Efficient search for prime--sieve of Eratosthenes

Sieve of Eratosthenes-Eratosthenes sieve, referred to as the sieve.

Idea: give the range of values to sieve N, find the prime number within . First use 2 to sift, that is, 2 left, the number of 2 is removed, and then the next prime, that is, 3 sieve, 3 left, the multiples of 3 is removed, then the next prime number of 5 sieve, 5 left, the multiples of 5 are removed; Keep repeating ... until it is less than equals.

As follows:

Example: A problem with searching for prime numbers on Leetcode (https://leetcode.com/problems/count-primes/).

Question:Count The number of prime numbers less than a non-negative number, n

PS: When the N=1500000,leetcode evaluation is 58ms, the operating efficiency is good in C + +.

int countprimes (int n) {    bool isprimes[n];    int primesnum=0;    for (int i=2; i<n; i++)        isprimes[i]=true;    Remove the times of primes.    for (int i=2; i<=sqrt (n), i++)        if (Isprimes[i]) for            (int j=2; j*i<=n; j + +)                Isprimes[j*i]=false;    The count of primes.    for (int i=2; i<n; i++)        if (Isprimes[i])            primesnum++;    return primesnum;}

Efficient search for prime--sieve of Eratosthenes