✨ feat(main.py)：重构主函数调用方式，将 app() 调用封装到 main() 函数中

✨ 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

solutions/0037.TruncatablePrimes/euler_37.py

@@ -0,0 +1,43 @@
+"""
+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