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189 lines
5.4 KiB
Python
Executable file
189 lines
5.4 KiB
Python
Executable file
"""#
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### 谜题描述
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Karafs is some kind of vegetable in shape of an 1 × h rectangle. Tavaspolis people love Karafs and they use Karafs in almost any kind of food. Tavas, himself, is crazy about Karafs.
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<image>
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Each Karafs has a positive integer height. Tavas has an infinite 1-based sequence of Karafses. The height of the i-th Karafs is si = A + (i - 1) × B.
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For a given m, let's define an m-bite operation as decreasing the height of at most m distinct not eaten Karafses by 1. Karafs is considered as eaten when its height becomes zero.
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Now SaDDas asks you n queries. In each query he gives you numbers l, t and m and you should find the largest number r such that l ≤ r and sequence sl, sl + 1, ..., sr can be eaten by performing m-bite no more than t times or print -1 if there is no such number r.
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Input
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The first line of input contains three integers A, B and n (1 ≤ A, B ≤ 106, 1 ≤ n ≤ 105).
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Next n lines contain information about queries. i-th line contains integers l, t, m (1 ≤ l, t, m ≤ 106) for i-th query.
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Output
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For each query, print its answer in a single line.
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Examples
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Input
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2 1 4
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1 5 3
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3 3 10
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7 10 2
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6 4 8
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Output
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4
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-1
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8
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-1
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Input
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1 5 2
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1 5 10
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2 7 4
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Output
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1
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2
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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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R=lambda:map(int,raw_input().split())
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a,b,n=R()
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for _ in range(n):
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l,t,m=R()
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lo,hi=0,10001000
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v=a+(l-1)*b
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while lo<hi:
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x=(lo+hi+1)/2
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if (v+v+(x-1)*b)*x>t*min(m,x)*2:
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hi=x-1
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else:
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lo=x
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print min(l+lo-1,(t-a)/b+1) if lo>0 else -1
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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 bootcamp import Basebootcamp
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def compute_r(A, B, l, t, m):
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v = A + (l - 1) * B
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if v > t:
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return -1
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lo, hi = 0, 10**8
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while lo < hi:
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mid = (lo + hi + 1) // 2
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sum_condition = (v + v + (mid - 1) * B) * mid
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right = t * min(m, mid) * 2
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if sum_condition > right:
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hi = mid - 1
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else:
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lo = mid
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if lo == 0:
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return -1
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max_r1 = l + lo - 1
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max_r2 = (t - A) // B + 1 if B != 0 else t
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r = min(max_r1, max_r2)
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return r if r >= l else -1
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class Ctavasandkarafsbootcamp(Basebootcamp):
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def __init__(self, A=None, B=None, **params):
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super().__init__(**params)
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# 调整参数范围，生成更合理的测试案例
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self.A = A if A is not None else random.randint(1, 100)
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self.B = B if B is not None else random.randint(1, 100)
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def case_generator(self):
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# 控制参数范围以提高案例质量
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for _ in range(100): # 防止无限循环
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l = random.randint(1, 50)
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s_l = self.A + (l-1)*self.B
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# 有50%概率生成有解案例
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if random.random() < 0.5:
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t = random.randint(s_l, s_l + 1000)
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else:
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t = random.randint(1, max(1, s_l - 1))
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m = random.randint(1, 100)
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expected_r = compute_r(self.A, self.B, l, t, m)
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# 确保生成的案例格式正确
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if expected_r != -1 or random.random() < 0.3: # 保留部分无解案例
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return {
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'A': self.A,
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'B': self.B,
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'l': l,
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't': t,
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'm': m,
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'expected_r': expected_r
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}
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# 保底返回一个无解案例
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l = random.randint(1, 50)
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return {
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'A': self.A,
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'B': self.B,
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'l': l,
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't': random.randint(1, 10),
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'm': random.randint(1, 10),
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'expected_r': -1
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}
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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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B = question_case['B']
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l = question_case['l']
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t = question_case['t']
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m = question_case['m']
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prompt = f"""你正在帮助SaDDas解决关于Ctavasandkarafs的查询问题。Ctavasandkarafs按照无限序列排列，第i个的高度为s_i = {A} + (i-1) × {B}。每次操作（m-bite）可以选择最多m个不同的未被吃掉的Ctavasandkarafs，每个减少1点高度。当一个Ctavasandkarafs的高度变为0时被吃掉，无法再被选择。
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给定查询参数：起始位置l={l}，最多允许t={t}次操作，每次操作最多选m={m}个Ctavasandkarafs。请找出最大的r满足以下条件：
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1. l ≤ r；
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2. 通过最多t次m-bite操作可以吃完第l到第r的所有Ctavasandkarafs。
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如果不存在这样的r，请输出-1。答案必须是整数，格式为[answer]答案[/answer]，例如：[answer]5[/answer]或[answer]-1[/answer]。
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当前问题参数：
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A = {A}
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B = {B}
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l = {l}
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t = {t}
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m = {m}
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注意：
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1. 最终答案必须满足：max(s_l,...,s_r) ≤ t 且总操作次数足够
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2. 答案只能放在[answer]标签内，其他位置将无法识别"""
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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'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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if not matches:
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return None
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last_match = matches[-1].strip()
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try:
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return int(last_match)
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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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return solution == identity['expected_r']
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