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202 lines
5.1 KiB
Python
Executable file
202 lines
5.1 KiB
Python
Executable file
"""#
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### 谜题描述
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Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer p (which may be positive, negative, or zero). To combine their tastes, they invented p-binary numbers of the form 2^x + p, where x is a non-negative integer.
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For example, some -9-binary (\"minus nine\" binary) numbers are: -8 (minus eight), 7 and 1015 (-8=2^0-9, 7=2^4-9, 1015=2^{10}-9).
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The boys now use p-binary numbers to represent everything. They now face a problem: given a positive integer n, what's the smallest number of p-binary numbers (not necessarily distinct) they need to represent n as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.
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For example, if p=0 we can represent 7 as 2^0 + 2^1 + 2^2.
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And if p=-9 we can represent 7 as one number (2^4-9).
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Note that negative p-binary numbers are allowed to be in the sum (see the Notes section for an example).
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Input
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The only line contains two integers n and p (1 ≤ n ≤ 10^9, -1000 ≤ p ≤ 1000).
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Output
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If it is impossible to represent n as the sum of any number of p-binary numbers, print a single integer -1. Otherwise, print the smallest possible number of summands.
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Examples
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Input
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24 0
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Output
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2
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Input
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24 1
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Output
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3
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Input
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24 -1
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Output
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4
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Input
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4 -7
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Output
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2
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Input
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1 1
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Output
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-1
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Note
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0-binary numbers are just regular binary powers, thus in the first sample case we can represent 24 = (2^4 + 0) + (2^3 + 0).
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In the second sample case, we can represent 24 = (2^4 + 1) + (2^2 + 1) + (2^0 + 1).
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In the third sample case, we can represent 24 = (2^4 - 1) + (2^2 - 1) + (2^2 - 1) + (2^2 - 1). Note that repeated summands are allowed.
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In the fourth sample case, we can represent 4 = (2^4 - 7) + (2^1 - 7). Note that the second summand is negative, which is allowed.
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In the fifth sample case, no representation is possible.
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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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n, p = map(int, raw_input().strip().split())
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k = 0
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x = n
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while 1:
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if (x <= 0):
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print -1
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exit()
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cnt = bin(x).count('1')
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if cnt <= k:
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if x >= k:
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print k
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exit()
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k += 1
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x -= p
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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 Cpbinarybootcamp(Basebootcamp):
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def __init__(self, max_n=10**9, min_p=-1000, max_p=1000):
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self.max_n = max_n
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self.min_p = min_p
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self.max_p = max_p
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def case_generator(self):
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# 生成保证有解的案例
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max_attempts = 100
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for _ in range(max_attempts):
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n = random.randint(1, self.max_n)
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p = random.randint(self.min_p, self.max_p)
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# 检查是否有可能的k值
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k_found = None
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x = n
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for k_candidate in range(0, 1000): # 限制k的范围以确保终止
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x_current = x - k_candidate * p
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if x_current <= 0:
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break
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if bin(x_current).count('1') <= k_candidate and x_current >= k_candidate:
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k_found = k_candidate
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break
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if k_found is not None:
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return {'n': n, 'p': p}
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# 回退到p=0的合法案例
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n = random.randint(1, self.max_n)
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return {'n': n, 'p': 0}
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@staticmethod
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def prompt_func(question_case) -> str:
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n = question_case['n']
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p = question_case['p']
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prompt = f"""Vasya和Petya定义了一种p-binary数，形式为2^x + p（x为非负整数）。给定n和p，求用最少数量的p-binary数相加得到n。如果无解，返回-1。
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输入:
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n = {n}
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p = {p}
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输出要求:
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- 答案必须为整数，格式如[answer]答案[/answer]，例如[answer]3[/answer]或[answer]-1[/answer]。
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- 确保答案为最小可能的数量。"""
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return prompt
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@staticmethod
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def extract_output(output):
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# 提取最后一个[answer]标签内容
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answers = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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if not answers:
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return None
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last_answer = answers[-1].strip()
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try:
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return int(last_answer)
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except ValueError:
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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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MAX_K_CHECK = 10000 # 防御性设计防止无限循环
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n = identity['n']
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p = identity['p']
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if solution == -1:
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# 检查是否存在有效k ≤ MAX_K_CHECK
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for k in range(0, MAX_K_CHECK + 1):
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x = n - k * p
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if x <= 0:
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continue
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if bin(x).count('1') <= k and x >= k:
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return False # 存在解但返回了-1，错误
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return True # 未找到有效k
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else:
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x = n - solution * p
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if x <= 0:
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return False
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return bin(x).count('1') <= solution and x >= solution
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