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