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276 lines
8 KiB
Python
Executable file
276 lines
8 KiB
Python
Executable file
"""#
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### 谜题描述
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Childan is making up a legendary story and trying to sell his forgery — a necklace with a strong sense of \"Wu\" to the Kasouras. But Mr. Kasoura is challenging the truth of Childan's story. So he is going to ask a few questions about Childan's so-called \"personal treasure\" necklace.
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This \"personal treasure\" is a multiset S of m \"01-strings\".
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A \"01-string\" is a string that contains n characters \"0\" and \"1\". For example, if n=4, strings \"0110\", \"0000\", and \"1110\" are \"01-strings\", but \"00110\" (there are 5 characters, not 4) and \"zero\" (unallowed characters) are not.
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Note that the multiset S can contain equal elements.
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Frequently, Mr. Kasoura will provide a \"01-string\" t and ask Childan how many strings s are in the multiset S such that the \"Wu\" value of the pair (s, t) is not greater than k.
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Mrs. Kasoura and Mr. Kasoura think that if s_i = t_i (1≤ i≤ n) then the \"Wu\" value of the character pair equals to w_i, otherwise 0. The \"Wu\" value of the \"01-string\" pair is the sum of the \"Wu\" values of every character pair. Note that the length of every \"01-string\" is equal to n.
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For example, if w=[4, 5, 3, 6], \"Wu\" of (\"1001\", \"1100\") is 7 because these strings have equal characters only on the first and third positions, so w_1+w_3=4+3=7.
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You need to help Childan to answer Mr. Kasoura's queries. That is to find the number of strings in the multiset S such that the \"Wu\" value of the pair is not greater than k.
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Input
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The first line contains three integers n, m, and q (1≤ n≤ 12, 1≤ q, m≤ 5⋅ 10^5) — the length of the \"01-strings\", the size of the multiset S, and the number of queries.
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The second line contains n integers w_1, w_2, …, w_n (0 ≤ w_i ≤ 100) — the value of the i-th caracter.
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Each of the next m lines contains the \"01-string\" s of length n — the string in the multiset S.
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Each of the next q lines contains the \"01-string\" t of length n and integer k (0≤ k≤ 100) — the query.
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Output
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For each query, print the answer for this query.
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Examples
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Input
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2 4 5
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40 20
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01
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01
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10
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11
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00 20
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00 40
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11 20
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11 40
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11 60
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Output
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2
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4
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2
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3
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4
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Input
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1 2 4
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100
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0
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1
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0 0
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0 100
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1 0
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1 100
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Output
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1
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2
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1
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2
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Note
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In the first example, we can get:
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\"Wu\" of (\"01\", \"00\") is 40.
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\"Wu\" of (\"10\", \"00\") is 20.
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\"Wu\" of (\"11\", \"00\") is 0.
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\"Wu\" of (\"01\", \"11\") is 20.
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\"Wu\" of (\"10\", \"11\") is 40.
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\"Wu\" of (\"11\", \"11\") is 60.
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In the first query, pairs (\"11\", \"00\") and (\"10\", \"00\") satisfy the condition since their \"Wu\" is not greater than 20.
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In the second query, all strings satisfy the condition.
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In the third query, pairs (\"01\", \"11\") and (\"01\", \"11\") satisfy the condition. Note that since there are two \"01\" strings in the multiset, the answer is 2, not 1.
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In the fourth query, since k was increased, pair (\"10\", \"11\") satisfies the condition too.
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In the fifth query, since k was increased, pair (\"11\", \"11\") satisfies the condition too.
