Find numbers with n-divisors in a given range

Given three integers a, b, n .Your task is to print number of numbers between a and b including them also which have n-divisors. A number is called n-divisor if it has total n divisors including 1 and itself.
Examples:

Input  : a = 1, b = 7, n = 2
Output : 4
There are four numbers with 2 divisors in 
range [1, 7]. The numbers are 2, 3, 5, and 7.

Naive Approach:
The naive approach is to check all the numbers between a and b how many of them are n-divisor number for doing this find out the number of each divisors for each number . If it is equal to n then it is a n-divisor number

Efficient Approach:
Any number can be written in the form of its prime factorization let the number be x and p1, p2..pm are the prime numbers which divide x so x = p1e1 * p2e2….pmem where e1, e2…em are the exponents of prime numbers p1, p2….pm. So the number of divisors of x will be (e1+1)*(e2+1)…*(em+1).
Now the second observation is for prime numbers greater than sqrt(x) their exponent cannot exceed 1. Let’s prove this by contradiction suppose there is a prime number P greater than sqrt(x) and its exponent E in prime factorization of x is greater than one (E >= 2) so P^E sqrt(x) so P^E > (sqrt(x))E and E >= 2 so PE will always be greater than x
Third observation is that number of prime numbers greater than sqrt(x) in the prime factorization of x will always be less than equal to 1. This can also be proved similarly by contradiction as above.

Now to solve this problem
Step 1: Apply sieve of erastothenes and calculate prime numbers upto sqrt(b).

Step 2: Traverse through each number from a to b and calculate exponents of each prime number in that number by repeatedly dividing that number by prime number and use the formula numberofdivisors(x) = (e1+1)*(e2+1)….(em+1).



Step 3: If after dividing by all the prime numbers less than equal to square root of that number if number > 1 this means there is a prime number greater than its square root which divides and its exponent will always be one as proved above.

C++

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// C++ program to count numbers with n divisors
#include<bits/stdc++.h>
using namespace std;
  
// applying sieve of erastothenes
void sieve(bool primes[], int x)
{
    primes[1] = false;
  
    // if a number is prime mark all its multiples
    // as non prime
    for (int i=2; i*i <= x; i++)
    {
        if (primes[i] == true)
        {
            for (int j=2; j*i <= x; j++)
                primes[i*j] = false;
        }
    }
}
  
// function that returns numbers of number that have
// n divisors in range from a to b. x is sqrt(b) + 1.
int nDivisors(bool primes[], int x, int a, int b, int n)
{
    // result holds number of numbers having n divisors
    int result = 0;
  
    // vector to hold all the prime numbers between 1
    // ans sqrt(b)
    vector <int> v;
    for (int i = 2; i <= x; i++)
        if (primes[i] == true)
            v.push_back (i);
  
    // Traversing all numbers in given range
    for (int i=a; i<=b; i++)
    {
        // initialising temp as i
        int temp = i;
  
        // total holds the number of divisors of i
        int total = 1;
        int j = 0;
  
        // we need to use that prime numbers that
        // are less than equal to sqrt(temp)
        for (int k = v[j]; k*k <= temp; k = v[++j])
        {
            // holds the exponent of k in prime
            // factorization of i
            int count = 0;
  
            // repeatedly divide temp by k till it is
            // divisble and accordingly increase count
            while (temp%k == 0)
            {
                count++;
                temp = temp/k;
            }
  
            // using the formula  no.of divisors =
            // (e1+1)*(e2+1)....
            total = total*(count+1);
        }
  
        // if temp is not equal to 1 then there is
        // prime number in prime factorization of i
        // greater than sqrt(i)
        if (temp != 1)
            total = total*2;
  
        // if i is a ndvisor number increase result
        if (total == n)
            result++;
    }
    return result;
}
  
// Returns count of numbers in [a..b] having
// n divisors.
int countNDivisors(int a, int b, int n)
{
    int x = sqrt(b) + 1;
  
    // primes[i] = true if i is a prime number
    bool primes[x];
  
    // initialising each number as prime
    memset(primes, true, sizeof(primes));
    sieve(primes, x);
  
    return nDivisors(primes, x, a, b, n);
}
  
// driver code
int main()
{
    int a = 1, b = 7, n = 2;
    cout << countNDivisors(a, b, n);
    return 0;
}

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Java

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// Java program to count numbers with n divisors 
import java.util.*;
  
class GFG{
// applying sieve of erastothenes 
static void sieve(boolean[] primes, int x) 
    primes[1] = true
  
    // if a number is prime mark all its multiples 
    // as non prime 
    for (int i=2; i*i <= x; i++) 
    
        if (primes[i] == false
        
            for (int j=2; j*i <= x; j++) 
                primes[i*j] = true
        
    
  
