to optimize the process efficiently.

Optimizing prefix sum calculation

I see that I can optimize the convolution by using a prefix sum technique for each residue class modulo p. Precomputing cumulative sums for each residue class allows for amortized O(1) calculations. This method is similar to the technique used for multiplying by (1xk)2(1 - x^k)^{-2} earlier. Now, I can apply it to the B_p(x) factor.

Re-Evaluating the DP formulation

The current approach seems overly complicated. Instead of directly comparing both sides of the sequence, a dynamic programming approach might be more efficient. For unimodal compositions, I want to focus on adjusting the gluing process, fixing the central plateau lengths, and considering sequences where the last element is exactly the peak. It looks like the current double-counting strategy needs refining.

Exploring possible optimizations

It looks like I’m going back and forth with various approaches. I keep trying to optimize this via dynamic programming and the residue trick, but it mostly gives me O(M) updates for each p — not ideal. Maybe the real breakthrough is simplifying the sums or using partitions more effectively. I keep thinking about the floor(M/p) trick and partitioning by q but it’s still a bit messy. Would rethinking the approach help?

Considering divide-and-conquer optimization

I’m trying to optimize using divide-and-conquer approach for prefix sums and convolutions, but it still feels like a bit too much. The goal is to efficiently update f(n) by computing S recursively and leveraging segment trees or NTT. But the complexity is still high, and I’m wondering whether this will ultimately be too slow to handle.

Exploring divide-and-conquer optimization

I’m exploring the possibility of using divide-and-conquer for polynomial convolution involving prefix sums. The idea is to optimize by building a segment tree and performing convolutions to compute contributions directly. The complexity looks manageable at O(M log^2 M), though implementing the triple NTT method for mod is quite intricate. However, I wonder whether there's a simpler approach using dynamic programming, especially considering the constraints.

Optimizing convolution and segment tree

To optimize convolutions, I propose using a divide-and-conquer approach with segment trees. The idea involves precomputing the necessary polynomials and building the product tree. By using NTT with Garner for modulus conversion, we can handle larger values within the constraints. I'll provide an outline and code using this method.

Thought for 3m 1s

Ответ

You’re running into the right idea (use the GF)

F(x)=p1(v=1p11(1xv)2)xp1xp,F(x)=\sum_{p\ge1}\Bigg(\prod_{v=1}^{p-1}\frac1{(1-x^v)^2}\Bigg)\cdot\frac{x^p}{1-x^p},

but the way you add each pp-block is O(M2)O(M^2) overall (and your update of QpQ_p is also O(M)O(M) per pp, so O(M2)O(M^2) total). With M3105M\le 3\cdot10^5 that will time out.

Below are two optimizations, in order of impact.


1) Do it in O~(M)\tilde O(M) multiplications via a divide-&-conquer (segment-tree) of polynomials

Key observation: define the (truncated) polynomials

  • Hk(x)=(1xk)2=c0(c+1)xckH_k(x)= (1-x^k)^{-2} = \sum_{c\ge 0}(c+1)\,x^{ck} (cut off at degree MM),
  • Gp(x)=xp1xp=c1xcpG_p(x)=\dfrac{x^p}{1-x^p} = \sum_{c\ge 1}x^{cp} (cut off at degree MM).

Then

F(x)=p=1M(v=1p1Hv(x))Gp(x)(truncated to degree M).F(x)=\sum_{p=1}^{M}\Big(\prod_{v=1}^{p-1}H_v(x)\Big)\cdot G_p(x)\quad\text{(truncated to degree }M\text{)}.

Instead of forming each prefix product and convolving it with GpG_p one-by-one, build a segment tree over p[1,M]p\in[1,M] and compute everything with O(M)O(M) polynomial multiplications (each is NTT/CRT-based, degrees trimmed to M\le M):

  • Build a product tree P\mathcal P: each node stores P[L,R](x)=v=LRHv(x)P_{[L,R]}(x)=\prod_{v=L}^R H_v(x).
  • Build a “weighted sum” tree D\mathcal D bottom-up with the identity D[L,R](x)  =  D[L,mid](x)  +  P[L,mid](x)D[mid+1,R](x),D_{[L,R]}(x) \;=\; D_{[L,mid]}(x)\;+\;P_{[L,mid]}(x)\cdot D_{[mid+1,R]}(x), with leaf D[p,p](x)=Gp(x)D_{[p,p]}(x)=G_p(x).
  • Then D[1,M](x)=p=1M(v=1p1Hv(x))Gp(x)=F(x)D_{[1,M]}(x)=\sum_{p=1}^{M}\Big(\prod_{v=1}^{p-1}H_v(x)\Big)\,G_p(x)=F(x).

