## Intuition The problem requires us to determine if a sequence of sorting operations can sort any arbitrary array of integers. The key insight is that for the sorting sequence to be correct, it must be able to sort any permutation of the array. This means that the sequence of operations should be able to sort any possible arrangement of the array elements. ## Approach 1. **Generate Permutations**: Since the maximum length of the array `n` is 40, we can generate all permutations of the array. This is feasible because the number of permutations is `n!`, which is manageable for `n = 40`. 2. **Simulate Sorting Operations**: For each permutation, simulate the sequence of sorting operations. Each operation sorts a specified subsequence of the array. 3. **Check Sortedness**: After applying all operations, check if the array is sorted. If any permutation remains unsorted after applying all operations, the algorithm is incorrect. 4. **Output Result**: If all permutations are sorted by the operations, print "ACCEPTED". Otherwise, print "REJECTED". ## Complexity - **Time Complexity**: The time complexity is `O(n! * k * n)`, where `n!` is the number of permutations, `k` is the number of sorting operations, and `n` is the maximum length of the subsequence to sort. - **Space Complexity**: The space complexity is `O(n)` for storing each permutation. ## Solution (C++) ```cpp #include using namespace std; bool isSorted(const vector& a) { for (int i = 1; i < a.size(); ++i) { if (a[i] < a[i - 1]) return false; } return true; } int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n, k; cin >> n >> k; vector> operations(k); for (int i = 0; i < k; ++i) { int q; cin >> q; operations[i].resize(q); for (int j = 0; j < q; ++j) { cin >> operations[i][j]; --operations[i][j]; // Convert to 0-based index } } vector a(n); iota(a.begin(), a.end(), 1); // Initialize a with [1, 2, ..., n] do { vector current = a; for (const auto& op : operations) { vector subseq; for (int index : op) { subseq.push_back(current[index]); } sort(subseq.begin(), subseq.end()); for (int j = 0; j < op.size(); ++j) { current[op[j]] = subseq[j]; } } if (!isSorted(current)) { cout << "REJECTED\n"; return 0; } } while (next_permutation(a.begin(), a.end())); cout << "ACCEPTED\n"; return 0; } ``` This solution systematically checks all permutations of the array to ensure that the sequence of sorting operations can sort any possible arrangement of the array. If any permutation is not sorted by the operations, the algorithm is deemed incorrect.

A Serious Referee

Problem Details