# Check if a number has an odd count of odd divisors and even count of even divisors

• Last Updated : 26 Mar, 2021

Given an integer N, the task is to check if N has an odd number of odd divisors and even number of even divisors.

Examples:

Input: N = 36
Output:  Yes
Explanation:
Divisors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36
Count of Odd Divisors(1, 3, 9) = 3 [Odd]
Count of Even Divisors(2, 4, 6, 12, 18, 36) = 6 [Even]

Input:  N  =  28
Output:  No

Naive Approach: The idea is to find the factors of the number N and count the odd factors of N and even factors of N. Finally, check if the count of odd factors is odd and count of even factors is even.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the``// above approach` `#include ``using` `namespace` `std;` `#define lli long long int` `// Function to find the count``// of even and odd factors of N``void` `checkFactors(lli N)``{``    ``lli ev_count = 0, od_count = 0;` `    ``// Loop runs till square root``    ``for` `(lli i = 1;``         ``i <= ``sqrt``(N) + 1; i++) {``        ``if` `(N % i == 0) {``            ``if` `(i == N / i) {``                ``if` `(i % 2 == 0)``                    ``ev_count += 1;``                ``else``                    ``od_count += 1;``            ``}``            ``else` `{``                ``if` `(i % 2 == 0)``                    ``ev_count += 1;``                ``else``                    ``od_count += 1;``                ``if` `((N / i) % 2 == 0)``                    ``ev_count += 1;``                ``else``                    ``od_count += 1;``            ``}``        ``}``    ``}` `    ``// Condition to check if the even``    ``// factors of the number N is``    ``// is even and count of``    ``// odd factors is odd``    ``if` `(ev_count % 2 == 0``        ``&& od_count % 2 == 1)``        ``cout << ``"Yes"` `<< endl;``    ``else``        ``cout << ``"No"` `<< endl;``}` `// Driver Code``int` `main()``{``    ``lli N = 36;``    ``checkFactors(N);``    ``return` `0;``}`

## Java

 `// Java implementation of the``// above approach``import` `java.util.*;` `class` `GFG{` `// Function to find the count``// of even and odd factors of N``static` `void` `checkFactors(``long` `N)``{``    ``long` `ev_count = ``0``, od_count = ``0``;` `    ``// Loop runs till square root``    ``for``(``long` `i = ``1``;``             ``i <= Math.sqrt(N) + ``1``; i++)``    ``{``        ``if` `(N % i == ``0``)``        ``{``            ``if` `(i == N / i)``            ``{``                ``if` `(i % ``2` `== ``0``)``                    ``ev_count += ``1``;``                ``else``                    ``od_count += ``1``;``            ``}``            ``else``            ``{``                ``if` `(i % ``2` `== ``0``)``                    ``ev_count += ``1``;``                ``else``                    ``od_count += ``1``;``                ``if` `((N / i) % ``2` `== ``0``)``                    ``ev_count += ``1``;``                ``else``                    ``od_count += ``1``;``            ``}``        ``}``    ``}` `    ``// Condition to check if the even``    ``// factors of the number N is``    ``// is even and count of``    ``// odd factors is odd``    ``if` `(ev_count % ``2` `== ``0` `&& od_count % ``2` `== ``1``)``        ``System.out.print(``"Yes"` `+ ``"\n"``);``    ``else``        ``System.out.print(``"No"` `+ ``"\n"``);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``long` `N = ``36``;``    ` `    ``checkFactors(N);``}``}` `// This code is contributed by amal kumar choubey`

## Python3

 `# Python3 implementation of the``# above approach` `# Function to find the count``# of even and odd factors of N``def` `checkFactors(N):``    ` `    ``ev_count ``=` `0``; od_count ``=` `0``;` `    ``# Loop runs till square root``    ``for` `i ``in` `range``(``1``, ``int``(``pow``(N, ``1` `/` `2``)) ``+` `1``):``        ``if` `(N ``%` `i ``=``=` `0``):``            ``if` `(i ``=``=` `N ``/` `i):``                ` `                ``if` `(i ``%` `2` `=``=` `0``):``                    ``ev_count ``+``=` `1``;``                ``else``:``                    ``od_count ``+``=` `1``;``                    ` `            ``else``:``                ``if` `(i ``%` `2` `=``=` `0``):``                    ``ev_count ``+``=` `1``;``                ``else``:``                    ``od_count ``+``=` `1``;``                ``if` `((N ``/` `i) ``%` `2` `=``=` `0``):``                    ``ev_count ``+``=` `1``;``                ``else``:``                    ``od_count ``+``=` `1``;``            ` `    ``# Condition to check if the even``    ``# factors of the number N is``    ``# is even and count of``    ``# odd factors is odd``    ``if` `(ev_count ``%` `2` `=``=` `0` `and``        ``od_count ``%` `2` `=``=` `1``):``        ``print``(``"Yes"` `+` `"");``    ``else``:``        ``print``(``"No"` `+` `"");` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``N ``=` `36``;` `    ``checkFactors(N);` `# This code is contributed by Princi Singh`

