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Program to find count of numbers having odd number of divisors in given range
• Difficulty Level : Easy
• Last Updated : 12 May, 2021

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```

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++

 `// 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

 `// 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

 `# 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#

 `// 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));``    ``}``}`

## PHP

 ``

## Javascript

 ``
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++

 `// 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

 `// 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

 `# 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#

 `// 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`

## Javascript

 ``
Output:
`3`

Time complexity: O(n * logn)