Given a range [L, R], the task is to find the numbers from the range which have the count of their divisors as even as well as prime.
Then, print the count of the numbers found. The values of L and R are less than 10^6 and L< R.
Examples:
Input: L=3, R=9
Output: Count = 3
Explanation: The numbers are 3, 5, 7
Input : L=3, R=17
Output : Count: 6
- The only number that is prime, as well as even, is ‘2’.
- So, we need to find all the numbers within the given range that have exactly 2 divisors,
i.e. prime numbers.
Naive Approach:
- Iterate through each number x in the range [L, R].
- For each number x, count the number of divisors it has.
- Check if the count of divisors is both even and prime. If it is, increment a counter.
- After iterating through all numbers in the range, return the final count.
Below is the implementation of the above approach:
C++
#include <iostream> using namespace std;
// Function to check if a number is prime bool isPrime( int num)
{ if (num < 2)
return false ;
for ( int i = 2; i * i <= num; i++) {
if (num % i == 0)
return false ;
}
return true ;
} // Function to count the divisors of a number int countDivisors( int num)
{ int count = 0;
for ( int i = 1; i <= num; i++) {
if (num % i == 0)
count++;
}
return count;
} // Function to count the numbers in the given range // with even prime divisors count int countNumbers( int L, int R)
{ int count = 0;
for ( int x = L; x <= R; x++) {
int divisors = countDivisors(x);
if (divisors % 2 == 0 && isPrime(divisors))
count++;
}
return count;
} int main()
{ int L = 3, R = 9;
int count = countNumbers(L, R);
cout << "Count: " << count << endl;
return 0;
} |
Java
public class GFG {
// Function to check if a number is prime
public static boolean isPrime( int num)
{
if (num < 2 )
return false ;
for ( int i = 2 ; i * i <= num; i++) {
if (num % i == 0 )
return false ;
}
return true ;
}
// Function to count the divisors of a number
public static int countDivisors( int num)
{
int count = 0 ;
for ( int i = 1 ; i <= num; i++) {
if (num % i == 0 )
count++;
}
return count;
}
// Function to count the numbers in the given range
// with even prime divisors count
public static int countNumbers( int L, int R)
{
int count = 0 ;
for ( int x = L; x <= R; x++) {
int divisors = countDivisors(x);
if (divisors % 2 == 0 && isPrime(divisors))
count++;
}
return count;
}
public static void main(String[] args)
{
int L = 3 , R = 9 ;
int count = countNumbers(L, R);
System.out.println( "Count: " + count);
}
} // This code is contributed by shivamgupta0987654321 |
Python3
# Function to check if a number is prime def is_prime(num):
if num < 2 :
return False
for i in range ( 2 , int (num * * 0.5 ) + 1 ):
if num % i = = 0 :
return False
return True
# Function to count the divisors of a number def count_divisors(num):
count = 0
for i in range ( 1 , num + 1 ):
if num % i = = 0 :
count + = 1
return count
# Function to count the numbers in the given range # with even prime divisors count def count_numbers(L, R):
count = 0
for x in range (L, R + 1 ):
divisors = count_divisors(x)
if divisors % 2 = = 0 and is_prime(divisors):
count + = 1
return count
L = 3
R = 9
count = count_numbers(L, R)
print ( "Count:" , count)
|
C#
using System;
class GFG {
// Function to check if a number is prime
static bool IsPrime( int num)
{
if (num < 2)
return false ;
for ( int i = 2; i * i <= num; i++) {
if (num % i == 0)
return false ;
}
return true ;
}
// Function to count the divisors of a number
static int CountDivisors( int num)
{
int count = 0;
for ( int i = 1; i <= num; i++) {
if (num % i == 0)
count++;
}
return count;
}
// Function to count the numbers in the given range
// with even prime divisors count
static int CountNumbers( int L, int R)
{
int count = 0;
for ( int x = L; x <= R; x++) {
int divisors = CountDivisors(x);
if (divisors % 2 == 0 && IsPrime(divisors))
count++;
}
return count;
}
static void Main()
{
int L = 3, R = 9;
int count = CountNumbers(L, R);
Console.WriteLine( "Count: " + count);
}
} // This code is contributed by shivamgupta0987654321 |
Javascript
// Function to check if a number is prime function isPrime(num) {
if (num < 2) {
return false ;
}
for (let i = 2; i * i <= num; i++) {
if (num % i === 0) {
return false ;
}
}
return true ;
} // Function to count the divisors of a number function countDivisors(num) {
let count = 0;
for (let i = 1; i <= num; i++) {
if (num % i === 0) {
count++;
}
}
return count;
} // Function to count the numbers in the given range // with an even count of prime divisors function countNumbers(L, R) {
let count = 0;
for (let x = L; x <= R; x++) {
let divisors = countDivisors(x);
if (divisors % 2 === 0 && isPrime(divisors)) {
count++;
}
}
return count;
} // Driver code const L = 3; const R = 9; const count = countNumbers(L, R); console.log( "Count: " + count);
|
Output
Count: 3
Time Complexity: O((R – L + 1) * sqrt(R)).
