# Jan 09-count Primes; Mathematics; optimization; Primes; DP;

Source: Internet
Author: User
Tags log log

The first method of the Sieve of Eratosthenes is one of the most efficient ways to find all prime numbers up to N.

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

Creates a Boolean array of length = N IsPrime, with each element initialized to true;

K = 2: (n-1), if the current number k is the prime, put K^k-(n-1)/k *k of the number of pairs of mappings = false;

Finally, count from 2 to n-1 IsPrime = = True.

Code:

public class Solution {
public int countprimes (int n) {
boolean[] Isprimes = new Boolean[n];
for (int i = 0; i < isprimes.length; i++) {
Isprimes[i] = true;
}

for (int k = 2; k <= (n-1)/k; k++) {
if (isprimes[k] = = True) {
for (int i = k; I <= (n-1)/k; i++) {
Isprimes[i*k] = false;
}
}
}
int count = 0;
for (int i = 2; i< isprimes.length; i++) {
if (isprimes[i] = = true) count++;
}
return count;
}
}

You can also use DP to solve.

public class Solution {
public int countprimes (int n) {

int count = 0;
int squareroot = 1;
int number = 2;

list<integer> list = new arraylist<> ();
for (int i = number; i < n; i++) {
Boolean isprime = true;
if (SquareRoot * SquareRoot < i) squareroot++;
for (int j = 0; J < list.size () && List.get (j) <= squareroot;j++) {
if (I%list.get (j) = = 0) {
IsPrime = false;
Break
}
}
if (IsPrime = = True) {
count++;
}
}
return count;
}
}

Runtime:o (N*SQRT (n)/log (n))

Finally, record the most primitive method of 0-sqrt (n):

public class Solution {
public int countprimes (int n) {

int count = 0;
int squareroot = 1;
int number = 2;

for (int i = number; i < n; i++) {
Boolean isprime = true;

for (int divisor = 2; divisor <= (int) (MATH.SQRT (i)); divisor++) {
if (I%divisor = = 0) {
IsPrime = false;
Break
}
}
if (IsPrime = = True) {
count++;
}
}
return count;
}
}

Jan 09-count Primes; Mathematics; optimization; Primes; DP;

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