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349 lines
10 KiB
Python
Executable file
349 lines
10 KiB
Python
Executable file
"""#
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### 谜题描述
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VK just opened its second HQ in St. Petersburg! Side of its office building has a huge string s written on its side. This part of the office is supposed to be split into m meeting rooms in such way that meeting room walls are strictly between letters on the building. Obviously, meeting rooms should not be of size 0, but can be as small as one letter wide. Each meeting room will be named after the substring of s written on its side.
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<image>
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For each possible arrangement of m meeting rooms we ordered a test meeting room label for the meeting room with lexicographically minimal name. When delivered, those labels got sorted backward lexicographically.
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What is printed on kth label of the delivery?
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Input
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In the first line, you are given three integer numbers n, m, k — length of string s, number of planned meeting rooms to split s into and number of the interesting label (2 ≤ n ≤ 1 000; 1 ≤ m ≤ 1 000; 1 ≤ k ≤ 10^{18}).
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Second input line has string s, consisting of n lowercase english letters.
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For given n, m, k there are at least k ways to split s into m substrings.
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Output
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Output single string – name of meeting room printed on k-th label of the delivery.
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Examples
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Input
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4 2 1
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abac
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Output
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aba
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Input
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19 5 1821
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aupontrougevkoffice
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Output
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au
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Note
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In the first example, delivery consists of the labels \"aba\", \"ab\", \"a\".
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In the second example, delivery consists of 3060 labels. The first label is \"aupontrougevkof\" and the last one is \"a\".
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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#include <bits/stdc++.h>
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using namespace std;
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using ll = long long;
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using ull = unsigned long long;
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constexpr int MOD = 1000 * 1000 * 1000 + 7;
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constexpr int ALPHABET_SIZE = 26;
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constexpr ll INF = 1e18;
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bool check(const string& s, ll k, vector<vector<ll>>& dp, vector<int>& cont,
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const string& cand, int m) {
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int n = s.length(), l = cand.length();
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if (n < l) {
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return false;
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}
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for (int i = 0; i < n; ++i) {
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int pos = i;
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while (pos < n && pos - i < l && s[pos] == cand[pos - i]) {
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++pos;
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}
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if ((pos < n && pos - i < l && s[pos] < cand[pos - i]) ||
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(pos == n && pos - i < l)) {
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cont[i] = -1;
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} else if (pos == n || pos - i == l) {
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cont[i] = pos;
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} else {
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cont[i] = pos + 1;
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}
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}
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for (int i = 0; i < n; ++i) {
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dp[i].assign(m, 0);
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}
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if (cont[0] != -1) {
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dp[cont[0] - 1][0] = 1;
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}
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for (int i = 0; i < n - 1; ++i) {
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for (int j = 0; j < m; ++j) {
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dp[i + 1][j] = min(k, dp[i][j] + dp[i + 1][j]);
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}
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if (cont[i + 1] == -1) {
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continue;
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}
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for (int j = 0; j + 1 < m; ++j) {
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dp[cont[i + 1] - 1][j + 1] =
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min(k, dp[cont[i + 1] - 1][j + 1] + dp[i][j]);
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}
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}
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return dp[n - 1][m - 1] > 0 && dp[n - 1][m - 1] >= k;
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}
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struct TreeNode {
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int have = 0;
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int interm = 0;
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array<int, ALPHABET_SIZE> nxt{};
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};
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string find_in_trie(const vector<TreeNode>& trie, int ind) {
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string ans;
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int have = 0;
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int cur = 0;
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while (cur != -1) {
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if (have + trie[cur].interm > ind) {
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return ans;
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}
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have += trie[cur].interm;
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for (int i = 0; i < ALPHABET_SIZE; ++i) {
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int nxt = trie[cur].nxt[i];
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if (nxt == -1) {
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continue;
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}
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if (have + trie[nxt].have > ind) {
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ans.push_back('a' + i);
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cur = nxt;
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break;
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} else {
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have += trie[nxt].have;
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}
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}
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}
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return ans;
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}
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int main() {
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ios_base::sync_with_stdio(false);
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int n, m;
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ll k;
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cin >> n >> m >> k;
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string s;
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cin >> s;
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vector<vector<ll>> dp(n, vector<ll>(m));
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vector<int> cont(n);
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vector<TreeNode> nodes;
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nodes.emplace_back();
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nodes.back().nxt.fill(-1);
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for (int i = 0; i < n; ++i) {
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int cur = 0;
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nodes[0].have += n - i;
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for (int j = i; j < n; ++j) {
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if (nodes[cur].nxt[s[j] - 'a'] == -1) {
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nodes[cur].nxt[s[j] - 'a'] = nodes.size();
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nodes.emplace_back();
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nodes.back().nxt.fill(-1);
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}
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cur = nodes[cur].nxt[s[j] - 'a'];
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nodes[cur].have += n - j;
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++nodes[cur].interm;
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}
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}
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int st = 0, e = (n * n + n) / 2;
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while (e - st > 1) {
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int mid = (st + e) / 2;
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string cand = find_in_trie(nodes, mid);
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if (check(s, k, dp, cont, cand, m)) {
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st = mid;
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} else {
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e = mid;
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}
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}
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cout << find_in_trie(nodes, st) << \"\n\";
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return 0;
