Symbol(clack:cancel)

solutions/0000_0029/0007.10001stPrime/euler_7.py

@@ -0,0 +1,43 @@
+"""
+By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
+What is the 10001st prime number?
+"""
+
+import time
+from math import log
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start = time.time()
+        result = func(*args, **kwargs)
+        end = time.time()
+        print(f"{func.__name__} took {end - start:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+@timer
+def sieve_nth_prime(n: int) -> int | None:
+    if n == 1:
+        return 2
+    # 估算上限：第n个质数约在 n*log(n) 附近
+    limit = int(n * (log(n) + log(log(n)))) + 10 if n > 6 else 20
+    sieve = [True] * limit
+    sieve[0:2] = [False, False]  # 0和1不是质数
+
+    count = 0
+    for p in range(2, limit):
+        if sieve[p]:
+            count += 1
+            if count == n:
+                return p
+            # 标记倍数
+            sieve[p * p : limit : p] = [False] * ((limit - 1 - p * p) // p + 1)
+
+    return None
+
+
+if __name__ == "__main__":
+    print(sieve_nth_prime(10001))