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Program to find count of numbers having odd number of divisors in given range
  • Difficulty Level : Easy
  • Last Updated : 10 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 <bits/stdc++.h>
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




<?php
// PHP 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]
function OddDivCount($a, $b)
{
    // variable to odd divisor count
    $res = 0;
     
    // iterate from a to b and count
    // their number of divisors
    for ($i = $a; $i <= $b; ++$i)
    {
 
        // variable to divisor count
        $divCount = 0;
        for ($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
$a = 1;
$b = 10;
echo OddDivCount($a, $b) ;
 
// This code is contributed
// by Shivi_Aggarwal
?>

Javascript




<script>
 
// Javascript 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]
function OddDivCount(a, b)
{
     
    // Variable to odd divisor count
    let res = 0;
     
    // Iterate from a to b and count their
    // number of divisors
    for(let i = a; i <= b; ++i)
    {
         
        // Variable to divisor count
        let divCount = 0;
        for(let 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
let a = 1, b = 10;
document.write(OddDivCount(a, b));
 
// This code is contributed by suresh07
 
</script>
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 <bits/stdc++.h>
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
Output: 
3

 

Time complexity: O(n * logn)
Please refer this article for an O(1) approach.
 

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