"""# ### 谜题描述 Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times: * A BC * B AC * C AB * AAA empty string Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source. Input The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'. The third line contains the number of queries Q (1 ≤ Q ≤ 105). The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times? Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U. It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|. Output Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise. Example Input AABCCBAAB ABCB 5 1 3 1 2 2 2 2 4 7 9 1 1 3 4 2 3 4 5 1 3 Output 10011 Note In the first query we can achieve the result, for instance, by using transitions . The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'. Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution. ```python #include #pragma GCC target(\"pclmul\") using ul = std::uint32_t; using ull = std::uint64_t; using li = std::int32_t; using ll = std::int64_t; using uss = std::uint8_t; const ul maxn = 1e5; char str[maxn + 2]; ul scb[maxn + 1]; ul sca[maxn + 1]; ul tcb[maxn + 1]; ul tca[maxn + 1]; ul q; int main() { std::scanf(\"%s\", str + 1); for (ul i = 1; str[i]; ++i) { if (str[i] == 'A') { sca[i] = sca[i - 1] + 1; scb[i] = scb[i - 1]; } else { scb[i] = scb[i - 1] + 1; } } std::scanf(\"%s\", str + 1); for (ul i = 1; str[i]; ++i) { if (str[i] == 'A') { tca[i] = tca[i - 1] + 1; tcb[i] = tcb[i - 1]; } else { tcb[i] = tcb[i - 1] + 1; } } std::scanf(\"%u\", &q); for (ul i = 1; i <= q; ++i) { ul a, b, c, d; std::scanf(\"%u%u%u%u\", &a, &b, &c, &d); ul sb = scb[b] - scb[a - 1]; ul tb = tcb[d] - tcb[c - 1]; ul sa = std::min(sca[b], b - a + 1); ul ta = std::min(tca[d], d - c + 1); if ((sb ^ tb) & 1) { std::putchar('0'); continue; } if (sa < ta) { std::putchar('0'); continue; } if ((sa - ta) % 3 == 0 && sb > tb) { std::putchar('0'); continue; } if ((sa - ta) % 3 != 0 && sb + 2 > tb) { std::putchar('0'); continue; } if (sb == 0 && tb != 0 && sa == ta) { std::putchar('0'); continue; } std::putchar('1'); } std::putchar('\n'); return 0; } ``` 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。 """ from bootcamp import Basebootcamp import random import re from bootcamp import Basebootcamp def compute_prefixes(s): sca = [0] * (len(s) + 1) scb = [0] * (len(s) + 1) for i in range(1, len(s) + 1): char = s[i-1] if char == 'A': sca[i] = sca[i-1] + 1 scb[i] = scb[i-1] else: sca[i] = sca[i-1] scb[i] = scb[i-1] + 1 return sca, scb def compute_query_result(sca, scb, tca, tcb, a, b, c, d): sb = scb[b] - scb[a-1] sa = min(sca[b], b - a + 1) tb = tcb[d] - tcb[c-1] ta = min(tca[d], d - c + 1) if (sb ^ tb) & 1: return '0' if sa < ta: return '0' if (sa - ta) % 3 == 0: if sb > tb: return '0' else: if (sb + 2) > tb: return '0' if sb == 0 and tb != 0 and sa == ta: return '0' return '1' class Dpickingstringsbootcamp(Basebootcamp): def __init__(self, max_s_length=10, max_t_length=10, max_queries=5, max_attempts=10): self.max_s_length = max_s_length self.max_t_length = max_t_length self.max_queries = max_queries self.max_attempts = max_attempts def case_generator(self): for _ in range(self.max_attempts): s = ''.join(random.choices(['A', 'B', 'C'], k=random.randint(1, self.max_s_length))) t = ''.join(random.choices(['A', 'B', 'C'], k=random.randint(1, self.max_t_length))) sca, scb = compute_prefixes(s) tca, tcb = compute_prefixes(t) a, b = 1, len(s) c, d = 1, len(t) if compute_query_result(sca, scb, tca, tcb, a, b, c, d) == '1': queries = [[a, b, c, d]] for _ in range(self.max_queries - 1): a_q = random.randint(1, len(s)) b_q = random.randint(a_q, len(s)) c_q = random.randint(1, len(t)) d_q = random.randint(c_q, len(t)) queries.append([a_q, b_q, c_q, d_q]) return { 'S': s, 'T': t, 'queries': queries } # Fallback if no valid case found s = ''.join(random.choices(['A', 'B', 'C'], k=random.randint(1, self.max_s_length))) t = ''.join(random.choices(['A', 'B', 'C'], k=random.randint(1, self.max_t_length))) queries = [] for _ in range(self.max_queries): a = random.randint(1, len(s)) b = random.randint(a, len(s)) c = random.randint(1, len(t)) d = random.randint(c, len(t)) queries.append([a, b, c, d]) return { 'S': s, 'T': t, 'queries': queries } @staticmethod def prompt_func(question_case): s = question_case['S'] t = question_case['T'] queries = question_case['queries'] prompt = f"""Alice有一个字符串S，内容为：{s}。Bob可以对该字符串的任意子串应用以下转换规则（次数不限、顺序不限）： - 将单个'A'替换为'BC'； - 将单个'B'替换为'AC'； - 将单个'C'替换为'AB'； - 将连续的三个'A'（即"AAA"）替换为空字符串（即删除）。 现有目标字符串T，内容为：{t}。你需要处理{len(queries)}个查询，每个查询询问是否存在一种可能，通过应用上述规则，将S的某个子串转换为T的对应子串。 每个查询的参数为四个整数a、b、c、d，表示：源子串是S的第a到第b个字符（包含两端，按1-based索引），目标子串是T的第c到第d个字符。要求判断是否可以进行这样的转换。 请依次对每个查询给出“是”（用'1'表示）或“否”（用'0'表示）的答案，并将所有答案按顺序组成一个连续的字符串，无需分隔符。 请将最终答案放置在[answer]和[/answer]标签之间。例如：[answer]101[/answer]。""" return prompt @staticmethod def extract_output(output): matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL) if not matches: return None return matches[-1].strip() @classmethod def _verify_correction(cls, solution, identity): s = identity['S'] t = identity['T'] queries = identity['queries'] sca, scb = compute_prefixes(s) tca, tcb = compute_prefixes(t) expected = [] for a, b, c, d in queries: res = compute_query_result(sca, scb, tca, tcb, a, b, c, d) expected.append(res) expected_str = ''.join(expected) return solution == expected_str