created and first commit(3 solusions)

solutions/0000.zero/euler_0.py

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+"""
+Project Euler Problem 0 :
+Find the sum of all the odd squares which do not exceed 764000.
+
+这是注册问题。
+"""
+
+UP_NUM = 764000
+
+
+def is_even(num: int) -> bool:
+    """Check if a number is even."""
+    return num & 1 == 0
+
+
+def main():
+    res: list[int] = []
+    for i in range(UP_NUM):
+        t = (i + 1) ** 2
+        if not is_even(t):
+            res.append(t)
+    print(sum(res))
+
+
+if __name__ == "__main__":
+    main()

solutions/0001.multiples/euler_1.py

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+"""
+Project Euler Problem 1 :
+Find the sum of all the multiples of 3 or 5 below 1000.
+"""
+
+UP_NUM = 1000
+
+
+def condition(num: int) -> bool:
+    if num % 3 == 0:
+        return True
+    elif num % 5 == 0:
+        return True
+    return False
+
+
+def main():
+    res = []
+    for i in range(1, UP_NUM):
+        if condition(i):
+            res.append(i)
+    print(sum(res))
+
+
+if __name__ == "__main__":
+    main()

solutions/0002.even_fibonacci/euler_2.py

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+"""
+Even Fibonacci Numbers
+
+Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
+1, 2, 3, 5, 8, 13, 21, 34, 55, 89
+By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
+"""
+
+UP_NUM = 4000000
+
+
+def is_even(num: int) -> bool:
+    """Check if a number is even."""
+    return num & 1 == 0
+
+
+"""
+下面这个方法实在太慢，计算400万项还会内存溢出。
+```python
+def fibonacci_sequence(n: int) -> list[int]:
+    a, b = 1, 2
+    res = [a, b]
+    while b <= n:
+        a, b = b, a + b
+        res.append(b)
+    return res
+
+
+def main():
+    fib_seq = fibonacci_sequence(UP_NUM)
+    even_sum = sum(num for num in fib_seq if is_even(num))
+    print(even_sum)
+```
+"""
+
+
+def even_fibonacci(limit: int) -> int:
+    a, b = 1, 2
+    even_sum = 0
+    while b <= limit:
+        if is_even(b):
+            even_sum += b
+        a, b = b, a + b
+    return even_sum
+
+
+if __name__ == "__main__":
+    print(even_fibonacci(UP_NUM))

solutions/0003.largestprime/euler_3.py

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+import random
+from math import gcd
+from typing import List, Set
+
+
+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 pollards_rho(n: int, max_iter: int = 100000) -> int:
+    """
+    Pollard's Rho 算法：返回n的一个非平凡因子
+
+    Args:
+        n: 待分解的合数
+        max_iter: 最大迭代次数防止无限循环
+
+    Returns:
+        n的一个因子（可能是质数也可能是合数）
+        若失败返回None
+    """
+    if n % 2 == 0:
+        return 2
+
+    # 随机生成多项式 f(x) = x^2 + c (mod n)
+    c = random.randrange(1, n - 1)
+
+    def f(x):
+        return (pow(x, 2, n) + c) % n
+
+    # Floyd 判圈算法
+    x = random.randrange(2, n - 1)
+    y = x
+    d = 1
+
+    iter_count = 0
+    while d == 1 and iter_count < max_iter:
+        x = f(x)  # 乌龟：走一步
+        y = f(f(y))  # 兔子：走两步
+        d = gcd(abs(x - y), n)
+        iter_count += 1
+
+    if d == n:
+        # 失败，尝试其他参数（递归或返回None）
+        return pollards_rho(n, max_iter) if max_iter > 1000 else None
+
+    return d
+
+
+def factorize(n: int | None) -> List[int | None]:
+    """
+    完整因数分解：递归分解所有质因数
+
+    Args:
+        n: 待分解的正整数
+
+    Returns:
+        质因数列表（可能含重复）
+    """
+    if n == 1:
+        return []
+
+    # 如果是质数，直接返回
+    if is_probable_prime(n):
+        return [n]
+
+    # 获取一个因子
+    factor = pollards_rho(n)
+
+    # 递归分解
+    return factorize(factor) + factorize(n // factor)
+
+
+def get_prime_factors(n: int) -> Set[int | None]:
+    """获取所有不重复的质因数"""
+    return set(factorize(n))