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158 lines
5.1 KiB
Python
Executable file
158 lines
5.1 KiB
Python
Executable file
"""#
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### 谜题描述
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One day in the IT lesson Anna and Maria learned about the lexicographic order.
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String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
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The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string \"aab\": \"a\", \"a\", \"aa\", \"ab\", \"aab\", \"b\"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework.
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Input
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The first line contains a non-empty string that only consists of small Latin letters (\"a\"-\"z\"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105).
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Output
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Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying \"No such line.\" (without the quotes).
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Examples
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Input
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aa
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2
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Output
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a
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Input
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abc
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5
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Output
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bc
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Input
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abab
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7
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Output
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b
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Note
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In the second sample before string \"bc\" follow strings \"a\", \"ab\", \"abc\", \"b\".
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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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from heapq import heappush
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from heapq import heappop
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from heapq import heapify
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S=raw_input()
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n=input()
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q=[]
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N=len(S)
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if N*(N+1)/2<n:
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print \"No such line.\"
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exit()
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i=0
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while i < N:
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q.append((S[i],i))
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i+=1
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heapify(q)
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while n:
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n-=1
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t=heappop(q)
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if not n :
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print t[0]
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if t[1]+1 < N:
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heappush(q,(t[0]+S[t[1]+1],t[1]+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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from heapq import heappush, heappop, heapify
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from bootcamp import Basebootcamp
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class Bstringbootcamp(Basebootcamp):
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def __init__(self, max_str_length=10, max_k=100000):
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self.max_str_length = max_str_length
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self.max_k = max_k
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def case_generator(self):
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# Generate random string (length >=1)
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length = random.randint(1, self.max_str_length)
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chars = [random.choice('abcdefghijklmnopqrstuvwxyz') for _ in range(length)]
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s = ''.join(chars)
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# Calculate total possible substrings
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total_substrings = length * (length + 1) // 2
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# Generate k with controlled distribution
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if random.random() < 0.7: # 70% valid k
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k = random.randint(1, min(total_substrings, self.max_k))
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else: # 30% invalid k
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k = random.randint(
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min(total_substrings + 1, self.max_k),
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self.max_k # Ensure k never exceeds problem constraints
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)
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return {'s': s, 'k': k}
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@staticmethod
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def prompt_func(question_case) -> str:
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s = question_case['s']
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k = question_case['k']
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prompt = (
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f"In an IT lesson, Anna and Maria learned about lexicographic order. Their homework is to find the k-th lexicographically smallest substring of a given string.\n\n"
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f"**Rules:**\n"
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f"1. All possible contiguous substrings are considered (including duplicates from different starting positions). For example, 'aa' has substrings: 'a'(pos0), 'a'(pos1), 'aa'.\n"
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f"2. Substrings are ordered lexicographically as defined by the '<' operator.\n"
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f"3. If there are fewer than k substrings, output \"No such line.\"\n\n"
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f"**Input:**\nString: {s}\nk: {k}\n\n"
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f"**Task:**\nOutput the k-th substring. Enclose your answer in [answer][/answer] tags.\n\n"
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f"**Example:**\nInput: aa\nk=2\nAnswer: [answer]a[/answer]"
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)
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return prompt
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@staticmethod
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def extract_output(output):
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import re
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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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s = identity['s']
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k = identity['k']
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# Edge case: k exceeds maximum possible
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total = len(s) * (len(s)+1) // 2
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if k > total or k > identity.get('max_k', 10**5):
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return solution == "No such line."
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# Heap-based k-th smallest calculation
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heap = [(s[i], i) for i in range(len(s))]
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heapify(heap)
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result = None
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for _ in range(k):
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curr, pos = heappop(heap)
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result = curr
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if pos + 1 < len(s):
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heappush(heap, (curr + s[pos+1], pos+1))
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return solution == result
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