Auxiliary Space: O(1)
An efficient approach:
- We have to count the prime numbers in range [L, R].
- First, create a sieve which will help in determining whether the number is prime or not in O(1) time.
- Then, create a prefix array to store the count of prime numbers where, element at index ‘i’ holds the count of the prime numbers from ‘1’ to ‘i’.
- Now, if we want to find the count of prime numbers in range [L, R], the count will be (sum[R] – sum[L-1])
- Finally, print the result i.e. (sum[R] – sum[L-1])
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
#define MAX 1000000 // stores whether the number is prime or not bool prime[MAX + 1];
// stores the count of prime numbers // less than or equal to the index int sum[MAX + 1];
// create the sieve void SieveOfEratosthenes()
{ // Create a boolean array "prime[0..n]" and initialize
// all the entries as true. A value in prime[i] will
// finally be false if 'i' is Not a prime, else true.
memset (prime, true , sizeof (prime));
memset (sum, 0, sizeof (sum));
prime[1] = false ;
for ( int p = 2; p * p <= MAX; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p]) {
// Update all multiples of p
for ( int i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
// stores the prefix sum of number
// of primes less than or equal to 'i'
for ( int i = 1; i <= MAX; i++) {
if (prime[i] == true )
sum[i] = 1;
sum[i] += sum[i - 1];
}
} // Driver code int main()
{ // create the sieve
SieveOfEratosthenes();
// 'l' and 'r' are the lower and upper bounds
// of the range
int l = 3, r = 9;
// get the value of count
int c = (sum[r] - sum[l - 1]);
// display the count
cout << "Count: " << c << endl;
return 0;
} |
Java
// Java implementation of the approach class GFG
{ static final int MAX= 1000000 ;
// stores whether the number is prime or not
static boolean []prime= new boolean [MAX + 1 ];
// stores the count of prime numbers
// less than or equal to the index
static int []sum= new int [MAX + 1 ];
// create the sieve
static void SieveOfEratosthenes()
{
// Create a boolean array "prime[0..n]" and initialize
// all the entries as true. A value in prime[i] will
// finally be false if 'i' is Not a prime, else true.
for ( int i= 0 ;i<=MAX;i++)
prime[i]= true ;
for ( int i= 0 ;i<=MAX;i++)
sum[i]= 0 ;
prime[ 1 ] = false ;
for ( int p = 2 ; p * p <= MAX; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p]) {
// Update all multiples of p
for ( int i = p * 2 ; i <= MAX; i += p)
prime[i] = false ;
}
}
// stores the prefix sum of number
// of primes less than or equal to 'i'
for ( int i = 1 ; i <= MAX; i++) {
if (prime[i] == true )
sum[i] = 1 ;
sum[i] += sum[i - 1 ];
}
}
// Driver code
public static void main(String []args)
{
// create the sieve
SieveOfEratosthenes();
// 'l' and 'r' are the lower and upper bounds
// of the range
int l = 3 , r = 9 ;
// get the value of count
int c = (sum[r] - sum[l - 1 ]);
// display the count
System.out.println( "Count: " + c);
}
} |
Python 3
# Python 3 implementation of the approach MAX = 1000000
# stores whether the number is prime or not prime = [ True ] * ( MAX + 1 )
# stores the count of prime numbers # less than or equal to the index sum = [ 0 ] * ( MAX + 1 )
# create the sieve def SieveOfEratosthenes():
prime[ 1 ] = False
p = 2
while p * p < = MAX :
# If prime[p] is not changed,
# then it is a prime
if (prime[p]):
# Update all multiples of p
i = p * 2
while i < = MAX :
prime[i] = False
i + = p
p + = 1
# stores the prefix sum of number
# of primes less than or equal to 'i'
for i in range ( 1 , MAX + 1 ):
if (prime[i] = = True ):
sum [i] = 1
sum [i] + = sum [i - 1 ]
# Driver code if __name__ = = "__main__" :
# create the sieve
SieveOfEratosthenes()
# 'l' and 'r' are the lower and
# upper bounds of the range
l = 3
r = 9
# get the value of count
c = ( sum [r] - sum [l - 1 ])
# display the count
print ( "Count:" , c)
# This code is contributed by ita_c |
C#
// C# implementation of the approach using System;
class GFG
{ static int MAX=1000000;
// stores whether the number is prime or not
static bool []prime= new bool [MAX + 1];
// stores the count of prime numbers
// less than or equal to the index
static int []sum= new int [MAX + 1];
// create the sieve
static void SieveOfEratosthenes()
{
// Create a boolean array "prime[0..n]" and initialize
// all the entries as true. A value in prime[i] will
// finally be false if 'i' is Not a prime, else true.
