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222 lines
6 KiB
Python
Executable file
222 lines
6 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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#include <bits/stdc++.h>
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const long long INF = 2000000005;
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const long long BIG_INF = 2000000000000000005;
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const long long mod = 1000000007;
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const long long P = 31;
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const long double PI = 3.141592653589793238462643;
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const double eps = 1e-9;
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using namespace std;
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vector<pair<long long, long long> > dir = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
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bool valid(long long x, long long y, long long n, long long m) {
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return x >= 0 && y >= 0 && x < n && y < m;
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}
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mt19937 rng(1999999973);
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signed main() {
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ios_base::sync_with_stdio(0);
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cin.tie(0);
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cout.tie(0);
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;
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long long n, p;
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cin >> n >> p;
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for (long long i = 1; i < 1000000; i++) {
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if (n - p * i > 0 && __builtin_popcountll(n - p * i) <= i &&
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n - p * i >= i) {
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cout << i;
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return 0;
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}
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}
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cout << -1;
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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 random
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import re
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class Apbinarybootcamp(Basebootcamp):
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def __init__(self, **params):
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self.params = {
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'p_min': -1000,
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'p_max': 1000,
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'n_min': 1,
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'n_max': 10**9,
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'edge_case_prob': 0.2 # 20%生成边界案例
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}
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self.params.update(params)
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def case_generator(self):
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if random.random() < self.params['edge_case_prob']:
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return self._generate_edge_case()
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p = random.randint(self.params['p_min'], self.params['p_max'])
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n = random.randint(self.params['n_min'], self.params['n_max'])
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return {'n': n, 'p': p}
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def _generate_edge_case(self): # 修正：添加正确的缩进
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"""生成边界测试案例"""
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edge_types = [
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{'p': 0}, # 常规二进制
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{'p': -1000, 'n': 10**9}, # 最小p值
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{'p': 1000, 'n': 1}, # 无解情况
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{'n': 1, 'p': 1}, # 样例5
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{'p': -1, 'n': 2**20 + 1} # 大数值案例
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]
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case = random.choice(edge_types)
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p = case.get('p', random.randint(-1000, 1000))
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n = case.get('n', random.randint(1, 10**9))
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return {'n': n, 'p': p}
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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"""你是一个数学问题解决者，需要帮助解决一个关于p-binary数的特殊求和问题。给定两个整数n和p，找出能够组成n的最少数量的p-binary数，若不可能则返回-1。
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**p-binary数**的定义为：2^x + p，其中x是非负整数。允许使用重复的p-binary数，并且允许和为负数的情况。
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**输入参数**:
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n = {n}
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p = {p}
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请确定所需的最少p-binary数的数量，并将答案放在[answer]标签内。例如：[answer]3[/answer]。
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**示例**:
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- 输入：n=24, p=0 → 输出：2（因为24=16+8=2^4+0 + 2^3+0）
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- 输入：n=4, p=-7 → 输出：2（因为4= (16-7) + (2-7)）
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请严格遵循输出格式，将最终答案放置在[answer]和[/answer]之间。"""
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return prompt
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\]\s*(\-?\d+)\s*\[/answer\]', output, re.IGNORECASE)
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return int(matches[-1]) if matches else 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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p = identity['p']
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correct = cls.solve(n, p)
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return solution == correct
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@staticmethod
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def solve(n, p):
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if p == 0: # 优化常规二进制情况
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if (n & (n-1)) == 0:
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return 1
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return bin(n).count('1')
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max_i = 10**6 if p < 0 else min(10**6, n//abs(p)+2)
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for i in range(1, max_i+1):
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s = n - p * i
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if s <= 0:
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continue
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if s.bit_length() > 60: # 处理极大数值溢出
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continue
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if bin(s).count('1') <= i and s >= i:
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return i
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return -1
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