diff --git a/solutions/0034/euler_34.py b/solutions/0034/euler_34.py
new file mode 100644
index 0000000..1481dd7
--- /dev/null
+++ b/solutions/0034/euler_34.py
@@ -0,0 +1,67 @@
+"""
+145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
+
+Find the sum of all numbers which are equal to the sum of the factorial of their digits.
+
+Note: As 1!=1 and 2!=2 are not sums they are not included.
+"""
+
+import time
+from functools import lru_cache
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.time()
+        result = func(*args, **kwargs)
+        end_time = time.time()
+        print(f"{func.__name__} time: {end_time - start_time:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+@lru_cache(maxsize=None)
+def factorial(n: int) -> int:
+    if n == 0:
+        return 1
+    elif n == 1:
+        return 1
+    elif n == 2:
+        return 2
+    elif n == 3:
+        return 6
+    elif n == 4:
+        return 24
+    elif n == 5:
+        return 120
+    elif n == 6:
+        return 720
+    elif n == 7:
+        return 5040
+    elif n == 8:
+        return 40320
+    elif n == 9:
+        return 362880
+    elif n == 10:
+        return 3628800
+    else:
+        return n * factorial(n - 1)
+
+
+def sum_digits(n: int) -> int:
+    tmp = str(n)
+    return sum(factorial(int(digit)) for digit in tmp)
+
+
+@timer
+def main() -> None:
+    res = list()
+    for i in range(12, 9999999):
+        if sum_digits(i) == i:
+            res.append(i)
+    print(sum(res))
+
+
+if __name__ == "__main__":
+    main()
diff --git a/solutions/0034/readme.md b/solutions/0034/readme.md
new file mode 100644
index 0000000..a3a88fa
--- /dev/null
+++ b/solutions/0034/readme.md
@@ -0,0 +1,8 @@
+参考 [Factorion](https://mathworld.wolfram.com/Factorion.html)
+
+---
+
+本来想的是，是否可以确定满足这个规则的数列上限。口算估计是999999，然后就是暴力计算了。
+看到结果，则比较让我惊讶，这样的数只有两个：145和40585。
+
+再加上看别人的思考，从我个人来说，可能不是所有问题都适合这个思路？
diff --git a/solutions/0035.CircularPrimes/euler_35.py b/solutions/0035.CircularPrimes/euler_35.py
new file mode 100644
index 0000000..96fd343
--- /dev/null
+++ b/solutions/0035.CircularPrimes/euler_35.py
@@ -0,0 +1,206 @@
+"""
+The number, 179, is called a circular prime because all rotations of the digits: 179, 791, and 917,
+are themselves prime.
+
+There are thirteen such primes below 100 : 2,3,5,7,11,13,17,31,37,71,73,79, and 97.
+
+How many circular primes are there below one million?
+"""
+
+import random
+import time
+from itertools import permutations
+from sqlite3.dbapi2 import Time
+from typing import Union
+
+# 预计算的小素数集合，用于快速排除
+_SMALL_PRIMES = frozenset((2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37))
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.time()
+        result = func(*args, **kwargs)
+        end_time = time.time()
+        print(f"{func.__name__} time: {end_time - start_time:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+def sieve_best(n: int) -> list[int]:
+    """返回 [2..n) 内所有素数"""
+    if n <= 2:
+        return []
+    size = (n - 1) // 2
+    comp = bytearray(size)
+    limit = int(n**0.5) // 2
+    for i in range(1, limit + 1):
+        if not comp[i]:
+            p = 2 * i + 1
+            s = (p * p - 1) // 2
+            comp[s::p] = b"\1" * ((size - s - 1) // p + 1)
+    res = [2, *(2 * i + 1 for i, v in enumerate(comp) if not v)]
+    res.sort()
+    return [i for i in res if i != 1]
+
+
+def is_circular(n: int, primes: list[int]) -> bool:
+    if n < 2:
+        return False
+    if n in [2, 3, 5, 7, 11]:
+        return True
+    digits = str(n)
+    allnum = [digits[i:] + digits[:i] for i in range(len(digits))]
+    for num in allnum:
+        if int(num) not in primes:
+            return False
+    return True
+
+
+def is_probable_prime(n: Union[int, str], k: int = 20) -> bool:
+    """
+    Miller-Rabin素性检测算法
+
+    参数:
+        n: 待检测的整数（支持int或数字字符串）
+        k: 测试次数，误差率约为4^(-k)，默认20次（误差率约9e-13）
+
+    返回:
+        bool: 如果n很可能是素数返回True，否则返回False
+    """
+    # 支持字符串输入
+    if isinstance(n, str):
+        n = int(n)
+
+    # 基本检查
+    if n < 2:
+        return False
+    if n in _SMALL_PRIMES:
+        return True
+
+    # 检查小素数的倍数
+    for p in _SMALL_PRIMES:
+        if n % p == 0:
+            return False
+
+    # 特殊情况：偶数（已排除2）
+    if n % 2 == 0:
+        return False
+
+    # 将 n-1 写成 2^s * d 的形式
+    d = n - 1
+    s = 0
+    while d % 2 == 0:
+        d //= 2
+        s += 1
+
+    # 边界：如果 n-1 可以直接作为基数，避免随机数生成问题
+    if n < 4:
+        return n in (2, 3)
+
+    # Miller-Rabin测试
+    for _ in range(k):
+        # 生成随机基数，范围 [2, n-2]
+        a = random.randrange(2, n - 1)
+
+        # 计算 x = a^d mod n
+        x = pow(a, d, n)
+
+        # 如果 x == 1 或 x == n-1，通过本轮测试
+        if x == 1 or x == n - 1:
+            continue
+
+        # 重复平方检查
+        for _ in range(s - 1):
+            x = pow(x, 2, n)
+            if x == n - 1:
+                break
+        else:
+            # 所有平方后都没有得到 n-1，n是合数
+            return False
+
+    return True
+
+
+# 优化的确定素数版本（适用于小整数）
+def is_prime_small(n: int) -> bool:
+    """确定性素数检测，适用于n < 2^64"""
+    if n < 2:
+        return False
+    if n % 2 == 0:
+        return n == 2
+    if n % 3 == 0:
+        return n == 3
+
+    # 检查小素数
+    for p in _SMALL_PRIMES:
+        if n % p == 0:
+            return n == p
+
+    # 6k±1 优化检查
+    i = 5
+    w = 2
+    while i * i <= n:
+        if n % i == 0:
+            return False
+        i += w
+        w = 6 - w  # 在2和4之间切换：5,7,11,13,17,19...
