✨ feat(solutions)：新增欧拉项目第25-27题解决方案

📝 docs(solutions)：添加Binet公式、循环节和质数生成多项式的数学原理文档

solutions/0026.ReciprocalCycles/euler_26.py

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+"""
+A unit fraction contains 1in the numerator.
+The decimal representation of the unit fractions with denominators 2 to 10 are given:
+
+1/2 = 0.5
+1/3 = 0.(3)
+1/4 = 0.25
+1/5 = 0.2
+1/6 = 0.1(6)
+1/7 = 0.(142857)
+1/8 = 0.125
+1/9 = 0.(1)
+1/10 = 0.1
+
+Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle.
+It can be seen that 1/7 has a 6-digit recurring cycle.
+
+Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
+"""
+
+import random
+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
+
+
+def repeating_cycle_fermat(d: int) -> int:
+    # 移除因子2和5
+    while d % 2 == 0:
+        d //= 2
+    while d % 5 == 0:
+        d //= 5
+
+    if d == 1:
+        return 0
+
+    # 费马小定理：对于与10互质的d，10^(φ(d)) ≡ 1 mod d
+    # 循环节长度一定是φ(d)的约数
+
+    # 计算欧拉函数φ(d)
+    phi = d
+    n = d
+    # 质因数分解求φ(d)
+    p = 2
+    while p * p <= n:
+        if n % p == 0:
+            phi -= phi // p
+            while n % p == 0:
+                n //= p
+        p += 1 if p == 2 else 2  # 从2开始，然后检查奇数
+
+    if n > 1:
+        phi -= phi // n
+
+    # 现在找φ的最小约数k，使得10^k ≡ 1 mod d
+    min_cycle = phi
+    for k in range(1, int(phi**0.5) + 1):
+        if phi % k == 0:
+            if pow(10, k, d) == 1:
+                return k
+            other = phi // k
+            if pow(10, other, d) == 1 and other < min_cycle:
+                min_cycle = other
+
+    return min_cycle
+
+
+@timer
+def main_searchall(n: int) -> None:
+    max_cycle = 0
+    max_d = 0
+    for d in range(1, n):
+        cycle = repeating_cycle_fermat(d)
+        if cycle > max_cycle:
+            max_cycle = cycle
+            max_d = d
+
+    print(f"{max_d} have {max_cycle} digits in its decimal fraction part.")
+
+
+def is_probable_prime(n: int, trials: int = 20) -> bool:
+    """Miller-Rabin素性测试（快速判断是否为质数）"""
+    if n < 2:
+        return False
+    if n in (2, 3):
+        return True
+    if n % 2 == 0:
+        return False
+
+    # 将 n-1 写成 d * 2^s 的形式
+    d = n - 1
+    s = 0
+    while d % 2 == 0:
+        d //= 2
+        s += 1
+
+    # 测试
+    for _ in range(trials):
+        a = random.randrange(2, n - 1)
+        x = pow(a, d, n)
+        if x == 1 or x == n - 1:
+            continue
+        for _ in range(s - 1):
+            x = pow(x, 2, n)
+            if x == n - 1:
+                break
+        else:
+            return False
+    return True
+
+
+@timer
+def main_by_prime(n: int) -> None:
+    for d in range(n, 0, -1):
+        max_cycle = d - 1
+        max_d = d
+        cycle = repeating_cycle_fermat(d)
+        if max_cycle == cycle:
+            print(f"{max_d} have {max_cycle} digits in its decimal fraction part.")
+            return None
+
+
+if __name__ == "__main__":
+    n = 10000
+    main_searchall(n)
+    main_by_prime(n)