for ( int i=0;i<=MAX;i++)
prime[i]= true ;
for ( int i=0;i<=MAX;i++)
sum[i]=0;
prime[1] = false ;
for ( int p = 2; p * p <= MAX; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p]) {
// Update all multiples of p
for ( int i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
// stores the prefix sum of number
// of primes less than or equal to 'i'
for ( int i = 1; i <= MAX; i++) {
if (prime[i] == true )
sum[i] = 1;
sum[i] += sum[i - 1];
}
}
// Driver code
public static void Main()
{
// create the sieve
SieveOfEratosthenes();
// 'l' and 'r' are the lower and upper bounds
// of the range
int l = 3, r = 9;
// get the value of count
int c = (sum[r] - sum[l - 1]);
// display the count
Console.WriteLine( "Count: " + c);
}
} |
Javascript
<script> // Javascript implementation of the approach var MAX = 1000000;
// stores whether the number is prime or not var prime = Array(MAX+1).fill( true );
// stores the count of prime numbers // less than or equal to the index var sum = Array(MAX+1).fill(0);
// create the sieve function SieveOfEratosthenes()
{ // Create a boolean array "prime[0..n]" and initialize
// all the entries as true. A value in prime[i] will
// finally be false if 'i' is Not a prime, else true.
prime[1] = false ;
for ( var p = 2; p * p <= MAX; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p]) {
// Update all multiples of p
for ( var i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
// stores the prefix sum of number
// of primes less than or equal to 'i'
for ( var i = 1; i <= MAX; i++) {
if (prime[i] == true )
sum[i] = 1;
sum[i] += sum[i - 1];
}
} // Driver code // create the sieve SieveOfEratosthenes(); // 'l' and 'r' are the lower and upper bounds // of the range var l = 3, r = 9;
// get the value of count var c = (sum[r] - sum[l - 1]);
// display the count document.write( "Count: " + c );
</script> |
PHP
<?php // PHP implementation of the approach $MAX = 100000;
// Create a boolean array "prime[0..n]" // and initialize all the entries as // true. A value in prime[i] will finally // be false if 'i' is Not a prime, else true. // stores whether the number // is prime or not $prime = array_fill (0, $MAX + 1, true);
// stores the count of prime numbers // less than or equal to the index $sum = array_fill (0, $MAX + 1, 0);
// create the sieve function SieveOfEratosthenes()
{ global $MAX , $sum , $prime ;
$prime [1] = false;
for ( $p = 2; $p * $p <= $MAX ; $p ++)
{
// If prime[p] is not changed,
// then it is a prime
if ( $prime [ $p ])
{
// Update all multiples of p
for ( $i = $p * 2; $i <= $MAX ; $i += $p )
$prime [ $i ] = false;
}
}
// stores the prefix sum of number
// of primes less than or equal to 'i'
for ( $i = 1; $i <= $MAX ; $i ++)
{
if ( $prime [ $i ] == true)
$sum [ $i ] = 1;
$sum [ $i ] += $sum [ $i - 1];
}
} // Driver code // create the sieve SieveOfEratosthenes(); // 'l' and 'r' are the lower // and upper bounds of the range $l = 3;
$r = 9;
// get the value of count $c = ( $sum [ $r ] - $sum [ $l - 1]);
// display the count echo "Count: " . $c . "\n" ;
// This code is contributed by mits ?> |
Output
Count: 3
Time Complexity: O(MAX3/2)
Auxiliary Space: O(MAX)