+
+    return True
+
+
+# 智能选择函数
+def is_prime(n: Union[int, str], k: int = 20) -> bool:
+    """
+    智能素数检测：对小整数使用确定性算法，对大整数使用Miller-Rabin
+
+    参数:
+        n: 待检测的整数
+        k: Miller-Rabin测试次数（仅对大整数有效）
+
+    返回:
+        bool: 如果是素数返回True
+    """
+    if isinstance(n, str):
+        n = int(n)
+
+    # 小整数使用确定性算法
+    if n < 2**64:
+        return is_prime_small(n)
+
+    # 大整数使用Miller-Rabin
+    return is_probable_prime(n, k)
+
+
+@timer
+def main_quick(max: int) -> None:
+    res = set()
+    for i in range(2, max):
+        if is_prime(i):
+            if i in res:
+                continue
+            else:
+                s = str(i)
+                allnums = [s[i:] + s[:i] for i in range(len(s))]
+                flag = 0
+                for j in allnums:
+                    num = int("".join(j))
+                    if is_prime(num):
+                        flag += 1
+                if flag == len(allnums):
+                    res.update(allnums)
+    print(len(res))
+
+
+@timer
+def main(n: int) -> None:
+    primes = sieve_best(n)
+    circular_primes = [p for p in primes if is_circular(p, primes)]
+    print(len(circular_primes))
+
+
+if __name__ == "__main__":
+    num = 1000000
+    main(num)
+    main_quick(num)
diff --git a/solutions/0035.CircularPrimes/euler_35_better.py b/solutions/0035.CircularPrimes/euler_35_better.py
new file mode 100644
index 0000000..6744499
--- /dev/null
+++ b/solutions/0035.CircularPrimes/euler_35_better.py
@@ -0,0 +1,80 @@
+import time
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.time()
+        result = func(*args, **kwargs)
+        end_time = time.time()
+        print(f"{func.__name__} took {end_time - start_time:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+# ---------- 1. PRIMES ----------
+MAX = 1_000_000
+is_p = bytearray(b"\1") * MAX  # 1 表示素数，0 表示合数（与原来相反，方便理解）
+is_p[0] = is_p[1] = 0  # 0和1不是素数
+
+# 优化：使用标准埃拉托斯特尼筛法
+for p in range(2, int(MAX**0.5) + 1):
+    if is_p[p]:
+        is_p[p * p : MAX : p] = b"\0" * ((MAX - p * p - 1) // p + 1)
+
+
+# ---------- 2. SCROLLING ----------
+def rotations(n: int) -> list[int]:
+    """返回 n 的所有旋转（数学移位版，无字符串开销）"""
+    if n < 10:
+        return [n]
+    s = str(n)
+    k = len(s)
+    res = []
+    for i in range(k):
+        # 使用切片旋转，避免重复计算
+        rotated = int(s[i:] + s[:i])
+        res.append(rotated)
+    return res
+
+
+# ---------- 3. MAIN ----------
+@timer
+def main(limit: int = MAX) -> int:
+    # 预过滤：只要含 0,2,4,5,6,8 就可以跳过（除了2和5本身）
+    bad_digits = set("024568")
+    total = 0
+    counted = set()  # 用于去重，避免重复计数
+
+    for p in range(2, limit):
+        if is_p[p]:  # p是素数
+            # 快速数字过滤（2和5是特例）
+            if p != 2 and p != 5:
+                s = str(p)
+                if not bad_digits.isdisjoint(s):
+                    continue
+
+            # 如果已经统计过，跳过
+            if p in counted:
+                continue
+
+            # 检查所有旋转
+            rs = rotations(p)
+
+            # 检查所有旋转数是否都是素数
+            valid = True
+            for r in rs:
+                if r >= limit or not is_p[r]:
+                    valid = False
+                    break
+
+            if valid:
+                # 如果是循环素数，统计并标记已计数
+                total += len(set(rs))  # 使用set避免重复旋转
+                counted.update(rs)
+
+    return total
+
+
+if __name__ == "__main__":
+    print(f"循环素数数量: {main()}")
diff --git a/solutions/0035.CircularPrimes/readme.md b/solutions/0035.CircularPrimes/readme.md
new file mode 100644
index 0000000..73160f8
--- /dev/null
+++ b/solutions/0035.CircularPrimes/readme.md
@@ -0,0 +1,6 @@
+# 判断素数
+
+这个问题也很有和判断素数，以及如何更好的记录判断结果。
+当然，如何更快计算出滚动的数字也很重要。
+
+个人感觉这是个算法问题更多的问题，数学占比并不多。