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197 lines
5.4 KiB
Python
Executable file
197 lines
5.4 KiB
Python
Executable file
"""#
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### 谜题描述
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In Python, code blocks don't have explicit begin/end or curly braces to mark beginning and end of the block. Instead, code blocks are defined by indentation.
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We will consider an extremely simplified subset of Python with only two types of statements.
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Simple statements are written in a single line, one per line. An example of a simple statement is assignment.
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For statements are compound statements: they contain one or several other statements. For statement consists of a header written in a separate line which starts with \"for\" prefix, and loop body. Loop body is a block of statements indented one level further than the header of the loop. Loop body can contain both types of statements. Loop body can't be empty.
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You are given a sequence of statements without indentation. Find the number of ways in which the statements can be indented to form a valid Python program.
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Input
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The first line contains a single integer N (1 ≤ N ≤ 5000) — the number of commands in the program. N lines of the program follow, each line describing a single command. Each command is either \"f\" (denoting \"for statement\") or \"s\" (\"simple statement\"). It is guaranteed that the last line is a simple statement.
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Output
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Output one line containing an integer - the number of ways the given sequence of statements can be indented modulo 109 + 7.
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Examples
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Input
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4
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s
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f
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f
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s
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Output
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1
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Input
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4
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f
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s
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f
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s
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Output
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2
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Note
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In the first test case, there is only one way to indent the program: the second for statement must be part of the body of the first one.
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simple statement
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for statement
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for statement
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simple statement
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In the second test case, there are two ways to indent the program: the second for statement can either be part of the first one's body or a separate statement following the first one.
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for statement
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simple statement
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for statement
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simple statement
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or
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for statement
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simple statement
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for statement
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simple statement
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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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mod = 10**9+7
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n = input()
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x = [1]
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for i in range(n):
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s = raw_input()
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if s=='f':
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x.insert(0,0)
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else:
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for i in range(len(x)-2,-1,-1):
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x[i] = (x[i]+x[i+1])%mod
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print x[0]
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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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from bootcamp import Basebootcamp
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import random
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import re
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class Cpythonindentationbootcamp(Basebootcamp):
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def __init__(self, min_n=1, max_n=10):
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if min_n < 1:
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raise ValueError("min_n must be at least 1")
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if max_n > 5000:
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raise ValueError("max_n cannot exceed 5000")
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self.min_n = max(1, min_n)
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self.max_n = min(5000, max(max_n, self.min_n))
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def case_generator(self):
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n = random.randint(self.min_n, self.max_n)
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commands = [random.choice(['f', 's']) for _ in range(n-1)]
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commands.append('s') # Ensure last command is 's'
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return {
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'n': n,
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'commands': commands
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}
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@staticmethod
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def prompt_func(question_case):
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input_lines = [str(question_case['n'])] + question_case['commands']
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input_example = '\n'.join(input_lines)
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prompt = f"""你是编程专家，需要解决一个关于Python缩进规则的谜题。请仔细阅读问题描述，并给出正确的答案。
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问题描述：
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在Python中，代码块由缩进定义，而没有显式的开始/结束符号。我们考虑一个极简化的Python子集，只有两种语句：简单语句（s）和for语句（f）。简单语句占据单独一行。for语句是复合语句，包含一个头部和一个循环体。循环体必须比for语句的头部缩进一级，且不能为空。
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给定一个由's'和'f'组成的命令序列，最后一个命令一定是's'。你需要计算所有可能的有效缩进方式的数量，结果对10^9+7取模。
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输入格式：
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第一行包含整数N（命令数）。随后N行，每行是'f'或's'。
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输出格式：
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输出一个整数，表示有效方式数取模后的结果。
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示例输入1：
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4
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s
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f
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f
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s
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示例输出1：
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1
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示例输入2：
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4
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f
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s
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f
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s
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示例输出2：
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2
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现在，请你解决以下输入案例：
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输入：
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{input_example}
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请将你的最终答案放在[answer]标签内，例如：[answer]123[/answer]。"""
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return prompt
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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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if not matches:
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return None
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last_match = matches[-1].strip()
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try:
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return int(last_match)
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except ValueError:
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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mod = 10**9 + 7
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commands = identity['commands']
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dp = [1]
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for stmt in commands:
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if stmt == 'f':
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dp.insert(0, 0)
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else:
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# Propagate the sum backwards
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for i in range(len(dp)-2, -1, -1):
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dp[i] = (dp[i] + dp[i+1]) % mod
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correct_answer = dp[0] % mod if dp else 0
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try:
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return int(solution) == correct_answer
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except (ValueError, TypeError):
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return False
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