# Program to find count of numbers having odd number of divisors in given range

Given two integers A and B. The task is to count how many numbers in the interval [ A, B ] have an odd number of divisors.

Examples:

Input : A = 1, B = 10
Output : 3

Input : A = 5, B = 15
Output : 1

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

Naive Approach :
The simple approach would be to iterate through all the numbers between range [A, B] and check if their number of divisors is odd.
Below is the implementation of the above idea :

 // C++ program to find count of numbers having // odd number of divisors in given range    #include using namespace std;    // Function to count numbers having odd // number of divisors in range [A, B] int OddDivCount(int a, int b) {     // variable to odd divisor count     int res = 0;     // iterate from a to b and count their     // number of divisors     for (int i = a; i <= b; ++i) {            // variable to divisor count         int divCount = 0;         for (int j = 1; j <= i; ++j) {             if (i % j == 0) {                 ++divCount;             }         }            // if count of divisor is odd         // then increase res by 1         if (divCount % 2) {             ++res;         }     }     return res; }    // Driver code int main() {     int a = 1, b = 10;     cout << OddDivCount(a, b) << endl;        return 0; }

 // Java program to find count of numbers having // odd number of divisors in given range    import java.io.*;    class GFG {     // Function to count numbers having odd     // number of divisors in range [A, B]     static int OddDivCount(int a, int b)     {         // variable to odd divisor count         int res = 0;         // iterate from a to b and count their         // number of divisors         for (int i = a; i <= b; ++i) {                // variable to divisor count             int divCount = 0;             for (int j = 1; j <= i; ++j) {                 if (i % j == 0) {                     ++divCount;                 }             }                // if count of divisor is odd             // then increase res by 1             if ((divCount % 2) != 0) {                 ++res;             }         }         return res;     }        // Driver code        public static void main(String[] args)     {            int a = 1, b = 10;         System.out.println(OddDivCount(a, b));     }     // This code is contributed by ajit. }

 # Python3 program to find count  # of numbers having odd number # of divisors in given range    # Function to count numbers  # having odd number of divisors  # in range [A, B] def OddDivCount(a, b):        # variable to odd divisor count     res = 0            # iterate from a to b and count      # their number of divisors     for i in range(a, b + 1) :            # variable to divisor count         divCount = 0         for j in range(1, i + 1) :             if (i % j == 0) :                 divCount += 1            # if count of divisor is odd         # then increase res by 1         if (divCount % 2) :             res += 1     return res    # Driver code if __name__ == "__main__":     a = 1     b = 10     print(OddDivCount(a, b))    # This code is contributed  # by ChitraNayal

 // C# program to find count of numbers having // odd number of divisors in given range using System;    class Geeks {        // Function to count numbers having odd     // number of divisors in range [A, B]     static int OddDivCount(int a, int b)     {         // variable to odd divisor count         int res = 0;         // iterate from a to b and count their         // number of divisors         for (int i = a; i <= b; ++i) {                // variable to divisor count             int divCount = 0;             for (int j = 1; j <= i; ++j) {                 if (i % j == 0) {                     ++divCount;                 }             }                // if count of divisor is odd             // then increase res by 1             if ((divCount % 2) != 0) {                 ++res;             }         }         return res;     }        // Driver code     public static void Main(String[] args)     {         int a = 1, b = 10;         Console.WriteLine(OddDivCount(a, b));     } }



Output:
3

Time complexity: O(n2)

Better Approach:
A number can be represented by the product of its prime factors with appropriate powers. Those powers can be used to get the number of factors an integer has. If the number is num and it can be represented as (ap1) * (bp2) * (cp3)
Then the count of factors of num are (p1 + 1) * (p2 + 1) * (p3 + 1)

Below is the implementation of the above approach:

