Symbol(clack:cancel)

solutions/0000_0029/0005.smallestMultiple/eular_5.py

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+"""
+2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
+What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
+"""
+
+import math
+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 is_probable_prime(n: int, trials: int = 10) -> 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
+
+
+def primes_up_to(n: int) -> list[int]:
+    if n < 2:
+        return []
+    if n == 2:
+        return [2]
+    if n == 3:
+        return [2, 3]
+
+    primes = [2, 3]
+    for i in range(5, n + 1, 2):
+        if is_probable_prime(i):
+            primes.append(i)
+
+    return primes
+
+
+@timer
+def smallest_multiple(n: int) -> int:
+    result = 1
+    for p in primes_up_to(n):
+        # 找出最大幂次
+        exp = int(math.log(n, p))
+        result *= p**exp
+    return result
+
+
+@timer
+def python_solution(n: int) -> int:
+    ls = list(range(1, n + 1))
+    return math.lcm(*ls)
+
+
+if __name__ == "__main__":
+    print(smallest_multiple(20))
+    print(python_solution(20))

solutions/0000_0029/0005.smallestMultiple/readme.md

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+想法来自： [A003418](https://oeis.org/A003418) 。
+
+对于 [a .. b] 这个连续整数区间，最小的可整除的整数，就是：
+
+**小于b的所有质数分别小于b的最大幂值的乘积** 。
+
+计算步骤就是：
+
+1. 找出不超过 $b$ 的所有质数 ${p}$
+2. 对每个质数 $p$ ，找出最大指数 $e$ 使得 $p^e \leq b$
+3. 将所有 $p^e$ 相乘