# Count the numbers < N which have equal number of divisors as K

Given two integers N and K, the task is to count all the numbers < N which have equal number of positive divisors as K.

Examples:

Input: n = 10, k = 5
Output: 3
2, 3 and 7 are the only numbers < 10 which have 2 divisors (equal to the number of divisors of 5)

Input: n = 500, k = 6
Output: 148

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Compute the number of divisor of each number < N and store the result in an array where arr[i] contains the number of divisors of i.
• Traverse arr[], if arr[i] = arr[K] then update count = count + 1.
• Print the value of count in the end.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of the ` `// divisors of a number ` `int` `countDivisors(``int` `n) ` `{ ` `    ``// Count the number of ` `    ``// 2s that divide n ` `    ``int` `x = 0, ans = 1; ` `    ``while` `(n % 2 == 0) { ` `        ``x++; ` `        ``n = n / 2; ` `    ``} ` `    ``ans = ans * (x + 1); ` ` `  `    ``// n must be odd at this point. ` `    ``// So we can skip one element ` `    ``for` `(``int` `i = 3; i <= ``sqrt``(n); i = i + 2) { ` ` `  `        ``// While i divides n ` `        ``x = 0; ` `        ``while` `(n % i == 0) { ` `            ``x++; ` `            ``n = n / i; ` `        ``} ` `        ``ans = ans * (x + 1); ` `    ``} ` ` `  `    ``// This condition is to ` `    ``// handle the case when ` `    ``// n is a prime number > 2 ` `    ``if` `(n > 2) ` `        ``ans = ans * 2; ` ` `  `    ``return` `ans; ` `} ` ` `  `int` `getTotalCount(``int` `n, ``int` `k) ` `{ ` `    ``int` `k_count = countDivisors(k); ` ` `  `    ``// Count the total elements ` `    ``// that have divisors exactly equal ` `    ``// to as that of k's ` `    ``int` `count = 0; ` `    ``for` `(``int` `i = 1; i < n; i++) ` `        ``if` `(k_count == countDivisors(i)) ` `            ``count++; ` ` `  `    ``// Exclude k from the result if it  ` `    ``// is smaller than n. ` `    ``if` `(k < n) ` `       ``count = count - 1; ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 500, k = 6; ` `    ``cout << getTotalCount(n, k); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` ` `  `public` `class` `GFG{ ` ` `  `    ``// Function to return the count of the  ` `    ``// divisors of a number  ` `    ``static` `int` `countDivisors(``int` `n)  ` `    ``{  ` `        ``// Count the number of  ` `        ``// 2s that divide n  ` `        ``int` `x = ``0``, ans = ``1``;  ` `        ``while` `(n % ``2` `== ``0``) {  ` `            ``x++;  ` `            ``n = n / ``2``;  ` `        ``}  ` `        ``ans = ans * (x + ``1``);  ` `     `  `        ``// n must be odd at this point.  ` `        ``// So we can skip one element  ` `        ``for` `(``int` `i = ``3``; i <= Math.sqrt(n); i = i + ``2``) {  ` `     `  `            ``// While i divides n  ` `            ``x = ``0``;  ` `            ``while` `(n % i == ``0``) {  ` `                ``x++;  ` `                ``n = n / i;  ` `            ``}  ` `            ``ans = ans * (x + ``1``);  ` `        ``}  ` `     `  `        ``// This condition is to  ` `        ``// handle the case when  ` `        ``// n is a prime number > 2  ` `        ``if` `(n > ``2``)  ` `            ``ans = ans * ``2``;  ` `     `  `        ``return` `ans;  ` `    ``}  ` `     `  `    ``static` `int` `getTotalCount(``int` `n, ``int` `k)  ` `    ``{  ` `        ``int` `k_count = countDivisors(k);  ` `     `  `        ``// Count the total elements  ` `        ``// that have divisors exactly equal  ` `        ``// to as that of k's  ` `        ``int` `count = ``0``;  ` `        ``for` `(``int` `i = ``1``; i < n; i++)  ` `            ``if` `(k_count == countDivisors(i))  ` `                ``count++;  ` `     `  `        ``// Exclude k from the result if it  ` `        ``// is smaller than n.  ` `        ``if` `(k < n)  ` `        ``count = count - ``1``;  ` `     `  `        ``return` `count;  ` `    ``}  ` `     `  `    ``// Driver code ` `     ``public` `static` `void` `main(String []args) ` `    ``{  ` `        ``int` `n = ``500``, k = ``6``;  ` `        ``System.out.println(getTotalCount(n, k));  ` `    ``}  ` `    ``// This code is contributed by Ryuga ` `} `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the count of  ` `# the divisors of a number ` `def` `countDivisors(n): ` `     `  `    ``# Count the number of 2s that divide n ` `    ``x, ans ``=` `0``, ``1` `    ``while` `(n ``%` `2` `=``=` `0``): ` `        ``x ``+``=` `1` `        ``n ``=` `n ``/` `2` `    ``ans ``=` `ans ``*` `(x ``+` `1``) ` ` `  `    ``# n must be odd at this point. ` `    ``# So we can skip one element ` `    ``for` `i ``in` `range``(``3``, ``int``(n ``*``*` `1` `/` `2``) ``+` `1``, ``2``): ` `         `  `        ``# While i divides n ` `        ``x ``=` `0` `        ``while` `(n ``%` `i ``=``=` `0``): ` `            ``x ``+``=` `1` `            ``n ``=` `n ``/` `i ` `        ``ans ``=` `ans ``*` `(x ``+` `1``) ` ` `  `    ``# This condition is to handle the  ` `    ``# case when n is a prime number > 2 ` `    ``if` `(n > ``2``): ` `        ``ans ``=` `ans ``*` `2` ` `  `    ``return` `ans ` ` `  `def` `getTotalCount(n, k): ` `    ``k_count ``=` `countDivisors(k) ` ` `  `    ``# Count the total elements that ` `    ``# have divisors exactly equal ` `    ``# to as that of k's ` `    ``count ``=` `0` `    ``for` `i ``in` `range``(``1``, n): ` `        ``if` `(k_count ``=``=` `countDivisors(i)): ` `            ``count ``+``=` `1` ` `  `    ``# Exclude k from the result if it  ` `    ``# is smaller than n. ` `    ``if` `(k < n): ` `        ``count ``=` `count ``-` `1` ` `  `    ``return` `count ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n, k ``=` `500``, ``6` `    ``print``(getTotalCount(n, k)) ` ` `  `# This code is contributed  ` `# by 29AjayKumar `