 // C++ implementation of the approach #include using namespace std;    // Function to return the count // of divisors of a number int divisor(int a) {     int div = 1, count = 0;     for (int i = 2; i <= sqrt(a); i++) {            // Count the powers of the current         // prime i which divides a         while (a % i == 0) {             count++;             a = a / i;         }            // Update the count of divisors         div = div * (count + 1);            // Reset the count         count = 0;     }        // If the remaining a is prime then a^1     // will be one of its prime factors     if (a > 1) {         div = div * (2);     }     return div; }    // Function to count numbers having odd // number of divisors in range [A, B] int OddDivCount(int a, int b) {     // To store the count of elements     // having odd number of divisors     int res = 0;        // Iterate from a to b and find the     // count of their divisors     for (int i = a; i <= b; ++i) {            // To store the count of divisors of i         int divCount = divisor(i);            // If the divisor count of i is odd         if (divCount % 2) {             ++res;         }     }     return res; }    // Driver code int main() {     int a = 1, b = 10;     cout << OddDivCount(a, b);        return 0; } // This code is contributed by jrolofmeister

 // Java implementation of the approach import java.util.*;    class GFG {    // Function to return the count // of divisors of a number static int divisor(int a) {     int div = 1, count = 0;     for (int i = 2; i <= Math.sqrt(a); i++)      {                    // Count the powers of the current         // prime i which divides a         while (a % i == 0)          {             count++;             a = a / i;         }            // Update the count of divisors         div = div * (count + 1);            // Reset the count         count = 0;     }        // If the remaining a is prime then a^1     // will be one of its prime factors     if (a > 1)      {         div = div * (2);     }     return div; }    // Function to count numbers having odd // number of divisors in range [A, B] static int OddDivCount(int a, int b) {     // To store the count of elements     // having odd number of divisors     int res = 0;        // Iterate from a to b and find the     // count of their divisors     for (int i = a; i <= b; ++i)     {            // To store the count of divisors of i         int divCount = divisor(i);            // If the divisor count of i is odd         if (divCount % 2 == 1)          {             ++res;         }     }     return res; }    // Driver code public static void main(String[] args)  {     int a = 1, b = 10;     System.out.println(OddDivCount(a, b)); } }     // This code is contributed by PrinciRaj1992

 # Python3 implementation of the approach    # Function to return the count # of divisors of a number def divisor(a):        div = 1;     count = 0;     for i in range(2, int(pow(a, 1 / 2)) + 1):            # Count the powers of the current         # prime i which divides a         while (a % i == 0):             count += 1;             a = a / i;                        # Update the count of divisors         div = div * (count + 1);            # Reset the count         count = 0;            # If the remaining a is prime then a^1     # will be one of its prime factors     if (a > 1):          div = div * (2);            return div;    # Function to count numbers having odd # number of divisors in range [A, B] def OddDivCount(a, b):        # To store the count of elements     # having odd number of divisors     res = 0;        # Iterate from a to b and find the     # count of their divisors     for i in range(a, b + 1):                    # To store the count of divisors of i         divCount = divisor(i);            # If the divisor count of i is odd         if (divCount % 2):              res += 1;        return res;    # Driver code if __name__ == '__main__':     a, b = 1, 10;     print(OddDivCount(a, b));    # This code is contributed by PrinciRaj1992

 // C# implementation of the approach using System; using System.Collections.Generic;        class GFG {    // Function to return the count // of divisors of a number static int divisor(int a) {     int div = 1, count = 0;     for (int i = 2;               i <= Math.Sqrt(a); i++)      {                    // Count the powers of the current         // prime i which divides a         while (a % i == 0)          {             count++;             a = a / i;         }            // Update the count of divisors         div = div * (count + 1);            // Reset the count         count = 0;     }        // If the remaining a is prime then a^1     // will be one of its prime factors     if (a > 1)      {         div = div * (2);     }     return div; }    // Function to count numbers having odd // number of divisors in range [A, B] static int OddDivCount(int a, int b) {     // To store the count of elements     // having odd number of divisors     int res = 0;        // Iterate from a to b and find the     // count of their divisors     for (int i = a; i <= b; ++i)     {            // To store the count of divisors of i         int divCount = divisor(i);            // If the divisor count of i is odd         if (divCount % 2 == 1)          {             ++res;         }     }     return res; }    // Driver code public static void Main(String[] args)  {     int a = 1, b = 10;     Console.WriteLine(OddDivCount(a, b)); } }     // This code is contributed by Princi Singh

Output:
3

Time complexity: O(n * logn)

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