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175 lines
5.4 KiB
Python
Executable file
175 lines
5.4 KiB
Python
Executable file
"""#
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### 谜题描述
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Many modern text editors automatically check the spelling of the user's text. Some editors even suggest how to correct typos.
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In this problem your task to implement a small functionality to correct two types of typos in a word. We will assume that three identical letters together is a typo (for example, word \"helllo\" contains a typo). Besides, a couple of identical letters immediately followed by another couple of identical letters is a typo too (for example, words \"helloo\" and \"wwaatt\" contain typos).
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Write a code that deletes the minimum number of letters from a word, correcting described typos in the word. You are allowed to delete letters from both ends and from the middle of the word.
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Input
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The single line of the input contains word s, its length is from 1 to 200000 characters. The given word s consists of lowercase English letters.
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Output
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Print such word t that it doesn't contain any typos described in the problem statement and is obtained from s by deleting the least number of letters.
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If there are multiple solutions, print any of them.
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Examples
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Input
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helloo
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Output
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hello
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Input
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woooooow
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Output
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woow
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Note
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The second valid answer to the test from the statement is \"heloo\".
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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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word1 = raw_input()
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ans = [\"!\", \"@\"]
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for char in word1:
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if char == ans[-1] and (char == ans[-2] or ans[-2] == ans[-3]):
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ans.pop()
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ans.append(char)
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ans_str = \"\"
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for char in ans[2:]:
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ans_str += char
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print ans_str
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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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import string
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from collections import deque
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from bootcamp import Basebootcamp
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def process_word(s):
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"""严格遵循题目参考代码的处理逻辑"""
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ans = ["!", "@"]
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for char in s:
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while len(ans) >= 3 and (char == ans[-1] == ans[-2] or char == ans[-1] and ans[-3] == ans[-2]):
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ans.pop()
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ans.append(char)
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return ''.join(ans[2:])
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class Cfixingtyposbootcamp(Basebootcamp):
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def __init__(self, min_length=6, max_length=200):
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self.min_length = max(6, min_length) # 确保最小有效长度
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self.max_length = min(max_length, 20000) # 遵循题目输入约束
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def case_generator(self):
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"""生成保证包含至少一个有效错误的测试用例"""
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error_type = random.choice([1, 2, 1, 2]) # 增加类型1的概率
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s = []
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# 生成基础字符流（保证无预存错误）
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base_chars = [
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c for c in random.choices(string.ascii_lowercase,
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k=random.randint(self.min_length-3, self.max_length))
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if len(s) < 1 or c != s[-1] # 防止自然产生连续对
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]
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# 插入错误模式
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if error_type == 1: # 三连字符
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insert_pos = random.randint(0, len(base_chars)-1)
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c = base_chars[insert_pos] if random.random() < 0.5 else random.choice(string.ascii_lowercase)
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error = [c]*3
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else: # 连续重复对
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pairs = [random.choice(string.ascii_lowercase) for _ in range(2)]
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error = pairs*2 if random.random() < 0.5 else [pairs[0]]*2 + [pairs[1]]*2
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# 将错误模式插入随机位置
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insert_pos = random.randint(0, len(base_chars))
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s = base_chars[:insert_pos] + error + base_chars[insert_pos:]
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# 转换为字符串并校验有效性
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s = ''.join(s[:self.max_length])
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t = process_word(s)
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# 确保生成有效错误
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if len(t) >= len(s): # 重新生成直到产生有效错误
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return self.case_generator()
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return {
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'input': s,
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'correct_length': len(t),
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'_ref_solution': t # 调试用
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}
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@staticmethod
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def prompt_func(question_case) -> str:
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return f"""作为文本处理专家，请修正以下单词中的打字错误：
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输入单词: {question_case['input']}
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修正规则：
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1. 删除最少数量的字符
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2. 不允许三个连续相同字母（如"baaaad"→"baad"）
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3. 不允许两组连续重复对相邻（如"wooooooow"→"woow"）
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请将最终答案放在[answer]标签内，例如：[answer]corrected[/answer]。"""
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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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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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# 基础格式检查
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if not isinstance(solution, str) or not solution.isalpha():
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return False
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# 长度校验
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if len(solution) != identity['correct_length']:
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return False
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# 子序列验证
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src = deque(identity['input'])
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try:
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for c in solution:
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while src.popleft() != c:
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pass
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except IndexError:
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return False
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# 错误模式检查
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# 三级联检查
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for i in range(len(solution)-2):
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if solution[i] == solution[i+1] == solution[i+2]:
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return False
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# 连续两对检查
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for i in range(len(solution)-3):
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if solution[i] == solution[i+1] and solution[i+2] == solution[i+3]:
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return False
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return True
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