# Divisors of n-square that are not divisors of n

Given an positive large integer n. Count the number of positive divisors of n2 which are not divisible by any divisor of n(1 <= n <= 1012).

```Input: 6
Output: 5
Explanation
Total divisors of 62 are 9 i.e.,
1, 2, 3, 4, 6, 9, 12, 18, 36
Total divisors of '6' are 4,
1, 2, 3, 6
Total divisor of '36' which are not
divisible by divisors of '6' are
'5' i.e., 4, 9, 12, 18, 36

Input: 8
Output: 3```

Simple approach is to traverse for every divisor of n2 and count only those divisors which are not divisor of ‘n’. Time complexity of this approach is O(n).
Efficient approach is to use prime factorization to count total divisors of n2. A number ‘n’ can be represented as product of primes . Refer this to understand more.

`Let  for some primes p1 and p2.Squaring both the sidesTotal factors of n2 will be,Total factors of 'n' will be,Difference between the two gives the requiredanswer`

## C++

 `// C++ program to count number of` `// divisors of n^2 which are not` `// divisible by divisor of n` `#include ` `using` `namespace` `std;`   `// Function to count divisors of n^2` `// having no factors of 'n'` `int` `factors(``long` `long` `n)` `{` `    ``unordered_map<``int``, ``int``> prime;` `    ``for` `(``int` `i = 2; i <= ``sqrt``(n); ++i) {` `        ``while` `(n % i == 0) {`   `            ``// Increment count of i-th prime divisor` `            ``++prime[i];`   `            ``// Find next  prime divisor` `            ``n = n / i;` `        ``}` `    ``}`   `    ``// Increment count if divisor still remains` `    ``if` `(n > 2)` `        ``++prime[n];`   `    ``// Initialize variable for counting the factors` `    ``// of  n^2 and n as ans1 and ans2 respectively` `    ``int` `ans1 = 1, ans2 = 1;`   `    ``// Range based for-loop` `    ``for` `(``auto` `it : prime) {`   `        ``// Use formula as discussed in above` `        ``ans1 *= 2 * it.second + 1;` `        ``ans2 *= it.second + 1;` `    ``}`   `    ``// return the difference of answers` `    ``return` `ans1 - ans2;` `}`   `// Driver code` `int` `main()` `{` `    ``long` `long` `n = 5;` `    ``cout << factors(n) << endl;` `    ``n = 8;` `    ``cout << factors(n);` `    ``return` `0;` `}`

## Java

 `// Java program to count number of` `// divisors of n^2 which are not` `// divisible by divisor of n` `import` `java.util.*;`   `class` `GFG ` `{`   `// Function to count divisors of n^2` `// having no factors of 'n'` `static` `int` `factors(``int` `n)` `{` `    ``HashMapprime = ``new` `HashMap();` `    ``for` `(``int` `i = ``2``; i <= Math.sqrt(n); ++i) ` `    ``{` `        ``while` `(n % i == ``0``) ` `        ``{`   `            ``// Increment count of i-th prime divisor` `            ``if` `(prime.containsKey(i)) ` `            ``{` `                ``prime.put(i, prime.get(i) + ``1``);` `            ``} ` `            `  `            ``else` `            ``{` `                ``prime.put(i, ``1``);` `            ``}`   `            ``// Find next prime divisor` `            ``n = n / i;` `        ``}` `    ``}`   `    ``// Increment count if divisor still remains` `    ``if` `(n > ``2``)` `    ``{` `        ``if``(prime.containsKey(n))` `        ``{` `            ``prime.put(n, prime.get(n) + ``1``);` `        ``}` `        ``else` `        ``{` `            ``prime.put(n, ``1``);` `        ``}` `    ``}`   `    ``// Initialize variable for counting the factors` `    ``// of n^2 and n as ans1 and ans2 respectively` `    ``int` `ans1 = ``1``, ans2 = ``1``;`   `    ``// Range based for-loop` `    ``for` `(Map.Entry it : prime.entrySet()) ` `    ``{`   `        ``// Use formula as discussed in above` `        ``ans1 *= ``2` `* it.getValue() + ``1``;` `        ``ans2 *= it.getValue() + ``1``;` `    ``}`   `    ``// return the difference of answers` `    ``return` `ans1 - ans2;` `}`   `// Driver