50题解决了，使用了动态上限，优化了搜索。

solutions/0050.ConsecutivePrimeSum/euler_50_better.py

@@ -10,7 +10,7 @@ Which prime, below one-million, can be written as the sum of the most consecutiv
 """
 
 import time
-from math import isqrt, log
+from math import isqrt, log, sqrt
 
 from bitarray import bitarray
 
@@ -88,26 +88,31 @@ def is_prime(num: int) -> bool:
     return True
 
 
+def get_bound_dynanic(N, iterations=3):
+    # 初始猜测（基于简化公式）
+    k = sqrt(2 * N / log(N)) if N > 100 else 10
+
+    # 牛顿法迭代（3次足够）
+    for _ in range(iterations):
+        ln_k = log(k)
+        # f(k) = k^2 * ln_k - 2N, f'(k) = k * (2*ln_k + 1)
+        k = (k**2 * (1 + ln_k) + 2 * N) / (k * (2 * ln_k + 1))
+
+    return int(k)  # 返回安全上界（如562）
+
+
 def get_bound(limit: int = 10**6) -> int:
     """
-    使用牛顿法估算筛法所需的素数上限。
+    估算素数个数上限。
     """
-
+    return int(isqrt(int(2 * limit / log(limit))) * 1.55)
-    def f(t: float) -> float:
-        return t**2 * (2 * log(t) - 1) - 2 * limit
-
-    def fp(t: float) -> float:
-        return 4 * t * log(t)
-
-    x, y = limit, limit + 2
-    while y - x > 1:
-        y, x = x, x - f(x) / fp(x)
-    return int(x * (log(x) + log(log(x) - 1)))
 
 
+@timer
 def max_primes_sum_best(limit: int = 10**6) -> tuple[int, int] | None:
     # === 1. 确定搜索范围 ===
-    bound = get_bound(limit)
+    bound = get_bound_dynanic(limit)
+    print(f"bound: {bound}")
     primes = n_primes(bound)
 
     # === 2. 构建前缀和数组 ===
@@ -157,7 +162,6 @@ def max_primes_sum_best(limit: int = 10**6) -> tuple[int, int] | None:
     return best_prime, best_length
 
 
-@timer
 def main():
     limit = int(input("limit:"))
     max_sum = max_primes_sum_best(limit)