## C#

 `// C# implementation of the approach  ` `using` `System; ` `                     `  `public` `class` `GFG{ ` `  `  `    ``// Function to return the count of the  ` `    ``// divisors of a number  ` `    ``static` `int` `countDivisors(``int` `n)  ` `    ``{  ` `        ``// Count the number of  ` `        ``// 2s that divide n  ` `        ``int` `x = 0, ans = 1;  ` `        ``while` `(n % 2 == 0) {  ` `            ``x++;  ` `            ``n = n / 2;  ` `        ``}  ` `        ``ans = ans * (x + 1);  ` `      `  `        ``// n must be odd at this point.  ` `        ``// So we can skip one element  ` `        ``for` `(``int` `i = 3; i <= Math.Sqrt(n); i = i + 2) {  ` `      `  `            ``// While i divides n  ` `            ``x = 0;  ` `            ``while` `(n % i == 0) {  ` `                ``x++;  ` `                ``n = n / i;  ` `            ``}  ` `            ``ans = ans * (x + 1);  ` `        ``}  ` `      `  `        ``// This condition is to  ` `        ``// handle the case when  ` `        ``// n is a prime number > 2  ` `        ``if` `(n > 2)  ` `            ``ans = ans * 2;  ` `      `  `        ``return` `ans;  ` `    ``}  ` `      `  `    ``static` `int` `getTotalCount(``int` `n, ``int` `k)  ` `    ``{  ` `        ``int` `k_count = countDivisors(k);  ` `      `  `        ``// Count the total elements  ` `        ``// that have divisors exactly equal  ` `        ``// to as that of k's  ` `        ``int` `count = 0;  ` `        ``for` `(``int` `i = 1; i < n; i++)  ` `            ``if` `(k_count == countDivisors(i))  ` `                ``count++;  ` `      `  `        ``// Exclude k from the result if it  ` `        ``// is smaller than n.  ` `        ``if` `(k < n)  ` `        ``count = count - 1;  ` `      `  `        ``return` `count;  ` `    ``}  ` `      `  `    ``// Driver code ` `     ``public` `static` `void` `Main() ` `    ``{  ` `        ``int` `n = 500, k = 6;  ` `        ``Console.WriteLine(getTotalCount(n, k));  ` `    ``}     ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## PHP

 ` 2 ` `    ``if` `(``\$n` `> 2) ` `        ``\$ans` `= ``\$ans` `* 2; ` ` `  `    ``return` `\$ans``; ` `} ` ` `  `function`  `getTotalCount(``\$n``, ``\$k``) ` `{ ` `    ``\$k_count` `= countDivisors(``\$k``); ` ` `  `    ``// Count the total elements ` `    ``// that have divisors exactly equal ` `    ``// to as that of k's ` `    ``\$count` `= 0; ` `    ``for` `(``\$i` `= 1; ``\$i` `< ``\$n``; ``\$i``++) ` `        ``if` `(``\$k_count` `== countDivisors(``\$i``)) ` `            ``\$count``++; ` ` `  `    ``// Exclude k from the result if it  ` `    ``// is smaller than n. ` `    ``if` `(``\$k` `< ``\$n``) ` `    ``\$count` `= ``\$count` `- 1; ` ` `  `    ``return` `\$count``; ` `} ` ` `  `// Driver code ` ` `  `    ``\$n` `= 500; ` `    ``\$k` `= 6; ` `    ``echo`  `getTotalCount(``\$n``, ``\$k``); ` ` `  `#This code is contributed by Sachin.. ` `?> `

Output:

```148
```

Optimization:
The above solution can be optimized using Sieve technique. Please refer Count number of integers less than or equal to N which has exactly 9 divisors for details.

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