Sidney Zhang
6fb29a863c
📝 docs(README.md)：更新项目说明并添加main.py使用指南 ⬆️ chore(pyproject.toml)：添加typer依赖以支持CLI功能 ✅ test(solutions)：添加0036号问题的解决方案文件
75 lines
1.9 KiB
Python
75 lines
1.9 KiB
Python
"""
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The decimal number, 585 = 1001001001_2 (binary), is palindromic in both bases.
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Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.
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(Please note that the palindromic number, in either base, may not include leading zeros.)
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"""
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import time
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def timer(func):
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def wrapper(*args, **kwargs):
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start_time = time.time()
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result = func(*args, **kwargs)
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end_time = time.time()
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print(f"{func.__name__} took time: {end_time - start_time:.6f} seconds")
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return result
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return wrapper
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def is_palindromic(num: int) -> bool:
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return str(num) == str(num)[::-1]
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@timer
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def main_simple(limit: int = 1000000) -> None:
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res = []
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for num in range(0, limit + 1):
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if is_palindromic(num):
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if is_palindromic(int(bin(num)[2:])):
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res.append(num)
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print(sum(res))
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def palindromes_of_length(length: int) -> list[int]:
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"""生成所有指定位数的回文数"""
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if length == 1:
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return list(range(10))
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result = []
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# 前半部分的起始和结束
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start = 10 ** ((length - 1) // 2)
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end = 10 ** ((length + 1) // 2)
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for half in range(start, end):
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s = str(half)
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if length % 2 == 0:
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# 偶数位：123 → 123321
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palindrome = s + s[::-1]
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else:
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# 奇数位：123 → 12321
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palindrome = s + s[-2::-1]
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result.append(int(palindrome))
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return result
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@timer
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def main_quick(limit: int = 1000000) -> None:
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res = []
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length = len(str(limit - 1))
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for num in range(1, length + 1):
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for palindrome in palindromes_of_length(num):
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if is_palindromic(int(bin(palindrome)[2:])):
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res.append(palindrome)
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print(sum(res))
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if __name__ == "__main__":
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limit = 1000000
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main_simple(limit)
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main_quick(limit)
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