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 Nvoid 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 Codeint main(){    lli N = 36;    checkFactors(N);    return 0;}

Java

 // Java implementation of the// above approachimport java.util.*; class GFG{ // Function to find the count// of even and odd factors of Nstatic 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 Codepublic 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 Ndef 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 Codeif __name__ == '__main__':         N = 36;     checkFactors(N); # This code is contributed by Princi Singh

C#

 // C# implementation of the// above approachusing System; class GFG{ // Function to find the count// of even and odd factors of Nstatic 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 Codepublic 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 squarebool 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 oddvoid checkFactors(lli N){    if (isPerfectSquare(N))        cout << "Yes" << endl;    else        cout << "No" << endl;} // Driver Codeint main(){    lli N = 36;     checkFactors(N);    return 0;}

Java

 // Java implementation of the above approachclass GFG{     // Function to check if the// number is a perfect squarestatic 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 oddstatic void checkFactors(int x){    if (isPerfectSquare(x))        System.out.print("Yes");    else        System.out.print("No");} // Driver codepublic static void main(String[] args){    int N = 36;     checkFactors(N);}} // This code is contributed by dewantipandeydp

Python3

 # Python3 implementation of the above approachimport math # Function to check if the# number is a perfect squaredef 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 odddef checkFactors(x):    if (isPerfectSquare(x)):        print("Yes");    else:        print("No"); # Driver codeif __name__ == '__main__':    N = 36;     checkFactors(N); # This code is contributed by sapnasingh4991

C#

 // C# implementation of the above approachusing System; class GFG{     // Function to check if the// number is a perfect squarestatic 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 oddstatic void checkFactors(int x){    if (isPerfectSquare(x))        Console.Write("Yes");    else        Console.Write("No");} // Driver codepublic static void Main(String[] args){    int N = 36;     checkFactors(N);}} // This code is contributed by Amit Katiyar

Javascript


Output:
Yes

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