// function that returns numbers of number that have 
// n divisors in range from a to b. x is sqrt(b) + 1. 
static int nDivisors(boolean[] primes, int x, int a, int b, int n) 
    // result holds number of numbers having n divisors 
    int result = 0
  
    // vector to hold all the prime numbers between 1 
    // ans sqrt(b) 
    ArrayList<Integer> v=new ArrayList<Integer>(); 
    for (int i = 2; i <= x; i++) 
        if (primes[i] == false
            v.add(i); 
      
    // Traversing all numbers in given range 
    for (int i=a; i<=b; i++) 
    
        // initialising temp as i 
        int temp = i; 
  
        // total holds the number of divisors of i 
        int total = 1
        int j = 0
  
        // we need to use that prime numbers that 
        // are less than equal to sqrt(temp)
        for (int k = v.get(j); k*k <= temp; k = v.get(++j))
        
            // holds the exponent of k in prime 
            // factorization of i 
            int count = 0
  
            // repeatedly divide temp by k till it is 
            // divisble and accordingly increase count 
            while (temp%k == 0
            
                count++; 
                temp = temp/k; 
            
  
            // using the formula no.of divisors = 
            // (e1+1)*(e2+1).... 
            total = total*(count+1);
              
        
  
        // if temp is not equal to 1 then there is 
        // prime number in prime factorization of i 
        // greater than sqrt(i) 
        if (temp != 1
            total = total*2
  
        // if i is a ndvisor number increase result 
        if (total == n) 
            result++; 
    
    return result; 
  
// Returns count of numbers in [a..b] having 
// n divisors. 
static int countNDivisors(int a, int b, int n) 
    int x = (int)Math.sqrt(b) + 1
  
    // primes[i] = true if i is a prime number 
    boolean[] primes=new boolean[x+1]; 
  
    // initialising each number as prime 
    sieve(primes, x); 
  
    return nDivisors(primes, x, a, b, n); 
  
// driver code 
public static void main(String[] args) 
    int a = 1, b = 7, n = 2
    System.out.println(countNDivisors(a, b, n)); 
   
}
// This code is contributed by mits

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Python3

# Python3 program to count numbers
# with n divisors
import math;

# applying sieve of erastothenes
def sieve(primes, x):
primes[1] = False;

# if a number is prime mark all
# its multiples as non prime
i = 2;
while (i * i <= x): if (primes[i] == True): j = 2; while (j * i <= x): primes[i * j] = False; j += 1; i += 1; # function that returns numbers of number # that have n divisors in range from a to b. # x is sqrt(b) + 1. def nDivisors(primes, x, a, b, n): # result holds number of numbers # having n divisors result = 0; # vector to hold all the prime # numbers between 1 and sqrt(b) v = []; for i in range(2, x + 1): if (primes[i]): v.append(i); # Traversing all numbers in given range for i in range(a, b + 1): # initialising temp as i temp = i; # total holds the number of # divisors of i total = 1; j = 0; # we need to use that prime numbers that # are less than equal to sqrt(temp) k = v[j]; while (k * k <= temp): # holds the exponent of k in prime # factorization of i count = 0; # repeatedly divide temp by k till it is # divisble and accordingly increase count while (temp % k == 0): count += 1; temp = int(temp / k); # using the formula no.of divisors = # (e1+1)*(e2+1).... total = total * (count + 1); j += 1; k = v[j]; # if temp is not equal to 1 then there is # prime number in prime factorization of i # greater than sqrt(i) if (temp != 1): total = total * 2; # if i is a ndvisor number # increase result if (total == n): result += 1; return result; # Returns count of numbers in [a..b] # having n divisors. def countNDivisors(a, b, n): x = int(math.sqrt(b) + 1); # primes[i] = true if i is a prime number # initialising each number as prime primes = [True] * (x + 1); sieve(primes, x); return nDivisors(primes, x, a, b, n); # Driver code a = 1; b = 7; n = 2; print(countNDivisors(a, b, n)); # This code is contributed by mits [tabby title="C#"]

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// C# program to count numbers with n divisors 
using System.Collections;
using System;
class GFG{
// applying sieve of erastothenes 
static void sieve(bool[] primes, int x) 
    primes[1] = true
  
    // if a number is prime mark all its multiples 
    // as non prime 
    for (int i=2; i*i <= x; i++) 
    
        if (primes[i] == false
        
            for (int j=2; j*i <= x; j++) 
                primes[i*j] = true
        
    
  
// function that returns numbers of number that have 
// n divisors in range from a to b. x is sqrt(b) + 1. 
static int nDivisors(bool[] primes, int x, int a, int b, int n) 
    // result holds number of numbers having n divisors 
    int result = 0; 
  
