你必须解释如何将 SEND -> 9567 和 COUPLE -> 653924

中提到的方法， itertools.permutations 因为 “禁止模块导入” :

import re
from typing import TypeVar
from collections.abc import Iterable

T = TypeVar('T')

# From https://docs.python.org/3/library/itertools.html#itertools.permutations.
def permutations(iterable: Iterable[T], r: int | None = None):
  # permutations('ABCD', 2) → AB AC AD BA BC BD CA CB CD DA DB DC
  # permutations(range(3)) → 012 021 102 120 201 210
  pool = tuple(iterable)
  n = len(pool)
  r = n if r is None else r
  if r > n:
    return
  indices = list(range(n))
  cycles = list(range(n, n-r, -1))
  yield tuple(pool[i] for i in indices[:r])
  while n:
    for i in reversed(range(r)):
      cycles[i] -= 1
      if cycles[i] == 0:
        indices[i:] = indices[i+1:] + indices[i:i+1]
        cycles[i] = n - i
      else:
        j = cycles[i]
        indices[i], indices[-j] = indices[-j], indices[i]
        yield tuple(pool[i] for i in indices[:r])
        break
    else:
        return

def extract_s1_s2_s3(puzzle: str) -> tuple[str, str, str] | None:
  """Extracts S1, S2, and S3 from the alphametic puzzle string.

  Args:
    puzzle: A string representing the alphametic equation
            (e.g., 'SEND + MORE = MONEY').
  Returns:
    A tuple (S1, S2, S3) if the puzzle format is valid, or None if invalid.
  """
  match = re.match(r'(\w+)\s*\+\s*(\w+)\s*=\s*(\w+)', puzzle)
  if not match:
    return None
  return match.groups()

def solve_alphametic(S1: str, S2: str, S3: str) -> tuple[int, int, int] | str:
  """Solves the alphametic problem S1 + S2 = S3.

  Args:
    S1: The first string operand in the equation.
    S2: The second string operand in the equation.
    S3: The result string in the equation.
  Returns:
    A tuple (N1, N2, N3) representing the solution if it exists,
    or 'UNSOLVABLE' otherwise.
  """
  letters = set(S1 + S2 + S3)
  digits = '0123456789'
  for perm in permutations(digits, len(letters)):
    mapping = {letter: digit for letter, digit in zip(letters, perm)}
    if mapping[S1[0]] == '0' or mapping[S2[0]] == '0' or mapping[S3[0]] == '0':
      continue  # Leading zeros are not allowed
    N1 = int(''.join(mapping[char] for char in S1))
    N2 = int(''.join(mapping[char] for char in S2))
    N3 = int(''.join(mapping[char] for char in S3))
    if N1 + N2 == N3:
      return N1, N2, N3
  return 'UNSOLVABLE'

def main() -> None:
  """Main function to execute the alphametic solver."""
  puzzle = input('Enter the alphametic puzzle (e.g., "SEND + MORE = MONEY"): ')
  parts = extract_s1_s2_s3(puzzle)
  if parts is None:
    print('Invalid puzzle format')
    return
  S1, S2, S3 = parts
  solution = solve_alphametic(S1, S2, S3)
  if solution == 'UNSOLVABLE':
    print(solution)
  else:
    a, b, c = solution
    print(f'{a} + {b} = {c}')

if __name__ == '__main__':
  main()

使用示例1：

Enter the alphametic puzzle (e.g., "SEND + MORE = MONEY"): SEND + MORE = MONEY
9567 + 1085 = 10652

示例用法2：

Enter the alphametic puzzle (e.g., "SEND + MORE = MONEY"): COUPLE + COUPLE = QUARTET
653924 + 653924 = 1307848

这可以使用 .join 函数来完成，该函数将循环遍历字符串中的每个字符，然后引用字母到数字字典映射并相应地转换输入字符串。

您可能需要更改 create_alphabet_to_number_map() 或 alphabet_to_number 自定义字母数字映射，因为我提供的示例只是 a=1、b=2，等等……

# Create a dictionary of your conversions 
def create_alphabet_to_number_map():
    # Create a map from alphabet letters A-J to corresponding numbers 0-9
    alphabet_to_number = {chr(65 + i): str(i) for i in range(26)}
    return alphabet_to_number

# Apply conversion function to every character 
def convert_string_using_map(input_string):
    # Create the alphabet-to-number map
    alphabet_to_number = create_alphabet_to_number_map()
    
    # Convert the input string using the map
    converted_string = ''.join([alphabet_to_number.get(char, char) for char in input_string])
    
    return converted_string


# Example usage
input_string = "Couple + Couple = Quartet"
converted_string = convert_string_using_map(input_string.upper())
print(f"Original String: {input_string}")
print(f"Converted String: {converted_string}")