Sidney Zhang
09ddf3f65c
✨ feat(euler_37.py)：添加截断质数问题的初始实现 ✨ feat(euler_37_primeclass.py)：添加基于质数生成器的截断质数完整解决方案 📝 docs(millerrabin_test.pdf)：添加 Miller-Rabin 测试的详细文档
44 lines
1.3 KiB
Python
44 lines
1.3 KiB
Python
"""
|
|
The number 3797 has an interesting property.
|
|
Being prime itself, it is possible to continuously remove digits from left to right,
|
|
and remain prime at each stage: 3797, 797, 97, and 7.
|
|
Similarly we can work from right to left: 3797, 379, 37, and 3.
|
|
|
|
Find the sum of the only eleven primes that are both truncatable from left to right and right to left.
|
|
|
|
NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
|
|
"""
|
|
|
|
from itertools import product
|
|
|
|
|
|
def combine_lists(a: list[int], b: list[int|None], c: list[int]) -> list[int]:
|
|
"""将三个列表的每个元素组合成数字"""
|
|
return [int(f"{x}{y}{z}") for x, y, z in product(a, b, c)]
|
|
|
|
|
|
def is_prime(n: int) -> bool:
|
|
"""判断一个数是否为素数"""
|
|
if n < 2:
|
|
return False
|
|
for i in range(2, int(n ** 0.5) + 1):
|
|
if n % i == 0:
|
|
return False
|
|
return True
|
|
|
|
|
|
def TruncatablePrime() -> list[int]:
|
|
begin = [2, 3, 5, 7]
|
|
end = [3, 7]
|
|
middle = [1, 3, 7, 9]
|
|
res = []
|
|
for length in range(2, 7):
|
|
if length - 2 == 0:
|
|
midb = []
|
|
else:
|
|
midb = product(middle, repeat=length - 2)
|
|
midb = [int("".join(map(str, x))) for x in midb]
|
|
nums = combine_lists(begin, midb, end)
|
|
for num in nums :
|
|
if is
|