## Intuition The problem involves transforming an array through a series of operations and counting the number of possible resulting arrays. The key insight is to realize that the operation allows us to create subsequences of the original array. The challenge is to count these subsequences efficiently for each possible value of `k`. ## Approach 1. **Understanding the Operation**: The operation allows us to remove elements from the array such that the remaining elements form a subsequence. The task is to count all possible subsequences for each `k` where `1 ≤ k ≤ ⌊(n-1)/2⌋`. 2. **Combinatorial Counting**: For a given `k`, the operation allows us to choose `k` pairs of elements to remove. The number of ways to choose `2k` elements out of `n` is given by the binomial coefficient `C(n, 2k)`. After choosing these pairs, the remaining elements can form any subsequence. 3. **Dynamic Programming for Binomial Coefficients**: Precompute binomial coefficients up to `n` using dynamic programming. This allows us to quickly compute `C(n, 2k)` for each test case. 4. **Modular Arithmetic**: Since the results can be large, use modular arithmetic with the modulus `998,244,353`. 5. **Iterate Over Test Cases**: For each test case, compute the number of valid arrays for each `k` using the precomputed binomial coefficients. ## Complexity - **Time Complexity**: `O(n + t * m)`, where `m` is the maximum value of `⌊(n-1)/2⌋` across all test cases. Precomputation of binomial coefficients is `O(n)`, and each test case requires `O(m)` operations. - **Space Complexity**: `O(n)` for storing factorials and inverse factorials. ## Solution (C++) ```cpp #include using namespace std; const int MOD = 998244353; const int MAXN = 1000000; vector fact(MAXN + 1), inv_fact(MAXN + 1); long long mod_exp(long long base, long long exp, long long mod) { long long result = 1; while (exp > 0) { if (exp % 2 == 1) { result = (result * base) % mod; } base = (base * base) % mod; exp /= 2; } return result; } void precompute_factorials() { fact[0] = 1; for (int i = 1; i <= MAXN; ++i) { fact[i] = fact[i - 1] * i % MOD; } inv_fact[MAXN] = mod_exp(fact[MAXN], MOD - 2, MOD); for (int i = MAXN - 1; i >= 0; --i) { inv_fact[i] = inv_fact[i + 1] * (i + 1) % MOD; } } long long binomial_coefficient(int n, int k) { if (k > n || k < 0) return 0; return fact[n] * inv_fact[k] % MOD * inv_fact[n - k] % MOD; } int main() { ios::sync_with_stdio(false); cin.tie(nullptr); precompute_factorials(); int t; cin >> t; while (t--) { int n; cin >> n; int m = (n - 1) / 2; vector result(m); for (int k = 1; k <= m; ++k) { result[k - 1] = binomial_coefficient(n, 2 * k); } for (int i = 0; i < m; ++i) { cout << result[i] << " "; } cout << "\n"; } return 0; } ``` This solution efficiently computes the number of possible arrays for each `k` using precomputed factorials and inverse factorials to handle large numbers modulo `998,244,353`.

2..3...4.... Wonderful! Wonderful!

Problem Details