## C#

 `// C# implementation of the``// above approach``using` `System;` `class` `GFG{` `// Function to find the count``// of even and odd factors of N``static` `void` `checkFactors(``long` `N)``{``    ``long` `ev_count = 0, od_count = 0;` `    ``// Loop runs till square root``    ``for``(``long` `i = 1;``             ``i <= Math.Sqrt(N) + 1; i++)``    ``{``        ``if` `(N % i == 0)``        ``{``            ``if` `(i == N / i)``            ``{``                ``if` `(i % 2 == 0)``                    ``ev_count += 1;``                ``else``                    ``od_count += 1;``            ``}``            ``else``            ``{``                ``if` `(i % 2 == 0)``                    ``ev_count += 1;``                ``else``                    ``od_count += 1;``                ``if` `((N / i) % 2 == 0)``                    ``ev_count += 1;``                ``else``                    ``od_count += 1;``            ``}``        ``}``    ``}` `    ``// Condition to check if the even``    ``// factors of the number N is``    ``// is even and count of``    ``// odd factors is odd``    ``if` `(ev_count % 2 == 0 && od_count % 2 == 1)``        ``Console.Write(``"Yes"` `+ ``"\n"``);``    ``else``        ``Console.Write(``"No"` `+ ``"\n"``);``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``long` `N = 36;``    ` `    ``checkFactors(N);``}``}` `// This code is contributed by Amit Katiyar `

## Javascript

 ``
Output:
`Yes`

Efficient Approach:  The key observation in the problem is that the number of odd divisors is odd and number of even divisors is even only in case of perfect squares. Hence, the best solution would be to check if the given number is a perfect square or not. If it’s a perfect square, then print “Yes” else print “No”.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the``// above approach` `#include ``using` `namespace` `std;` `#define lli long long int` `// Function to check if the``// number is a perfect square``bool` `isPerfectSquare(``long` `double` `x)``{``    ``long` `double` `sr = ``sqrt``(x);` `    ``// If square root is an integer``    ``return` `((sr - ``floor``(sr)) == 0);``}` `// Function to check``// if count of even divisors is even``// and count of odd divisors is odd``void` `checkFactors(lli N)``{``    ``if` `(isPerfectSquare(N))``        ``cout << ``"Yes"` `<< endl;``    ``else``        ``cout << ``"No"` `<< endl;``}` `// Driver Code``int` `main()``{``    ``lli N = 36;` `    ``checkFactors(N);``    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG{``    ` `// Function to check if the``// number is a perfect square``static` `boolean` `isPerfectSquare(``double` `x)``{``    ` `    ``// Find floating point value of``    ``// square root of x.``    ``double` `sr = Math.sqrt(x);` `    ``// If square root is an integer``    ``return` `((sr - Math.floor(sr)) == ``0``);``}` `// Function to check if count of``// even divisors is even and count``// of odd divisors is odd``static` `void` `checkFactors(``int` `x)``{``    ``if` `(isPerfectSquare(x))``        ``System.out.print(``"Yes"``);``    ``else``        ``System.out.print(``"No"``);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``36``;` `    ``checkFactors(N);``}``}` `// This code is contributed by dewantipandeydp`

## Python3

 `# Python3 implementation of the above approach``import` `math` `# Function to check if the``# number is a perfect square``def` `isPerfectSquare(x):` `    ``# Find floating povalue of``    ``# square root of x.``    ``sr ``=` `pow``(x, ``1` `/` `2``);` `    ``# If square root is an integer``    ``return` `((sr ``-` `math.floor(sr)) ``=``=` `0``);` `# Function to check if count of``# even divisors is even and count``# of odd divisors is odd``def` `checkFactors(x):``    ``if` `(isPerfectSquare(x)):``        ``print``(``"Yes"``);``    ``else``:``        ``print``(``"No"``);` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``N ``=` `36``;` `    ``checkFactors(N);` `# This code is contributed by sapnasingh4991`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG{``    ` `// Function to check if the``// number is a perfect square``static` `bool` `isPerfectSquare(``double` `x)``{``    ` `    ``// Find floating point value of``    ``// square root of x.``    ``double` `sr = Math.Sqrt(x);` `    ``// If square root is an integer``    ``return` `((sr - Math.Floor(sr)) == 0);``}` `// Function to check if count of``// even divisors is even and count``// of odd divisors is odd``static` `void` `checkFactors(``int` `x)``{``    ``if` `(isPerfectSquare(x))``        ``Console.Write(``"Yes"``);``    ``else``        ``Console.Write(``"No"``);``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `N = 36;` `    ``checkFactors(N);``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``
Output:
`Yes`

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