Every internal node performs one polynomial multiplication for the product tree and one for the D-tree, both truncated to degree MM. With an NTT this is O(Mlog2M)O(M\log^2 M) total and is easily fast enough.

Implementation tips:

  • Leaves:
    • HkH_k is very sparse: set poly[0]+=1, poly[k]+=2, poly[2k]+=3, … until degree >M>M.
    • GpG_p is a simple 0/1 periodic vector: set poly[cp]=1 for cp in {p,2p,… ≤ M}.
  • Always truncate to degree MM after each multiply.
  • Use NTT(s). Since the modulus is 109+710^9+7 (not NTT-friendly), do MTT/CRT:
    • run NTT under two or three friendly primes (e.g., 998244353, 1004535809, 469762049),
    • Garner / CRT the result back mod 109+710^9+7.
  • For very small degrees, switch to a naive O(nm)O(nm) multiply to reduce constants.
  • Pre-allocate buffers; reuse vectors to avoid allocations.

This completely eliminates your O(M2)O(M^2) loops; you never form QpQ_p explicitly and you never add “multiples of pp” with nested loops—the tree does the algebra for all pp at once.

Sketch (only the structure; drop-in NTT is standard and omitted for brevity):

cpp
// Build leaves H_k and G_p up to degree M. // Product tree: P[node] = P[left]*P[right] (truncate to deg M) // D-tree: D[node] = D[left] + P[left] * D[right] (truncate to deg M) // Answer poly = D[root]; print coeffs 1..M. struct Poly { vector<int> a; }; // coeffs mod 1e9+7 via CRT in mul() Poly mul(const Poly& A, const Poly& B, int M); // NTT+CRT, truncate to M void add_inplace(Poly& A, const Poly& B); // (mod 1e9+7), resize as needed void trim(Poly& A, int M); vector<Poly> Hleaf(M+1), Gleaf(M+1); void build_H_G(int M) { for (int k = 1; k <= M; ++k) { auto& H = Hleaf[k].a; H.assign(M+1, 0); for (int t = 0, deg = 0; deg <= M; ++t, deg = t*k) { if (deg > M) break; H[deg] = (H[deg] + (t+1)) % MOD; // (1 - x^k)^{-2} } auto& G = Gleaf[k].a; G.assign(M+1, 0); for (int deg = k; deg <= M; deg += k) G[deg] = 1; // x^k/(1-x^k) } } struct Node { Poly P, D; }; vector<Node> seg(4*(M+5)); void build(int idx, int L, int R, int M) { if (L == R) { seg[idx].P.a = Hleaf[L].a; trim(seg[idx].P, M); seg[idx].D.a = Gleaf[L].a; trim(seg[idx].D, M); return; } int mid = (L+R)>>1; build(idx<<1, L, mid, M); build(idx<<1|1, mid+1, R, M); seg[idx].P = mul(seg[idx<<1].P, seg[idx<<1|1].P, M); Poly rightWeighted = mul(seg[idx<<1].P, seg[idx<<1|1].D, M); seg[idx].D = seg[idx<<1].D; add_inplace(seg[idx].D, rightWeighted); trim(seg[idx].D, M); }

After build(1,1,M,M), seg[1].D.a[n] (for 1 ≤ n ≤ M) are your answers.


2) If you must keep your current shape, at least drop a log-factor (still too slow at 3e5 though)

Right now you do:

cpp
for (int m = p; m <= M; m += p) for (int N = m; N <= M; ++N) f[N] += Q[N - m];

Rewrite that using residue-class prefix sums (for each r in [0..p-1]):

  • let a[i] = Q[r + i*p] for i ≥ 0,
  • prefix pref[i] = pref[i-1] + a[i],
  • then for j ≥ 1, add pref[j-1] to f[r + j*p].

This makes the B_p update O(M)O(M) (down from O(MlogM)O(M\log M) per pp), but the whole algorithm remains O(M2)O(M^2) and will still TLE at the given limit. Use it only as an incremental improvement if you can’t implement #1 yet.


Complexity & practicality

  • Segment-tree D&C + NTT/CRT: ~O(Mlog2M)O(M \log^2 M) polynomial time with small constants; passes comfortably for M=3105M=3\cdot10^5.
  • Memory: store a few hundred MB at most if you aggressively trim to degree MM and reuse buffers.

If you want, I can share a full drop-in NTT+CRT template wired to the tree above; it’s a bit long for a single message, but that’s the exact code pattern to make this pass.

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