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167 lines
5.1 KiB
Python
Executable file
167 lines
5.1 KiB
Python
Executable file
"""#
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### 谜题描述
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You've got an array, consisting of n integers a1, a2, ..., an. Also, you've got m queries, the i-th query is described by two integers li, ri. Numbers li, ri define a subsegment of the original array, that is, the sequence of numbers ali, ali + 1, ali + 2, ..., ari. For each query you should check whether the corresponding segment is a ladder.
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A ladder is a sequence of integers b1, b2, ..., bk, such that it first doesn't decrease, then doesn't increase. In other words, there is such integer x (1 ≤ x ≤ k), that the following inequation fulfills: b1 ≤ b2 ≤ ... ≤ bx ≥ bx + 1 ≥ bx + 2... ≥ bk. Note that the non-decreasing and the non-increasing sequences are also considered ladders.
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Input
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The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of array elements and the number of queries. The second line contains the sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 109), where number ai stands for the i-th array element.
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The following m lines contain the description of the queries. The i-th line contains the description of the i-th query, consisting of two integers li, ri (1 ≤ li ≤ ri ≤ n) — the boundaries of the subsegment of the initial array.
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The numbers in the lines are separated by single spaces.
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Output
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Print m lines, in the i-th line print word \"Yes\" (without the quotes), if the subsegment that corresponds to the i-th query is the ladder, or word \"No\" (without the quotes) otherwise.
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Examples
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Input
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8 6
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1 2 1 3 3 5 2 1
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1 3
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2 3
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2 4
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8 8
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1 4
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5 8
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Output
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Yes
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Yes
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No
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Yes
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No
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Yes
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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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n,m=R()
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a=R()
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q=[i for i in range(n)]
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w=[i for i in range(n)]
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for i in reversed(range(n-1)):
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if (a[i]<=a[i+1]):q[i]=q[i+1]
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if (a[i]>=a[i+1]):w[i]=w[i+1]
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for i in xrange(m):
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l,r=R()
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if (w[q[l-1]]>=r-1):
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print \"Yes\"
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else:print \"No\"
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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 re
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import random
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from bootcamp import Basebootcamp
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class Cladderbootcamp(Basebootcamp):
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def __init__(self, n=8, m=6, max_val=10**9, seed=None):
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super().__init__()
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self.n = n
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self.m = m
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self.max_val = max_val
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self.seed = seed
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if seed is not None:
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random.seed(seed)
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def case_generator(self):
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if self.seed is not None:
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random.seed(self.seed)
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# 生成随机数组时允许非单调序列（含波峰波谷）
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a = [random.randint(1, self.max_val) for _ in range(self.n)]
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n = self.n
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# 预处理数组
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q = list(range(n)) # 每个位置能向右扩展到的最远非递减位置
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w = list(range(n)) # 从q[i]开始能向右扩展到的最远非递增位置
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for i in reversed(range(n-1)): # 从倒数第二位向前遍历
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if a[i] <= a[i+1]:
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q[i] = q[i+1]
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if a[i] >= a[i+1]:
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w[i] = w[i+1]
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# 生成多样化查询（包含全段、单元素、随机范围）
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queries = []
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answers = []
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for _ in range(self.m):
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# 生成合法的1-based区间
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li = random.randint(1, n)
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ri = random.randint(li, n)
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queries.append([li, ri])
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# 转换为0-based索引
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l_0, r_0 = li-1, ri-1
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peak_pos = q[l_0]
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final_pos = w[peak_pos]
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answers.append("Yes" if final_pos >= r_0 else "No")
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return {
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'n': self.n,
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'm': self.m,
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'a': a,
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'queries': queries,
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'answers': answers
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}
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@staticmethod
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def prompt_func(question_case) -> str:
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a_str = ' '.join(map(str, question_case['a']))
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queries = [f"{q[0]} {q[1]}" for q in question_case['queries']]
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return f"""You need to determine if subarrays form ladder sequences.
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**Rules:**
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1. A ladder first non-decreases then non-increases (can be entirely non-decreasing or non-increasing)
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2. Array indices are 1-based
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**Input:**
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- Array: {a_str}
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- Queries:
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{chr(10).join(queries)}
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**Output Format:**
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Put each answer (Yes/No) on separate lines within [answer] tags like:
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[answer]
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Yes
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No
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...
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[/answer]"""
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@staticmethod
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def extract_output(output):
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# 匹配最后一个完整的答案块
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matches = re.findall(
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r'\[answer\](.*?)\[\/answer\]',
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output,
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flags=re.DOTALL|re.IGNORECASE
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)
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if not matches:
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return None
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answers = []
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for line in matches[-1].strip().splitlines():
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line = line.strip()
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if re.fullmatch(r'yes', line, re.I):
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answers.append("Yes")
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elif re.fullmatch(r'no', line, re.I):
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answers.append("No")
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return answers if len(answers) > 0 else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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# solution需要与预设答案完全匹配
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return solution == identity.get('answers', [])
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