Contar primos en rangos

Dado un rango [L, R], necesitamos encontrar el número total de números primos en el rango [L, R] donde 0 <= L <= R < 10000. Considere que hay una gran cantidad de consultas para rangos diferentes
Ejemplos: 
 

Input : Query 1 : L = 1, R = 10
        Query 2 : L = 5, R = 10
Output : 4
         2
Explanation
Primes in the range L = 1 to R = 10 are 
{2, 3, 5, 7}. Therefore for query, answer 
is 4 {2, 3, 5, 7}.
For the second query, answer is 2 {5, 7}.

Una solución simple es hacer lo siguiente para cada consulta [L, R]. Atraviesa de L a R, verifica si el número actual es primo . Si es así, incremente el conteo. Finalmente, devuelve el conteo.
Una solución eficiente es usar la Tamiz de Eratóstenes para encontrar todos los números primos hasta el límite dado. Luego calculamos una array de prefijos para almacenar recuentos hasta cada valor antes del límite. Una vez que tenemos una array de prefijos, podemos responder consultas en tiempo O (1). Solo necesitamos devolver prefijo[R] – prefijo[L-1]. 
 

// CPP program to answer queries for count of
// primes in given range.
#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 10000;
 
// prefix[i] is going to store count of primes
// till i (including i).
int prefix[MAX + 1];
 
void buildPrefix()
{
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    bool prime[MAX + 1];
    memset(prime, true, sizeof(prime));
 
    for (int p = 2; p * p <= MAX; p++) {
 
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) {
 
            // Update all multiples of p
            for (int i = p * 2; i <= MAX; i += p)
                prime[i] = false;
        }
    }
 
    // Build prefix array
    prefix[0] = prefix[1] = 0;
    for (int p = 2; p <= MAX; p++) {
        prefix[p] = prefix[p - 1];
        if (prime[p])
            prefix[p]++;
    }
}
 
// Returns count of primes in range from L to
// R (both inclusive).
int query(int L, int R)
{
    return prefix[R] - prefix[L - 1];
}
 
// Driver code
int main()
{
    buildPrefix();
 
    int L = 5, R = 10;
    cout << query(L, R) << endl;
 
    L = 1, R = 10;
    cout << query(L, R) << endl;
 
    return 0;
}

// Java program to answer queries for
// count of primes in given range.
import java.util.*;
 
class GFG {
     
static final int MAX = 10000;
 
// prefix[i] is going to store count
// of primes till i (including i).
static int prefix[] = new int[MAX + 1];
 
static void buildPrefix() {
     
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    boolean prime[] = new boolean[MAX + 1];
    Arrays.fill(prime, true);
 
    for (int p = 2; p * p <= MAX; p++) {
 
    // If prime[p] is not changed, then
    // it is a prime
    if (prime[p] == true) {
 
        // Update all multiples of p
        for (int i = p * 2; i <= MAX; i += p)
        prime[i] = false;
    }
    }
 
    // Build prefix array
    prefix[0] = prefix[1] = 0;
    for (int p = 2; p <= MAX; p++) {
    prefix[p] = prefix[p - 1];
    if (prime[p])
        prefix[p]++;
    }
}
 
// Returns count of primes in range
// from L to R (both inclusive).
static int query(int L, int R)
{
    return prefix[R] - prefix[L - 1];
}
 
// Driver code
public static void main(String[] args) {
     
    buildPrefix();
    int L = 5, R = 10;
    System.out.println(query(L, R));
 
    L = 1; R = 10;
    System.out.println(query(L, R));
}
}
 
// This code is contributed by Anant Agarwal.

# Python3 program to answer queries for
# count of primes in given range.
MAX = 10000
 
# prefix[i] is going to
# store count of primes
# till i (including i).
prefix =[0]*(MAX + 1)
 
def buildPrefix():
     
    # Create a boolean array value in
    # prime[i] will "prime[0..n]". A
    # finally be false if i is Not a
    # prime, else true.
    prime = [1]*(MAX + 1)
 
    p = 2
    while(p * p <= MAX):
 
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p] == 1):
 
            # Update all multiples of p
            i = p * 2
            while(i <= MAX):
                prime[i] = 0
                i += p
        p+=1
 
    # Build prefix array
    # prefix[0] = prefix[1] = 0;
    for p in range(2,MAX+1):
        prefix[p] = prefix[p - 1]
        if (prime[p]==1):
            prefix[p]+=1
 
# Returns count of primes
# in range from L to
# R (both inclusive).
def query(L, R):
    return prefix[R]-prefix[L - 1]
 
# Driver code
if __name__=='__main__':
    buildPrefix()
 
    L = 5
    R = 10
    print(query(L, R))
 
