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234 lines
7.3 KiB
Python
Executable file
234 lines
7.3 KiB
Python
Executable file
"""#
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### 谜题描述
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Can the greatest common divisor and bitwise operations have anything in common? It is time to answer this question.
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Suppose you are given a positive integer a. You want to choose some integer b from 1 to a - 1 inclusive in such a way that the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers a ⊕ b and a \> \& \> b is as large as possible. In other words, you'd like to compute the following function:
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$$$f(a) = max_{0 < b < a}{gcd(a ⊕ b, a \> \& \> b)}.$$$
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Here ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR), and \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).
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The greatest common divisor of two integers x and y is the largest integer g such that both x and y are divided by g without remainder.
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You are given q integers a_1, a_2, …, a_q. For each of these integers compute the largest possible value of the greatest common divisor (when b is chosen optimally).
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Input
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The first line contains an integer q (1 ≤ q ≤ 10^3) — the number of integers you need to compute the answer for.
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After that q integers are given, one per line: a_1, a_2, …, a_q (2 ≤ a_i ≤ 2^{25} - 1) — the integers you need to compute the answer for.
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Output
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For each integer, print the answer in the same order as the integers are given in input.
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Example
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Input
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3
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2
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3
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5
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Output
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3
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1
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7
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Note
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For the first integer the optimal choice is b = 1, then a ⊕ b = 3, a \> \& \> b = 0, and the greatest common divisor of 3 and 0 is 3.
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For the second integer one optimal choice is b = 2, then a ⊕ b = 1, a \> \& \> b = 2, and the greatest common divisor of 1 and 2 is 1.
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For the third integer the optimal choice is b = 2, then a ⊕ b = 7, a \> \& \> b = 0, and the greatest common divisor of 7 and 0 is 7.
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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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#!/usr/bin/env python
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\"\"\"
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This file is part of https://github.com/Cheran-Senthil/PyRival.
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Copyright 2019 Cheran Senthilkumar <hello@cheran.io>
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\"\"\"
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from __future__ import division, print_function
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import cmath
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import itertools
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import math
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import operator as op
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# import random
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import sys
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from atexit import register
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from bisect import bisect_left, bisect_right
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# from collections import Counter, defaultdict, deque
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# from copy import deepcopy
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# from decimal import Decimal
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# from difflib import SequenceMatcher
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# from functools import reduce
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# from heapq import heappop, heappush
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from io import BytesIO, FileIO, StringIO
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if sys.version_info[0] < 3:
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class dict(dict):
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\"\"\"dict() -> new empty dictionary\"\"\"
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def items(self):
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\"\"\"D.items() -> a set-like object providing a view on D's items\"\"\"
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return dict.iteritems(self)
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def keys(self):
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\"\"\"D.keys() -> a set-like object providing a view on D's keys\"\"\"
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return dict.iterkeys(self)
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def values(self):
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\"\"\"D.values() -> an object providing a view on D's values\"\"\"
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return dict.itervalues(self)
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input = raw_input
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range = xrange
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filter = itertools.ifilter
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map = itertools.imap
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zip = itertools.izip
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INP_FILE = 0
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OUT_FILE = 1
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if sys.version_info[0] < 3:
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sys.stdin = BytesIO(FileIO(INP_FILE).read())
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sys.stdout = BytesIO()
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register(lambda: FileIO(OUT_FILE, 'w').write(sys.stdout.getvalue()))
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else:
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sys.stdin = StringIO(FileIO(INP_FILE).read().decode())
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sys.stdout = StringIO()
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register(lambda: FileIO(OUT_FILE, 'w').write(sys.stdout.getvalue().encode()))
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input = lambda: sys.stdin.readline().rstrip('\r\n')
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def main():
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q = int(input())
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n = [1, 1, 5, 1, 21, 1, 85, 73, 341, 89, 1365, 1, 5461, 4681, 21845, 1, 87381, 1, 349525, 299593, 1398101, 178481, 5592405, 1082401, 22369621]
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for _ in range(q):
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a = int(input())
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m = ((1 << a.bit_length()) - 1)
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if m == a:
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print(n[a.bit_length() - 2])
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else:
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print(m)
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if __name__ == '__main__':
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main()
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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 random
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import re
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class Cmeaninglessoperationsbootcamp(Basebootcamp):
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n = [
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1, 1, 5, 1, 21, 1, 85, 73, 341, 89, 1365, 1, 5461, 4681, 21845,
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1, 87381, 1, 349525, 299593, 1398101, 178481, 5592405, 1082401, 22369621
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]
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def __init__(self, all_ones_prob=0.5, min_k=2, max_k=25):
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"""
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Parameters:
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all_ones_prob: 生成全1二进制数的概率 (0-1)
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min_k: 最小二进制位数范围 (2-25)
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max_k: 最大二进制位数范围 (2-25)
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"""
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self.all_ones_prob = min(max(all_ones_prob, 0.0), 1.0)
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self.min_k = max(2, min_k)
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self.max_k = min(25, max_k)
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def case_generator(self):
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"""生成保证有效的测试用例"""
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if random.random() < self.all_ones_prob:
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# 生成全1二进制数 (格式: 2^k - 1)
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k = random.randint(self.min_k, self.max_k)
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return {"a": (1 << k) - 1}
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else:
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# 生成非全1数，确保至少有一个0位
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min_valid_k = max(self.min_k, 3) # 保证k>=3才有非全1解
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k = random.randint(min_valid_k, self.max_k)
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# 生成有效范围内数字
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a_min = (1 << (k-1)) + 1
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a_max = (1 << k) - 2
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a = random.randint(a_min, a_max)
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# 二次验证数字有效性
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while bin(a).count('0') == 0 or a.bit_length() != k:
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a = random.randint(a_min, a_max)
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return {"a": a}
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@staticmethod
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def prompt_func(question_case) -> str:
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a = question_case["a"]
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return f"""你需要解决一个基于位运算和最大公约数（GCD）的数学谜题。
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给定正整数 a={a}，请找到一个整数 b（1 ≤ b < a），使得以下两个值的GCD最大：
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1. a XOR b（按位异或）
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2. a AND b（按位与）
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请通过以下步骤解决：
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1. 分析不同b值对应的计算结果
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2. 找出使GCD最大的最优b值
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3. 计算并返回最大GCD值
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示例：
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当a=5时，选择b=2：
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- XOR: 5 ^ 2 = 7
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- AND: 5 & 2 = 0
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- GCD(7, 0) = 7
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请将最终答案放在[answer]和[/answer]标签之间，例如：
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[answer]7[/answer]"""
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@staticmethod
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def extract_output(output):
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# 提取最后一个符合格式的答案
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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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a = identity["a"]
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bit_len = a.bit_length()
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# 全1二进制数验证逻辑
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if (a & (a + 1)) == 0:
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# 参考数组索引计算
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correct_index = bit_len - 2
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if 0 <= correct_index < len(cls.n):
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correct = cls.n[correct_index]
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else:
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# 超出数组范围时使用数学公式计算
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correct = (1 << (bit_len - 1)) - 1
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else:
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# 非全1数逻辑
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correct = (1 << bit_len) - 1
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try:
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return int(solution) == correct
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except ValueError:
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return False
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