    // vector to hold all the prime numbers between 1 
    // ans sqrt(b) 
    ArrayList v=new ArrayList(); 
    for (int i = 2; i <= x; i++) 
        if (primes[i] == false
            v.Add(i); 
      
    // Traversing all numbers in given range 
    for (int i=a; i<=b; i++) 
    
        // initialising temp as i 
        int temp = i; 
  
        // total holds the number of divisors of i 
        int total = 1; 
        int j = 0; 
  
        // we need to use that prime numbers that 
        // are less than equal to sqrt(temp)
        for (int k = (int)v[j]; k*k <= temp; k = (int)v[++j])
        
            // holds the exponent of k in prime 
            // factorization of i 
            int count = 0; 
  
            // repeatedly divide temp by k till it is 
            // divisble and accordingly increase count 
            while (temp%k == 0) 
            
                count++; 
                temp = temp/k; 
            
  
            // using the formula no.of divisors = 
            // (e1+1)*(e2+1).... 
            total = total*(count+1);
              
        
  
        // if temp is not equal to 1 then there is 
        // prime number in prime factorization of i 
        // greater than sqrt(i) 
        if (temp != 1) 
            total = total*2; 
  
        // if i is a ndvisor number increase result 
        if (total == n) 
            result++; 
    
    return result; 
  
// Returns count of numbers in [a..b] having 
// n divisors. 
static int countNDivisors(int a, int b, int n) 
    int x = (int)Math.Sqrt(b) + 1; 
  
    // primes[i] = true if i is a prime number 
    bool[] primes=new bool[x+1]; 
  
    // initialising each number as prime 
    sieve(primes, x); 
  
    return nDivisors(primes, x, a, b, n); 
  
// driver code 
public static void Main() 
    int a = 1, b = 7, n = 2; 
    Console.WriteLine(countNDivisors(a, b, n)); 
   
}
// This code is contributed by mits

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PHP

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<?php
// PHP program to count numbers with n divisors 
  
// applying sieve of erastothenes 
function sieve(&$primes, $x
    $primes[1] = false; 
  
    // if a number is prime mark all
    // its multiples as non prime 
    for ($i = 2; $i * $i <= $x; $i++) 
    
        if ($primes[$i] == true) 
        
            for ($j = 2; $j * $i <= $x; $j++) 
                $primes[$i * $j] = false; 
        
    
  
// function that returns numbers of number 
// that have n divisors in range from a to
// b. x is sqrt(b) + 1. 
function nDivisors($primes, $x, $a, $b, $n
    // result holds number of numbers
    // having n divisors 
    $result = 0; 
  
    // vector to hold all the prime numbers 
    // between 1 ans sqrt(b) 
    $v = array(); 
    for ($i = 2; $i <= $x; $i++) 
        if ($primes[$i] == true) 
            array_push($v, $i); 
  
    // Traversing all numbers in given range 
    for ($i = $a; $i <= $b; $i++) 
    
        // initialising temp as i 
        $temp = $i
  
        // total holds the number of 
        // divisors of i 
        $total = 1; 
        $j = 0; 
  
        // we need to use that prime numbers that 
        // are less than equal to sqrt(temp) 
        for ($k = $v[$j]; 
             $k * $k <= $temp; $k = $v[++$j]) 
        
            // holds the exponent of k in 
            // prime factorization of i 
            $count = 0; 
  
            // repeatedly divide temp by k till 
            // it is divisble and accordingly 
            // increase count 
            while ($temp % $k == 0) 
            
                $count++; 
                $temp = (int)($temp / $k); 
            
  
            // using the formula no.of divisors = 
            // (e1+1)*(e2+1).... 
            $total = $total * ($count + 1); 
        
  
        // if temp is not equal to 1 then there is 
        // prime number in prime factorization of i 
        // greater than sqrt(i) 
        if ($temp != 1) 
            $total = $total * 2; 
  
        // if i is a ndvisor number increase result 
        if ($total == $n
            $result++; 
    
    return $result
  
// Returns count of numbers in [a..b] 
// having n divisors. 
function countNDivisors($a, $b, $n
    $x = (int)(sqrt($b) + 1); 
  
    // primes[i] = true if i is a prime number 
    // initialising each number as prime 
    $primes = array_fill(0, $x + 1, true);
    sieve($primes, $x); 
  
    return nDivisors($primes, $x, $a, $b, $n); 
  
// Driver code 
$a = 1;
$b = 7;
$n = 2; 
print(countNDivisors($a, $b, $n)); 
  
// This code is contributed by mits
?>

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Output:

4

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