    L = 1
    R = 10
    print(query(L, R))
 
# This code is contributed by mits.

// C# program to answer
// queries for count of
// primes in given range.
using System;
 
class GFG
{
static int MAX = 10000;
 
// prefix[i] is going
// to store count of
// primes till i (including i).
static int[] prefix = new int[MAX + 1];
 
static void buildPrefix()
{
     
    // Create a boolean array
    // "prime[0..n]". A value
    // in prime[i] will finally
    // be false if i is Not a
    // prime, else true.
    bool[] prime = new bool[MAX + 1];
 
    for (int p = 2;
             p * p <= MAX; p++)
    {
 
    // If prime[p] is
    // not changed, then
    // it is a prime
    if (prime[p] == false)
    {
 
        // Update all
        // multiples of p
        for (int i = p * 2;
                 i <= MAX; i += p)
        prime[i] = true;
    }
    }
 
    // Build prefix array
    prefix[0] = prefix[1] = 0;
    for (int p = 2; p <= MAX; p++)
    {
        prefix[p] = prefix[p - 1];
        if (prime[p] == false)
            prefix[p]++;
    }
}
 
// Returns count of primes
// in range from L to R
// (both inclusive).
static int query(int L, int R)
{
    return prefix[R] -
           prefix[L - 1];
}
 
// Driver code
public static void Main()
{
    buildPrefix();
    int L = 5, R = 10;
    Console.WriteLine(query(L, R));
 
    L = 1; R = 10;
    Console.WriteLine(query(L, R));
}
}
 
// This code is contributed
// by mits.

<?php
// PHP program to answer queries for
// count of primes in given range.
$MAX = 10000;
 
// prefix[i] is going to
// store count of primes
// till i (including i).
$prefix = array_fill(0, ($MAX + 1), 0);
 
function buildPrefix()
{
    global $MAX, $prefix;
     
    // Create a boolean array value in
    // prime[i] will "prime[0..n]". A
    // finally be false if i is Not a
    // prime, else true.
    $prime = array_fill(0, ($MAX + 1), true);
 
    for ($p = 2;
         $p * $p <= $MAX; $p++)
    {
 
        // If prime[p] is not changed,
        // then it is a prime
        if ($prime[$p] == true)
        {
 
            // Update all multiples of p
            for ($i = $p * 2;
                 $i <= $MAX; $i += $p)
                $prime[$i] = false;
        }
    }
 
    // Build prefix array
    // $prefix[0] = $prefix[1] = 0;
    for ($p = 2; $p <= $MAX; $p++)
    {
        $prefix[$p] = $prefix[$p - 1];
        if ($prime[$p])
            $prefix[$p]++;
    }
}
 
// Returns count of primes
// in range from L to
// R (both inclusive).
function query($L, $R)
{
    global $prefix;
    return $prefix[$R] -
           $prefix[$L - 1];
}
 
// Driver code
buildPrefix();
 
$L = 5;
$R = 10;
echo query($L, $R) . "\n";
 
$L = 1;
$R = 10;
echo query($L, $R) . "\n";
 
// This code is contributed by mits.
?>

<script>
 
// Javascript program to answer queries for
// count of primes in given range.
 
let MAX = 10000;
   
// prefix[i] is going to store count
// of primes till i (including i).
let prefix = [];
   
function buildPrefix() {
       
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    let prime = [];
    for (let p = 1; p <= MAX +1; p++) {
        prime[p] = true;
    }
   
    for (let p = 2; p * p <= MAX; p++) {
   
    // If prime[p] is not changed, then
    // it is a prime
    if (prime[p] == true) {
   
        // Update all multiples of p
        for (let i = p * 2; i <= MAX; i += p)
        prime[i] = false;
    }
    }
   
    // Build prefix array
    prefix[0] = prefix[1] = 0;
    for (let p = 2; p <= MAX; p++) {
    prefix[p] = prefix[p - 1];
    if (prime[p])
        prefix[p]++;
    }
}
   
// Returns count of primes in range
// from L to R (both inclusive).
function query(L, R)
{
    return prefix[R] - prefix[L - 1];
}
 
// driver program
 
    buildPrefix();
    let L = 5, R = 10;
    document.write(query(L, R) + "<br/>");
   
    L = 1; R = 10;
    document.write(query(L, R));
 
</script>

Producción:  

2
4

Complejidad del tiempo: O(n*log(log(n)))

Espacio Auxiliar: O(n)

Aquí, n es el tamaño de la array principal, que es MAX aquí

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