✨ 新增欧拉第41、42题Python解法

solutions/0041.PandigitalPrime/euler_41.py

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+"""
+We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once.
+For example, 2143 is a 4-digit pandigital and is also prime.
+
+What is the largest n-digit pandigital prime that exists?
+"""
+
+import random
+import time
+from functools import lru_cache
+from itertools import permutations
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.perf_counter()
+        result = func(*args, **kwargs)
+        end_time = time.perf_counter()
+        print(f"Execution time: {end_time - start_time} seconds")
+        return result
+
+    return wrapper
+
+
+def miller_rabin_test(n: int, k: int = 10) -> bool:
+    if n < 2:
+        return False
+    if n == 2 or n == 3:
+        return True
+    if n % 2 == 0 or n % 3 == 0:
+        return False
+    r, s = 0, n - 1
+    while s % 2 == 0:
+        r += 1
+        s //= 2
+    for _ in range(k):
+        a = random.randrange(2, n - 1)
+        x = pow(a, s, n)
+        if x == 1 or x == n - 1:
+            continue
+        for _ in range(r - 1):
+            x = pow(x, 2, n)
+            if x == n - 1:
+                break
+        else:
+            return False
+    return True
+
+
+@lru_cache(maxsize=128)
+def is_even_or_five_ending(num: int) -> bool:
+    """快速检查是否为偶数或以5结尾"""
+    return num % 2 == 0 or num % 5 == 0
+
+
+@timer
+def main() -> None:
+    str_nums = "987654321"
+    stop = False
+    for i in range(9, 0, -1):
+        if sum(int(digit) for digit in str_nums[-i:]) % 3 == 0:
+            continue
+        for perm in permutations(str_nums[-i:], i):
+            num = int("".join(perm))
+            if is_even_or_five_ending(num):
+                continue
+            if miller_rabin_test(num):
+                print(num)
+                stop = True
+                break
+        if stop:
+            break
+
+
+if __name__ == "__main__":
+    main()

solutions/0042.CodedTriangleNum/euler_42.py

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+"""
+The nth term of the sequence of triangle numbers is given by, $t_n = \frac{n(n+1)}{2}$  ;
+so the first ten triangle numbers are:
+    1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
+By converting each letter in a word to a number corresponding to its alphabetical position
+and adding these values we form a word value. For example, the word value for SKY is 19+11+25 = 55 = t_10 .
+If the word value is a triangle number then we shall call the word a triangle word.
+
+Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand
+common English words, how many are triangle words?
+"""
+
+import time
+from pathlib import Path
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.perf_counter()
+        result = func(*args, **kwargs)
+        end_time = time.perf_counter()
+        print(f"Time taken by {func.__name__}: {end_time - start_time:.8f} seconds")
+        return result
+
+    return wrapper
+
+
+def word_to_num(txt: str) -> int:
+    """
+    将单词转换为数字，每个字母对应其在字母表中的位置（A=1, B=2, ..., Z=26）
+
+    参数:
+        txt: 要转换的单词
+
+    返回:
+        单词的数字表示
+
+    示例:
+        >>> word_to_num("SKY")
+        55
+    """
+    return sum(ord(c) - ord("A") + 1 for c in txt)
+
+
+def is_triangle_number(n: int) -> bool:
+    """
+    判断一个数是否为三角形数
+
+    参数:
+        n: 要判断的数
+
+    返回:
+        True 如果 n 是三角形数，否则 False
+
+    示例:
+        >>> is_triangle_number(10)
+        True
+        >>> is_triangle_number(11)
+        False
+    """
+    # 使用二次方程求解，如果结果是整数则为三角形数
+    # 三角形数公式：n = k(k+1)/2
+    # 解得 k = (-1 + sqrt(1 + 8n)) / 2
+    k = (-1 + (1 + 8 * n) ** 0.5) / 2
+    return k.is_integer()
+
+
+@timer
+def main() -> None:
+    with open(Path(__file__).parent / "0042_words.txt", "r") as file:
+        words = file.read().replace('"', "").split(",")
+
+    triangle_words = [word for word in words if is_triangle_number(word_to_num(word))]
+    print(f"Number of triangle words: {len(triangle_words)}")
+
+
+if __name__ == "__main__":
+    main()