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253 lines
7.1 KiB
Python
Executable file
253 lines
7.1 KiB
Python
Executable file
"""#
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### 谜题描述
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The only difference between easy and hard versions is the length of the string. You can hack this problem only if you solve both problems.
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Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
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* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
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* The number of zeroes in t is the maximum possible.
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A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
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If there are multiple substrings which satisfy the conditions, output any.
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Input
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The first line contains a binary string of length not more than 2\: 000.
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Output
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Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
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Examples
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Input
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110
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Output
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010
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Input
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010
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Output
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010
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Input
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0001111
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Output
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0000000
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Input
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0111001100111011101000
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Output
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0011001100001011101000
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Note
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In the first example:
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* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
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* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
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* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
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* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
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The second example is similar to the first one.
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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 __future__ import division, print_function
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def main():
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s = list(map(int, input()))
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n = len(s)
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t = [0] * n
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st = []
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for i in range(n):
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if s[i] == 1:
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st.append(i)
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else:
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if st:
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v = st.pop()
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t[v] = 1
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print(*t, sep='')
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INF = float('inf')
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MOD = 10**9 + 7
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import os, sys
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from atexit import register
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from io import BytesIO
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import itertools
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if sys.version_info[0] < 3:
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input = raw_input
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range = xrange
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filter = itertools.ifilter
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map = itertools.imap
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zip = itertools.izip
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if \"LOCAL_\" in os.environ:
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debug_print = print
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else:
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sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
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sys.stdout = BytesIO()
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register(lambda: os.write(1, sys.stdout.getvalue()))
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input = lambda: sys.stdin.readline().rstrip('\r\n')
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debug_print = lambda *x, **y: None
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def input_as_list():
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return list(map(int, input().split()))
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def array_of(f, *dim):
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return [array_of(f, *dim[1:]) for _ in range(dim[0])] if dim else f()
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main()
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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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class D1kirkandabinarystringeasyversionbootcamp(Basebootcamp):
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def __init__(self, min_length=3, max_length=2000):
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"""
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初始化训练场参数，设置生成字符串的最小和最大长度。
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添加特殊场景生成概率参数，增强测试覆盖率
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"""
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self.min_length = min_length
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self.max_length = max_length
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self.special_case_prob = 0.3 # 30%概率生成特殊场景案例
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def case_generator(self):
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"""
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生成二进制字符串s，包含随机、全0、全1、交替模式等多种场景
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根据算法生成对应的正确解t
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"""
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n = random.randint(self.min_length, self.max_length)
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# 30%概率生成特殊模式字符串
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if random.random() < self.special_case_prob:
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pattern = random.choice(['all_zero', 'all_one', 'alternate'])
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if pattern == 'all_zero':
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s = '0' * n
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elif pattern == 'all_one':
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s = '1' * n
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else: # alternate模式
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s = ('01' * (n//2 + 1))[:n]
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else:
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s = ''.join(random.choices(['0', '1'], k=n))
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t = self.compute_t(s)
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return {"s": s, "t": t}
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@staticmethod
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def compute_t(s_str):
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"""
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根据参考算法计算给定二进制字符串s的正确解t。
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添加类型转换确保处理字节型数据
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"""
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s = list(map(int, s_str))
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n = len(s)
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t = [0] * n
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stack = []
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for i in range(n):
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if s[i] == 1:
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stack.append(i)
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else:
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if stack:
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idx = stack.pop()
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t[idx] = 1
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return ''.join(map(str, t))
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@staticmethod
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def prompt_func(question_case) -> str:
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"""
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增强问题描述的规则说明，补充边界条件说明
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"""
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s = question_case['s']
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prompt = f"""你是一个二进制字符串处理专家，请解决以下问题：
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给定一个二进制字符串s，构造一个新的二进制字符串t，满足下列条件：
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1. 对于所有可能的子区间[l, r] (1 ≤ l ≤ r ≤ n)，s的子串s[l..r]的最长非递减子序列(LNDS)长度必须严格等于t对应子串t[l..r]的LNDS长度
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2. 在满足条件1的前提下，t中0的数量要尽可能多
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附加规则说明：
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- 子序列不需要连续，但元素必须保持原始顺序
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- 当s全为0时，t必须等于s（此时已是最优解）
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- 当s全为1时，t可以是全0（观察示例验证此情况）
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- 对于形如"01"交替的字符串，需要确保每个1的位置被正确处理
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输入s为：{s}
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你的输出必须是长度与s相同（{len(s)}位）的二进制字符串。请将最终答案放在[answer]和[/answer]标签之间，例如：[answer]010[/answer]。答案必须完全匹配正则表达式：^[01]+$"""
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return prompt
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@staticmethod
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def extract_output(output):
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"""
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强化提取逻辑，增加格式校验和非法字符过滤
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"""
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# 匹配最后一个有效答案块
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matches = re.findall(r'\[answer\](.*?)\[\/answer\]', output, re.DOTALL | re.IGNORECASE)
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if not matches:
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return None
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# 提取并清理答案
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raw_answer = matches[-1].strip().replace('\n', '').replace(' ', '')
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# 过滤非法字符
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filtered = re.sub(r'[^01]', '', raw_answer)
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return filtered if len(filtered) == len(raw_answer) and filtered else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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"""
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增加solution格式校验，确保与s等长
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"""
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if not solution or len(solution) != len(identity['s']):
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return False
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if not re.match('^[01]+$', solution):
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return False
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return solution == identity['t']
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