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}
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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from bootcamp import Basebootcamp
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import re
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import math
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from itertools import combinations
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class Faupontrougebootcamp(Basebootcamp):
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class TreeNode:
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def __init__(self):
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self.have = 0
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self.interm = 0
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self.nxt = [-1] * 26 # For each lowercase letter a-z
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def __init__(self, **params):
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# Default parameters with dynamic generation capability
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self.n = params.get('n', 4)
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self.m = params.get('m', 2)
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self.k = params.get('k', 1)
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self.s = params.get('s', 'abac')
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# Ensure parameters are valid
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self.n = len(self.s)
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if not 1 <= self.m <= self.n:
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raise ValueError("m must be between 1 and string length")
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max_k = math.comb(self.n-1, self.m-1)
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if self.k > max_k:
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raise ValueError(f"k exceeds maximum possible value {max_k}")
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def case_generator(self):
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"""Generate a valid puzzle case ensuring adequate combinations"""
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# Generate random case parameters with constraints
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import random
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import string
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# Random string length between 4-8 for manageability
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n = random.randint(4, 8)
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s = ''.join(random.choices(string.ascii_lowercase, k=n))
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# Ensure m is at least 1 and less than n
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m = random.randint(1, min(3, n-1))
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# Calculate maximum possible k
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max_k = math.comb(n-1, m-1)
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k = random.randint(1, max(1, max_k//2)) # Use conservative k
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return {
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'n': n,
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'm': m,
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'k': k,
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's': s
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}
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@staticmethod
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def prompt_func(question_case) -> str:
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params = question_case
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return f"""VK's building has string '{params['s']}' (length {params['n']}). Split it into {params['m']} rooms. For each split, take the lex-smallest substring. Sort these minimal substrings in reverse lex order. What's the {params['k']}-th element?
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Output must be wrapped in [answer] and [/answer] tags. Example: [answer]abc[/answer]"""
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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return matches[-1].strip() if matches else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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try:
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# Reimplement C++ logic for validation
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s = identity['s']
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m = identity['m']
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k = identity['k']
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n = len(s)
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# Build trie structure
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nodes = [cls.TreeNode()]
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for i in range(n):
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cur = 0
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nodes[cur].have += (n - i)
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for j in range(i, n):
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c = ord(s[j]) - ord('a')
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if nodes[cur].nxt[c] == -1:
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nodes.append(cls.TreeNode())
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nodes[cur].nxt[c] = len(nodes) - 1
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nodes[-1].nxt = [-1]*26
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cur = nodes[cur].nxt[c]
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nodes[cur].have += (n - j)
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nodes[cur].interm += 1
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# Binary search with DP verification
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left, right = 0, (n * (n + 1)) // 2
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answer = ""
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while left < right:
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mid = (left + right + 1) // 2
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candidate = cls.find_in_trie(nodes, mid)
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if cls.check_valid(s, k, candidate, m):
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left = mid
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answer = candidate
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else:
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right = mid - 1
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return solution == answer
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except:
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return False
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@classmethod
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def find_in_trie(cls, nodes, idx):
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result = []
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cur = 0
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have = 0
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while True:
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have += nodes[cur].interm
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if have > idx:
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return ''.join(result)
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found = False
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for i in range(26):
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next_node = nodes[cur].nxt[i]
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if next_node == -1:
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continue
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if have + nodes[next_node].have > idx:
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result.append(chr(ord('a') + i))
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cur = next_node
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found = True
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break
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else:
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have += nodes[next_node].have
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if not found:
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break
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return ''.join(result)
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@classmethod
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def check_valid(cls, s, k, candidate, m):
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n = len(s)
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l = len(candidate)
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cont = [-1] * n
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# Precompute continuation points
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for i in range(n):
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pos = i
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while pos < n and pos - i < l and s[pos] == candidate[pos - i]:
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pos += 1
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if pos < n and pos - i < l and s[pos] < candidate[pos - i]:
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cont[i] = -1
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elif pos == n or pos - i == l:
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cont[i] = pos
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else:
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cont[i] = pos + 1 if pos < n else -1
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# DP table initialization
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dp = [[0]*m for _ in range(n)]
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if cont[0] != -1 and cont[0] <= n:
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end = cont[0] - 1
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if end < n:
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dp[end][0] = 1
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for i in range(n-1):
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for j in range(m):
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dp[i+1][j] = min(k, dp[i+1][j] + dp[i][j])
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if cont[i+1] == -1:
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continue
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for j in range(m-1):
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if dp[i][j] == 0:
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continue
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next_pos = cont[i+1] - 1
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if next_pos >= n or j+1 >= m:
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continue
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dp[next_pos][j+1] = min(k, dp[next_pos][j+1] + dp[i][j])
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return dp[n-1][m-1] >= k
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