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internbootcamp/bootcamp/cpalindromifier/cpalindromifier.py
Executable file
289
internbootcamp/bootcamp/cpalindromifier/cpalindromifier.py
Executable file
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"""#
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### 谜题描述
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Ringo found a string s of length n in his [yellow submarine](https://www.youtube.com/watch?v=m2uTFF_3MaA). The string contains only lowercase letters from the English alphabet. As Ringo and his friends love palindromes, he would like to turn the string s into a palindrome by applying two types of operations to the string.
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The first operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_2s_3 … s_i (i - 1 characters) reversed to the front of s.
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The second operation allows him to choose i (2 ≤ i ≤ n-1) and to append the substring s_i s_{i + 1}… s_{n - 1} (n - i characters) reversed to the end of s.
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Note that characters in the string in this problem are indexed from 1.
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For example suppose s=abcdef. If he performs the first operation with i=3 then he appends cb to the front of s and the result will be cbabcdef. Performing the second operation on the resulted string with i=5 will yield cbabcdefedc.
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Your task is to help Ringo make the entire string a palindrome by applying any of the two operations (in total) at most 30 times. The length of the resulting palindrome must not exceed 10^6
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It is guaranteed that under these constraints there always is a solution. Also note you do not have to minimize neither the number of operations applied, nor the length of the resulting string, but they have to fit into the constraints.
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Input
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The only line contains the string S (3 ≤ |s| ≤ 10^5) of lowercase letters from the English alphabet.
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Output
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The first line should contain k (0≤ k ≤ 30) — the number of operations performed.
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Each of the following k lines should describe an operation in form L i or R i. L represents the first operation, R represents the second operation, i represents the index chosen.
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The length of the resulting palindrome must not exceed 10^6.
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Examples
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Input
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abac
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Output
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2
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R 2
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R 5
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Input
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acccc
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Output
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2
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L 4
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L 2
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Input
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hannah
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Output
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0
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Note
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For the first example the following operations are performed:
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abac → abacab → abacaba
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The second sample performs the following operations: acccc → cccacccc → ccccacccc
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The third example is already a palindrome so no operations are required.
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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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def check(c):
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for i in xrange(len(c)):
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if c[len(c) - i - 1] != c[i]:
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print i
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return False
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return True
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def process(st):
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print 5
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st = st + st[1:-1][::-1]
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print st
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a, b, H, b = st[0], st[-1], st[2:-1], st[-1]
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print a, b, H, b
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print b+a+b+H+b
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print b+a+b+H+b+H+b+a
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print b+a+b+H+b+H+b+a+b
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c = b+a+b+H+b+H+b+a+b
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print check(c)
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def ans(st):
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print 4
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st = st + st[1:-1][::-1]
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a, b, H, b = st[0], st[-1], st[2:-1], st[-1]
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#print a, b, H, b
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print 'R', 2
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#print b+a+b+H+b
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print 'L', 2
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#print b+a+b+H+b+H+b+a
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print 'R', 2
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#print b+a+b+H+b+H+b+a+b
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print 'R', len(b+a+b+H+b+H+b+a) - 1
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#c = b+a+b+H+b+H+b+a+b
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#print c
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#print check(c)
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def main():
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t = input()
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for _ in xrange(t):
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n = input()
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process(n)
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if __name__ == '__main__':
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#main()
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s = raw_input()
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#process(s)
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ans(s)
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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 string
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from bootcamp import Basebootcamp
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class Cpalindromifierbootcamp(Basebootcamp):
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def __init__(self, min_length=3, max_length=100, palindrome_prob=0.2):
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"""
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参数说明:
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min_length: 生成字符串最小长度 (≥3)
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max_length: 生成字符串最大长度 (≤1e5)
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palindrome_prob: 直接生成回文字符串的概率
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"""
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self.min_length = max(3, min_length)
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self.max_length = min(100000, max_length) # 遵守题目上限
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self.palindrome_prob = palindrome_prob
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def case_generator(self):
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"""生成满足题目长度限制的字符串"""
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length = random.randint(self.min_length, self.max_length)
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if random.random() < self.palindrome_prob:
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s = self._generate_palindrome(length)
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else:
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s = ''.join(random.choices(string.ascii_lowercase, k=length))
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# 确保生成的字符串不是回文
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while s == s[::-1]:
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s = ''.join(random.choices(string.ascii_lowercase, k=length))
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return {'s': s}
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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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return f"""Ringo found a string s consisting of lowercase letters. He wants to turn it into a palindrome using at most 30 operations. The allowed operations are:
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1. L i: Prepend the reversed substring from the 2nd to i-th character (2 ≤ i ≤ current length -1)
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2. R i: Append the reversed substring from the i-th to second-last character (2 ≤ i ≤ current length -1)
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The final string must be a palindrome with length ≤ 1,000,000.
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Current string: {s}
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Provide your solution in the format:
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k
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op1 i1
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op2 i2
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...
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Enclose your answer within [answer] and [/answer] tags. For example:
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[answer]
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2
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R 2
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R 5
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[/answer]"""
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@staticmethod
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def extract_output(output):
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# 支持嵌套标签的情况，查找最后一个匹配的标签对
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start_tag = '[answer]'
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end_tag = '[/answer]'
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last_start = output.rfind(start_tag)
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if last_start == -1:
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return None
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last_end = output.find(end_tag, last_start)
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if last_end == -1:
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return None
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return output[last_start+len(start_tag):last_end].strip()
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@classmethod
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def _verify_correction(cls, solution, identity):
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try:
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lines = solution.strip().split('\n')
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if not lines:
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return False
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# 解析操作数量
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try:
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k = int(lines[0].strip())
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except ValueError:
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return False
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# 验证操作数量范围
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if k < 0 or k > 30:
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return False
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# 验证操作行数匹配
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if len(lines[1:]) != k:
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return False
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operations = []
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current = identity['s']
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total_len = len(current)
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for line in lines[1:k+1]:
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parts = line.strip().split()
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if len(parts) != 2:
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return False
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op, i_str = parts
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if op not in ('L', 'R'):
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return False
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# 验证i的有效性
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try:
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i = int(i_str)
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except ValueError:
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return False
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m = len(current)
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if i < 2 or i > m - 1:
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return False # 违反操作规则
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# 执行操作（验证阶段不实际执行，只检查有效性）
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if op == 'L':
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new_part_len = i - 1 # 新增字符数
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else:
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new_part_len = (m - 1) - (i - 1) + 1 # 原题中i到n-1的子串长度
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total_len += new_part_len
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if total_len > 1e6:
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return False
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operations.append((op, i))
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# 实际操作验证
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current = identity['s']
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for op, i in operations:
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m = len(current)
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if op == 'L':
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# 取s[1:i] (1-based到i-1) 对应Python的[1-1:i-1] => [0:i-1]
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part = current[0:i-1] # s_2到s_i的原始部分
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prepend = part[::-1]
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current = prepend + current
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else:
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# 取s[i-1:m-1] (原题i到n-1的Python表示)
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part = current[i-1:m-1]
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append_part = part[::-1]
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current = current + append_part
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return current == current[::-1]
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except Exception as e:
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return False
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@staticmethod
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def _generate_palindrome(length):
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"""生成随机回文字符串的优化版本"""
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half = (length + 1) // 2
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chars = [random.choice(string.ascii_lowercase) for _ in range(half)]
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return ''.join(chars + chars[:-1 if length % 2 else None][::-1])
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