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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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const int maxn = 5e5 + 5;
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int w[15];
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int a[maxn];
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int n;
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int change(char *s) {
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int res = 0, tmp = 1;
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for (int i = n - 1; ~i; i--) {
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if (s[i] == '1') res += tmp;
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tmp *= 2;
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}
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return res;
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}
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int sum[maxn / 5][105];
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int cnt[maxn / 5];
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int get2(int x, int t) {
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if (x & (1 << t))
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return 1;
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else
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return 0;
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}
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int main() {
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int m, q;
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char s[15];
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cin >> n >> m >> q;
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for (int i = 0; i < n; i++) cin >> w[i];
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for (int i = 0; i < m; i++) {
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scanf(\"%s\", s);
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a[i] = change(s);
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cnt[a[i]]++;
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}
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sort(a, a + m);
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int siz = unique(a, a + m) - a;
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int mx = (1 << n) - 1;
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int tmp;
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for (int i = 0; i <= mx; i++) {
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for (int j = 0; j < siz; j++) {
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tmp = 0;
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for (int t = 0; t < n; t++) {
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if (get2(i, t) ^ get2(a[j], t) == 0) {
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tmp += w[n - t - 1];
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}
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}
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if (tmp <= 100) sum[i][tmp] += cnt[a[j]];
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}
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for (int j = 1; j <= 100; j++) sum[i][j] += sum[i][j - 1];
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}
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int k, id;
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while (q--) {
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scanf(\"%s%d\", s, &k);
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id = change(s);
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printf(\"%d\n\", sum[id][k]);
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}
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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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import random
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import re
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from collections import defaultdict
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from bootcamp import Basebootcamp
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class Dthewubootcamp(Basebootcamp):
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def __init__(self, **params):
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self.n = params.get('n', random.randint(2, 4))
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self.m = params.get('m', random.randint(5, 20))
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self.n = min(max(self.n, 1), 12) # 确保n在合法范围
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self.m = min(max(self.m, 1), 10000)
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def case_generator(self):
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MAX_ATTEMPTS = 20
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n = self.n # 定义局部变量避免self引用错误
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w = [random.randint(0, 100) for _ in range(n)]
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s_multiset = defaultdict(int)
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for _ in range(self.m):
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s = ''.join(random.choices('01', k=n))
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s_multiset[s] += 1
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valid_k = None
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for _ in range(MAX_ATTEMPTS):
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t = ''.join(random.choices('01', k=n))
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wu_counts = defaultdict(int)
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for s, cnt in s_multiset.items():
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xor = int(s, 2) ^ int(t, 2)
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# 修正变量作用域错误（原错误行）
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wu = sum(w[n - i - 1] for i in range(n) if (xor & (1 << i)) == 0)
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wu_counts[wu] += cnt
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sorted_wu = sorted(wu_counts.keys())
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if not sorted_wu:
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valid_k = 0
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break
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for candidate in [sorted_wu[len(sorted_wu)//2],
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sorted_wu[0] + (sorted_wu[-1]-sorted_wu[0])//3,
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sorted_wu[-1]]:
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total = sum(cnt for wu, cnt in wu_counts.items() if wu <= candidate)
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if 0 < total < self.m:
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valid_k = candidate
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break
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if valid_k is not None:
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break
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if valid_k is None:
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valid_k = random.choice(sorted_wu) if sorted_wu else 0
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return {
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'n': n,
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'w': w,
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's_multiset': dict(s_multiset),
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't': t,
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'k': valid_k
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}
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@staticmethod
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def prompt_func(question_case) -> str:
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problem_desc = [
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"## 谜题背景",
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"你需要帮助Childan验证'Dthewu'值阈值条件下的项链数量。每个01字符串的长度为n，每个位置的权重不同。",
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"\n## 参数说明",
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f"字符串长度 n = {question_case['n']}",
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f"权重数组 w = {question_case['w']}",
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"\n## 多集合S内容（字符串:出现次数）",
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*[f"- {s} × {cnt}" for s, cnt in question_case['s_multiset'].items()],
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"\n## 当前查询",
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f"目标字符串 t = {question_case['t']}",
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f"阈值 k = {question_case['k']}",
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"\n请计算满足条件的字符串总数，并将最终数值用[answer]标签包裹。"
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]
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return '\n'.join(problem_desc)
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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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if not matches:
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return None
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try:
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return int(matches[-1].strip())
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except (ValueError, TypeError):
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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n = identity['n']
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w = identity['w']
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s_multiset = identity['s_multiset']
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t_str = identity['t']
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k = identity['k']
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t_int = int(t_str, 2)
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correct = 0
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for s_str, count in s_multiset.items():
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s_int = int(s_str, 2)
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xor = s_int ^ t_int
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wu = sum(w[n - i - 1] for i in range(n) if (xor & (1 << i)) == 0)
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if wu <= k:
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correct += count
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return solution == correct
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