solutions of problem 21 and problem 20

solutions/0021.AmicableNumbers/euler_21.py

@@ -0,0 +1,96 @@
+"""
+Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
+If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.
+
+For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, and 110; therefore d(220) = 284.
+The proper divisors of 284 are 1, 2, 4, 71, and 142; so d(284) = 220.
+
+Evaluate the sum of all the amicable numbers under 10000.
+"""
+
+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:.8f} seconds")
+        return result
+
+    return wrapper
+
+
+def sieve_primes(limit):
+    """埃拉托斯特尼筛法生成质数列表"""
+    is_prime = [True] * (limit + 1)
+    is_prime[0:2] = [False, False]
+    primes = []
+    for i in range(2, limit + 1):
+        if is_prime[i]:
+            primes.append(i)
+            for j in range(i * i, limit + 1, i):
+                is_prime[j] = False
+    return primes
+
+
+def prime_factors_with_sieve(n, primes):
+    """使用预计算的质数列表进行分解"""
+    factors = {}
+    temp = n
+    for p in primes:
+        if p * p > temp:
+            break
+        while temp % p == 0:
+            factors[p] = factors.get(p, 0) + 1
+            temp //= p
+    if temp > 1:
+        factors[temp] = factors.get(temp, 0) + 1
+    return factors
+
+
+def get_divisors_optimized(n):
+    """优化版本：预计算质数加速分解"""
+    if n < 2:
+        return [1] if n == 1 else []
+
+    # 计算需要的质数范围
+    limit = int(n**0.5) + 1
+    primes = sieve_primes(limit)
+
+    # 质因数分解
+    factors = prime_factors_with_sieve(n, primes)
+
+    # 生成除数
+    divisors = [1]
+    for p, exp in factors.items():
+        powers = [p**e for e in range(exp + 1)]
+        divisors = [a * b for a in divisors for b in powers]
+
+    return sorted(divisors)
+
+
+def sum_of_divisors(n):
+    """计算一个数的所有真因子之和"""
+    return sum(get_divisors_optimized(n)) - n
+
+
+def find_amicable_numbers(limit):
+    """查找小于给定上限的全部亲和数对"""
+    amicable_pairs = set()
+    for a in range(2, limit):
+        b = sum_of_divisors(a)
+        if a != b and sum_of_divisors(b) == a:
+            amicable_pairs.add((min(a, b), max(a, b)))
+    return amicable_pairs
+
+
+@timer
+def sum_of_amicable_numbers(limit):
+    """计算小于给定上限的全部亲和数之和"""
+    return sum(a + b for a, b in find_amicable_numbers(limit))
+
+
+if __name__ == "__main__":
+    print(sum_of_amicable_numbers(10000))