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// Source : https://leetcode.com/problems/count-primes/

// Author : Hao Chen

// Date : 2015-06-09

/**********************************************************************************

*

* Description:

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

*

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

*

* Let's start with a isPrime function. To determine if a number is prime, we need to check if

* it is not divisible by any number less than n. The runtime complexity of isPrime function

* would be O(n) and hence counting the total prime numbers up to n would be O(n2). Could we do better?

*

* As we know the number must not be divisible by any number > n / 2, we can immediately cut the total

* iterations half by dividing only up to n / 2. Could we still do better?

*

* Let's write down all of 12's factors:

*

* 2 × 6 = 12

* 3 × 4 = 12

* 4 × 3 = 12

* 6 × 2 = 12

*

* As you can see, calculations of 4 × 3 and 6 × 2 are not necessary. Therefore, we only need to consider

* factors up to √n because, if n is divisible by some number p, then n = p × q and since p ≤ q, we could derive that p ≤ √n.

*

* Our total runtime has now improved to O(n1.5), which is slightly better. Is there a faster approach?

*

* public int countPrimes(int n) {

* int count = 0;

* for (int i = 1; i < n; i++) {

* if (isPrime(i)) count++;

* }

* return count;

* }

*

* private boolean isPrime(int num) {

* if (num <= 1) return false;

* // Loop's ending condition is i * i <= num instead of i <= sqrt(num)

* // to avoid repeatedly calling an expensive function sqrt().

* for (int i = 2; i * i <= num; i++) {

* if (num % i == 0) return false;

* }

* return true;

* }

*

* The Sieve of Eratosthenes is one of the most efficient ways to find all prime numbers up to n.

* But don't let that name scare you, I promise that the concept is surprisingly simple.

*

* [Sieve of Eratosthenes](http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes)

*

* [https://leetcode.com/static/images/solutions/Sieve_of_Eratosthenes_animation.gif]

* [http://commons.wikimedia.org/wiki/File:Sieve_of_Eratosthenes_animation.gif]

*

* Sieve of Eratosthenes: algorithm steps for primes below 121. "Sieve of Eratosthenes Animation"()

* by SKopp is licensed under CC BY 2.0.

*

* * [Skoop](http://de.wikipedia.org/wiki/Benutzer:SKopp)

* * [CC BY 2.0](http://creativecommons.org/licenses/by/2.0/)

*

* We start off with a table of n numbers. Let's look at the first number, 2. We know all multiples of 2

* must not be primes, so we mark them off as non-primes. Then we look at the next number, 3. Similarly,

* all multiples of 3 such as 3 × 2 = 6, 3 × 3 = 9, ... must not be primes, so we mark them off as well.

* Now we look at the next number, 4, which was already marked off. What does this tell you? Should you

* mark off all multiples of 4 as well?

*

* 4 is not a prime because it is divisible by 2, which means all multiples of 4 must also be divisible

* by 2 and were already marked off. So we can skip 4 immediately and go to the next number, 5. Now,

* all multiples of 5 such as 5 × 2 = 10, 5 × 3 = 15, 5 × 4 = 20, 5 × 5 = 25, ... can be marked off.

* There is a slight optimization here, we do not need to start from 5 × 2 = 10. Where should we start marking off?

*

* In fact, we can mark off multiples of 5 starting at 5 × 5 = 25, because 5 × 2 = 10 was already marked off

* by multiple of 2, similarly 5 × 3 = 15 was already marked off by multiple of 3. Therefore, if the current

* number is p, we can always mark off multiples of p starting at p2, then in increments of p: p2 + p, p2 + 2p, ...

* Now what should be the terminating loop condition?

*

* It is easy to say that the terminating loop condition is p n, which is certainly correct but not efficient.

* Do you still remember Hint #3?

*

* Yes, the terminating loop condition can be p n, as all non-primes ≥ √n must have already been marked off.

* When the loop terminates, all the numbers in the table that are non-marked are prime.

*

* The Sieve of Eratosthenes uses an extra O(n) memory and its runtime complexity is O(n log log n).

* For the more mathematically inclined readers, you can read more about its algorithm complexity on

* [Wikipedia](http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes#Algorithm_complexity).

*

* public int countPrimes(int n) {

* boolean[] isPrime = new boolean[n];

* for (int i = 2; i < n; i++) {

* isPrime[i] = true;

* }

* // Loop's ending condition is i * i < n instead of i < sqrt(n)

* // to avoid repeatedly calling an expensive function sqrt().

* for (int i = 2; i * i < n; i++) {

* if (!isPrime[i]) continue;

* for (int j = i * i; j < n; j += i) {

* isPrime[j] = false;

* }

* }

* int count = 0;

* for (int i = 2; i < n; i++) {

* if (isPrime[i]) count++;

* }

* return count;

* }

*

*

**********************************************************************************/

#include <stdlib.h>

#include <iostream>

#include <vector>

using namespace std;

int countPrimes(int n) {

vector<bool> isPrimer(n, true);

for(int i=2; i*i<n; i++){

if (isPrimer[i]){

for(int j=i*i; j<n; j+=i){

isPrimer[j] = false;

}

}

}

int cnt = 0;

for(int i=2; i<n; i++){

if (isPrimer[i]) {

//cout << i << ", ";

cnt++;

}

}

return cnt;

}

int main(int argc, char**argv)

{

int n = 100;

if (argc>1){

n = atoi(argv[1]);

}

cout << endl << n << " : " << countPrimes(n) << endl;

return 0;

}