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Count the number of prime numbers less than a non-negative number, n

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References: How Many Primes Are There?

Sieve of Eratosthenes

Credits: Special thanks to @mithmatt for adding this problem and creating all test cases.


使用埃拉托斯特尼筛法, 具体步骤是


A prime number is a natural number that has exactly two distinct natural number divisors: 1 and itself.

To find all the prime numbers less than or equal to a given integer n by Eratosthenes' method:

Create a list of consecutive integers from 2 through n: (2, 3, 4, ..., n). Initially, let p equal 2, the first prime number. Starting from p, enumerate its multiples by counting to n in increments of p, and mark them in the list (these will be 2p, 3p, 4p, ... ; the p itself should not be marked). Find the first number greater than p in the list that is not marked. If there was no such number, stop. Otherwise, let p now equal this new number (which is the next prime), and repeat from step 3.

Input: an integer n > 1
Let A be an array of Boolean values, indexed by integers 2 to n,
initially all set to true.
 for i = 2, 3, 4, ..., not exceeding √n:
  if A[i] is true:
    for j = i2, i2+i, i2+2i, i2+3i, ..., not exceeding n :
      A[j] := false
Output: all i such that A[i] is true.


int countPrimes(int n)
	vector<bool> flags(n, true);
	flags[0] = false;
	flags[1] = false;
	int sqr = sqrt(n - 1);
	for (int i = 2; i <= sqr; ++i) {
		if (flags[i]) {
			for (int j = i * i; j < n; j += i)
				flags[j] = false;
	return count(flags.begin(), flags.end(), true);


参见How Many Primes Are There?.


  1. pi(x)约等于`x / (logx - 1)
  2. 第n个素数的值约为nlogn
  3. x是素数的概率约为1/logx
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