code` `public` `static` `void` `main(String[] args) ` `{` `    ``int` `n = ``5``;` `    ``System.out.println(factors(n));` `    ``n = ``8``;` `    ``System.out.println(factors(n));` `}` `}`   `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 program to count number of` `# divisors of n^2 which are not` `# divisible by divisor of n` `import` `math as mt`   `# Function to count divisors of n^2` `# having no factors of 'n'` `def` `factors(n):`   `    ``prime ``=` `dict``()` `    ``for` `i ``in` `range``(``2``, mt.ceil(mt.sqrt(n ``+` `1``))):` `        ``while` `(n ``%` `i ``=``=` `0``):` `            `  `            ``# Increment count of i-th ` `            ``# prime divisor` `            ``if` `i ``in` `prime.keys():` `                ``prime[i] ``+``=` `1` `            ``else``:` `                ``prime[i] ``=` `1` `                `  `            ``# Find next prime divisor` `            ``n ``=` `n ``/``/` `i` `        `  `    ``# Increment count if divisor` `    ``# still remains` `    ``if` `(n > ``2``):` `        ``if` `n ``in` `prime.keys():` `            ``prime[n] ``+``=` `1` `        ``else``:` `            ``prime[n] ``=` `1`   `    ``# Initialize variable for counting ` `    ``# the factors of n^2 and n as ans1 ` `    ``# and ans2 respectively` `    ``ans1 ``=` `1` `    ``ans2 ``=` `1`   `    ``# Range based for-loop` `    ``for` `it ``in` `prime:`   `        ``# Use formula as discussed in above` `        ``ans1 ``*``=` `2` `*` `prime[it] ``+` `1` `        ``ans2 ``*``=` `prime[it] ``+` `1` `    `  `    ``# return the difference of answers` `    ``return` `ans1 ``-` `ans2`   `# Driver code` `n ``=` `5` `print``(factors(n))` `n ``=` `8` `print``(factors(n))`   `# This code is contributed by ` `# Mohit kumar 29`

## C#

 `// C# program to count number of` `// divisors of n^2 which are not` `// divisible by divisor of n` `using` `System;` `using` `System.Collections.Generic;` `    `  `class` `GFG ` `{`   `// Function to count divisors of n^2` `// having no factors of 'n'` `static` `int` `factors(``int` `n)` `{` `    ``Dictionary<``int``,` `               ``int``> prime = ``new` `Dictionary<``int``,` `                                           ``int``>();` `    ``for` `(``int` `i = 2; i <= Math.Sqrt(n); ++i) ` `    ``{` `        ``while` `(n % i == 0) ` `        ``{`   `            ``// Increment count of i-th prime divisor` `            ``if` `(prime.ContainsKey(i)) ` `            ``{` `                ``prime[i] = prime[i] + 1;` `            ``} ` `            `  `            ``else` `            ``{` `                ``prime.Add(i, 1);` `            ``}`   `            ``// Find next prime divisor` `            ``n = n / i;` `        ``}` `    ``}`   `    ``// Increment count if divisor still remains` `    ``if` `(n > 2)` `    ``{` `        ``if``(prime.ContainsKey(n))` `        ``{` `            ``prime[n] = prime[n] + 1;` `        ``}` `        ``else` `        ``{` `            ``prime.Add(n, 1);` `        ``}` `    ``}`   `    ``// Initialize variable for counting the factors` `    ``// of n^2 and n as ans1 and ans2 respectively` `    ``int` `ans1 = 1, ans2 = 1;`   `    ``// Range based for-loop` `    ``foreach``(KeyValuePair<``int``, ``int``> it ``in` `prime) ` `    ``{`   `        ``// Use formula as discussed in above` `        ``ans1 *= 2 * it.Value + 1;` `        ``ans2 *= it.Value + 1;` `    ``}`   `    ``// return the difference of answers` `    ``return` `ans1 - ans2;` `}`   `// Driver code` `public` `static` `void` `Main(String[] args) ` `{` `    ``int` `n = 5;` `    ``Console.WriteLine(factors(n));` `    ``n = 8;` `    ``Console.WriteLine(factors(n));` `}` `}`   `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

```1
3```

Time complexity: O(log(n))
Auxiliary space: O(n) as using unordered_map

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