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 : 3Input : 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 Beginner { // 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> |
3
Time complexity: O(n2)
Auxiliary Space: O(1)
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 |
Javascript
<script> // Javascript implementation of the approach // Function to return the count // of divisors of a number function divisor(a) { let div = 1, count = 0; for (let i = 2; i <= Math.sqrt(a); i++) { // Count the powers of the current // prime i which divides a while (a % i == 0) { count++; a = parseInt(a / i, 10); } // 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] function OddDivCount(a, b) { // To store the count of elements // having odd number of divisors let res = 0; // Iterate from a to b and find the // count of their divisors for (let i = a; i <= b; ++i) { // To store the count of divisors of i let divCount = divisor(i); // If the divisor count of i is odd if (divCount % 2 == 1) { ++res; } } return res; } let a = 1, b = 10; document.write(OddDivCount(a, b)); </script> |
3
Time complexity: O(n * logn)
Auxiliary Space: O(1)
Please refer this article for an O(1) approach.