1- 问题描述
Count the number of prime numbers less than a non-negative number, n
2- 算法思想
给出要筛数值的范围 $n$,找出 $sqrt{n}$ 以内的素数 $p_{1}, p_{2}, cdots, p_{k}$。先用2去筛,即把2留下,把2的倍数剔除掉;再用下一个素数,也就是3筛,把3留下,把3的倍数剔除掉;接下去用下一个素数5筛,把5留下,把5的倍数剔除掉;不断重复下去......。
3- Python实现
1 from math import sqrt 2 3 def countPrimes(n): 4 if n in (0, 1, 2): 5 return 0 6 boards = [True] * n 7 boards[0] = boards[1] = False 8 for i in range(2, int(sqrt(n)) + 1): 9 for j in range(i * i, n, i): 10 boards[j] = False 11 primes = [i for i in boards if i] 12 return len(primes)
4- 实例