50题解决中，已有简单解法，在尝试更快的版本。

solutions/0050.ConsecutivePrimeSum/euler_50.py

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+"""
+The prime 41, can be written as the sum of six consecutive primes:
+    41 = 2 + 3 + 5 + 7 + 11 + 13
+This is the longest sum of consecutive primes that adds to a prime below one-hundred.
+
+The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms,
+and is equal to 953.
+
+Which prime, below one-million, can be written as the sum of the most consecutive primes?
+"""
+
+import time
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.time()
+        result = func(*args, **kwargs)
+        end_time = time.time()
+        elapsed_time = end_time - start_time
+        print(f"{func.__name__} time: {elapsed_time:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+def primes_list(limit: int = 10**6) -> list[int]:
+    if limit < 2:
+        return []
+
+    is_prime = bytearray(b"\x01") * limit
+    is_prime[0:2] = b"\x00\x00"  # 更简洁的写法，同时置0和1
+
+    # 埃氏筛，使用切片赋值加速
+    # int(limit**0.5) 等价于 isqrt(limit)（Python 3.8+ 可用 math.isqrt）
+    for i in range(2, int(limit**0.5) + 1):
+        if is_prime[i]:
+            start = i * i
+            # 计算切片长度：从 start 到 limit，步长 i 的元素个数
+            count = (limit - start + i - 1) // i  # 等效于 ceil((limit-start)/i)
+            if count > 0:
+                is_prime[start:limit:i] = b"\x00" * count
+
+    return [i for i in range(limit) if is_prime[i]]
+
+
+def max_primes_sum(limit: int = 10**6) -> tuple:
+    primes = primes_list(limit)
+    prime_set = set(primes)
+
+    # 计算从2开始连续加素数，最多能加多少个不超过limit
+    max_possible_len = 0
+    temp_sum = 0
+    for p in primes:
+        temp_sum += p
+        if temp_sum >= limit:
+            break
+        max_possible_len += 1
+
+    # 从最长可能长度往下枚举
+    for length in range(max_possible_len, 0, -1):
+        # 滑动窗口：计算连续length个素数的和
+        window_sum = sum(primes[:length])
+        if window_sum >= limit:
+            continue
+
+        # 滑动窗口遍历所有起始位置
+        for start in range(len(primes) - length):
+            if window_sum in prime_set:
+                return window_sum, length
+            # 滑动：减去头部，加上尾部
+            window_sum += primes[start + length] - primes[start]
+            if window_sum >= limit:
+                break
+
+    return 0, 0
+
+
+@timer
+def main():
+    limit = int(input("limit:"))
+    max_sum, max_length = max_primes_sum(limit)
+    print(f"max primt: {max_sum}, max_length: {max_length}")
+
+
+if __name__ == "__main__":
+    main()