Given two numbers a and b, and a number k which is odd. The task is to find all the numbers between a and b (both inclusive) having exactly k divisors.
Examples:
Input : a = 2, b = 49, k = 3 Output: 4 // Between 2 and 49 there are four numbers // with three divisors // 4 (Divisors 1, 2, 4), 9 (Divisors 1, 3, 9), // 25 (Divisors 1, 5, 25} and 49 (1, 7 and 49) Input : a = 1, b = 100, k = 9 Output: 2 // between 1 and 100 there are 36 (1, 2, 3, 4, 6, 9, 12, 18, 36) // and 100 (1, 2, 4, 5, 10, 20, 25, 50, 100) having exactly 9 // divisors
This problem has simple solution, here we are given that k is odd and we know that only perfect square numbers have odd number of divisors , so we just need to check all perfect square numbers between a and b, and calculate divisors of only those perfect square numbers.
C++
// C++ program to count numbers with k odd// divisors in a range.#include<bits/stdc++.h>using namespace std;// Utility function to check if number is// perfect square or notbool isPerfect(int n){ int s = sqrt(n); return (s*s == n);}// Utility Function to return count of divisors// of a numberint divisorsCount(int n){ // Note that this loop runs till square root int count=0; for (int i=1; i<=sqrt(n)+1; i++) { if (n%i==0) { // If divisors are equal, count it // only once if (n/i == i) count += 1; // Otherwise print both else count += 2; } } return count;}// Function to calculate all divisors having// exactly k divisors between a and bint kDivisors(int a,int b,int k){ int count = 0; // Initialize result // calculate only for perfect square numbers for (int i=a; i<=b; i++) { // check if number is perfect square or not if (isPerfect(i)) // total divisors of number equals to // k or not if (divisors(i) == k) count++; } return count;}// Driver program to run the caseint main(){ int a = 2, b = 49, k = 3; cout << kDivisors(a, b, k); return 0;} |
Java
// Java program to count numbers// with k odd divisors in a range.import java.io.*;import java.math.*;class GFG { // Utility function to check if // number is perfect square or not static boolean isPerfect(int n) { int s = (int)(Math.sqrt(n)); return (s*s == n); } // Utility Function to return // count of divisors of a number static int divisorsCount(int n) { // Note that this loop // runs till square root int count=0; for (int i = 1; i <= Math.sqrt(n) + 1; i++) { if (n % i == 0) { // If divisors are equal, // count it only once if (n / i == i) count += 1; // Otherwise print both else count += 2; } } return count; } // Function to calculate all // divisors having exactly k // divisors between a and b static int kDivisors(int a,int b,int k) { // Initialize result int count = 0; // calculate only for // perfect square numbers for (int i = a; i <= b; i++) { // check if number is // perfect square or not if (isPerfect(i)) // total divisors of number // equals to k or not if (divisorsCount(i) == k) count++; } return count; } // Driver program to run the case public static void main(String args[]) { int a = 21, b = 149, k = 333; System.out.println(kDivisors(a, b, k)); }}// This code is contributed by Nikita Tiwari. |
Python3
# Python3 program to count numbers# with k odd divisors in a range.import math# Utility function to check if number # is perfect square or notdef isPerfect(n) : s = math.sqrt(n) return (s * s == n)# Utility Function to return # count of divisors of a numberdef divisorsCount(n) : # Note that this loop runs till # square root count = 0 for i in range(1, (int)(math.sqrt(n) + 2)) : if (n % i == 0) : # If divisors are equal, # count it only once if (n // i == i) : count = count + 1 # Otherwise print both else : count = count + 2 return count # Function to calculate all divisors having# exactly k divisors between a and bdef kDivisors(a, b, k) : count = 0 # Initialize result # calculate only for perfect square numbers for i in range(a, b + 1) : # check if number is perfect square or not if (isPerfect(i)) : # total divisors of number equals to # k or not if (divisorsCount(i) == k) : count = count + 1 return count# Driver program to run the casea = 2b = 49k = 3print(kDivisors(a, b, k))# This code is contributed by Nikita Tiwari. |
C#
// C# program to count numbers with// k odd divisors in a range.using System;class GFG { // Utility function to check if number // is perfect square or not static bool isPerfect(int n) { int s = (int)(Math.Sqrt(n)); return (s * s == n); } // Utility Function to return // count of divisors of a number static int divisorsCount(int n) { // Note that this loop // runs till square root int count=0; for (int i = 1; i <= Math.Sqrt(n) + 1; i++) { if (n % i == 0) { // If divisors are equal, // count it only once if (n / i == i) count += 1; // Otherwise print both else count += 2; } } return count; } // Function to calculate all // divisors having exactly k // divisors between a and b static int kDivisors(int a, int b, int k) { // Initialize result int count = 0; // calculate only for // perfect square numbers for (int i = a; i <= b; i++) { // check if number is // perfect square or not if (isPerfect(i)) // total divisors of number // equals to k or not if (divisorsCount(i) == k) count++; } return count; } // Driver Code public static void Main(String []args) { int a = 21, b = 149, k = 333; Console.Write(kDivisors(a, b, k)); }}// This code is contributed by Nitin Mittal. |
PHP
<?php// PHP program to count numbers// with k odd divisors in a range.// function to check if number is// perfect square or notfunction isPerfect($n){ $s = sqrt($n); return ($s * $s == $n);}// Function to return count// of divisors of a numberfunction divisorsCount($n){ // Note that this loop // runs till square root $count = 0; for ($i = 1; $i <= sqrt($n) + 1; $i++) { if ($n % $i == 0) { // If divisors are equal, // count it only once if ($n / $i == $i) $count += 1; // Otherwise print both else $count += 2; } } return $count;}// Function to calculate // all divisors having// exactly k divisors // between a and bfunction kDivisors($a, $b, $k){ // Initialize result $count = 0; // calculate only for // perfect square numbers for ($i = $a; $i <= $b; $i++) { // check if number is // perfect square or not if (isPerfect($i)) // total divisors of // number equals to // k or not if (divisorsCount($i) == $k) $count++; } return $count;} // Driver Code $a = 2; $b = 49; $k = 3; echo kDivisors($a, $b, $k); // This code is contributed by nitin mittal. ?> |
Javascript
<script>// JavaScript program to count numbers with// k odd divisors in a range.// Utility function to check if number // is perfect square or notfunction isPerfect(n){ var s = parseInt((Math.sqrt(n))); return (s * s == n);}// Utility Function to return // count of divisors of a numberfunction divisorsCount(n){ // Note that this loop // runs till square root var count=0; for (var i = 1; i <= parseInt(Math.sqrt(n)) + 1; i++){ if (n % i == 0) { // If divisors are equal, // count it only once if (parseInt(n / i) == i) count += 1; // Otherwise print both else count += 2; }} return count;}// Function to calculate all // divisors having exactly k // divisors between a and bfunction kDivisors(a, b, k){ // Initialize result var count = 0; // calculate only for // perfect square numbers for(var i = a; i <= b; i++) { // check if number is // perfect square or not if (isPerfect(i)) { // total divisors of number // equals to k or not if (divisorsCount(i)==k) { count++; } } } return count;}// Driver Codevar a = 2, b = 49, k = 3;document.write(kDivisors(a, b, k));</script> |
Output:
4
Time Complexity: O(nsqrtn) , where n is the range of a and b
Auxiliary Space: O(1)
This problem can be solved more efficiently. Please refer method 2 of below post for an efficient solution.
Number of perfect squares between two given numbers
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