Find numbers with n-divisors in a given range

23 Jun, 2022
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 numberEfficient 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 eratosthenes 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++
Java
Python3
C#
PHP
Javascript
// C++ program to count numbers with n divisors#include<bits/stdc++.h>using namespace std; // applying sieve of eratosthenesvoid 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*j <= 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            // divisible 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 codeint main(){    int a = 1, b = 7, n = 2;    cout << countNDivisors(a, b, n);    return 0;}
// Java program to count numbers with n divisorsimport java.util.*; class GFG{// applying sieve of eratosthenesstatic 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            // divisible 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 codepublic 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
# Python3 program to count numbers# with n divisorsimport math; # applying sieve of eratosthenesdef 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            # divisible 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 ndivisor 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 codea = 1;b = 7;n = 2;print(countNDivisors(a, b, n)); # This code is contributed by mits
// C# program to count numbers with n divisorsusing System.Collections;using System;class GFG{// applying sieve of eratosthenesstatic 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            // divisible 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 ndivisor 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 codepublic static void Main(){    int a = 1, b = 7, n = 2;    Console.WriteLine(countNDivisors(a, b, n));  }}// This code is contributed by mits
<?php// PHP program to count numbers with n divisors // applying sieve of eratosthenesfunction 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            $count = 0;             // repeatedly divide temp by k till            // it is divisible and accordingly            // increase count            while ($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 = $total * 2;         // if i is a n divisor number increase result        if ($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?>
<script> // Javascript program to count numbers with n divisors // applying sieve of eratosthenesfunction sieve(primes, x){    primes[1] = true;     // if a number is prime mark all its multiples    // as non prime    for (var i=2; i*i <= x; i++)    {        if (primes[i] == false)        {            for (var 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.function nDivisors(primes, x,  a, b, n){    // result holds number of numbers having n divisors    var result = 0;     // vector to hold all the prime numbers between 1    // ans sqrt(b)    var v = [];    for (var i = 2; i <= x; i++)        if (primes[i] == false)            v.push(i);         // Traversing all numbers in given range    for (var i=a; i<=b; i++)    {        // initialising temp as i        var temp = i;         // total holds the number of divisors of i        var total = 1;        var j = 0;         // we need to use that prime numbers that        // are less than equal to sqrt(temp)        for (var k = v[j]; k*k <= temp; k = v[++j])        {            // holds the exponent of k in prime            // factorization of i            var count = 0;             // repeatedly divide temp by k till it is            // divisible and accordingly increase count            while (temp%k == 0)            {                count++;                temp = parseInt(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 ndivisor number increase result        if (total == n)            result++;    }    return result;} // Returns count of numbers in [a..b] having// n divisors.function countNDivisors(a, b, n){    var x = parseInt(Math.sqrt(b)) + 1;     // primes[i] = true if i is a prime number    var primes = Array(x+1).fill(false);     // initialising each number as prime    sieve(primes, x);     return nDivisors(primes, x, a, b, n);} // driver codevar a = 1, b = 7, n = 2;document.write(countNDivisors(a, b, n)); // This code is contributed by rutvik_56.</script>
Output:  
4
Time Complexity: O(n) Auxiliary Space: O(n) 
This article is contributed by Aarti_Rathi and Ayush Jha. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
Mithun Kumar
ManasChhabra2
rutvik_56
pravesh25pandey
codewithshinchan
divisibility
divisors
sieve
Mathematical
Mathematical
sieve
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.