Given N numbers and two numbers L and R., the task is to print the count of numbers in the range [L, R] which are divisible by all numbers of the array.
Examples:
Input: a[] = {1, 4, 2], L = 1, R = 10
Output : 2
In range [1, 10], the numbers 4 and 8 are divisible all array elements.Input : a[] = {1, 3, 2], L = 7, R = 11
Output : 0
A naive approach is to iterate from L to R and count the numbers which are divisible by all of the array elements.
Time Complexity: O((R-L) * N)
An efficient approach is to find the LCM of N numbers and then count the numbers that are divisible by LCM in range L and R. The numbers divisible by LCM till R are R/LCM. So using exclusion principle, the count will be (R/LCM – L-1/LCM).
Below is the implementation of the above approach.
C++
// C++ program to count numbers in a range // that are divisible by all array elements #include <bits/stdc++.h> using namespace std; // Function to find the lcm of array int findLCM(int arr[], int n) { int lcm = arr[0]; // Iterate in the array for (int i = 1; i < n; i++) { // Find lcm lcm = (lcm * arr[i]) / __gcd(arr[i], lcm); } return lcm; } // Function to return the count of numbers int countNumbers(int arr[], int n, int l, int r) { // Function call to find the // LCM of N numbers int lcm = findLCM(arr, n); // Return the count of numbers int count = (r / lcm) - ((l - 1) / lcm); } // Driver Code int main() { int arr[] = { 1, 4, 2 }; int n = sizeof(arr) / sizeof(arr[0]); int l = 1, r = 10; cout << countNumbers(arr, n, l, r); return 0; } |
chevron_right
filter_none
Java
// Java program to count numbers in a range // that are divisible by all array elements class GFG { // Function to calculate gcd static int __gcd(int a, int b) { // Everything divides 0 if (a == 0 || b == 0) return 0; // base case if (a == b) return a; // a is greater if (a > b) return __gcd(a - b, b); return __gcd(a, b - a); } // Function to find the lcm of array static int findLCM(int arr[], int n) { int lcm = arr[0]; // Iterate in the array for (int i = 1; i < n; i++) { // Find lcm lcm = (lcm * arr[i]) / __gcd(arr[i], lcm); } return lcm; } // Function to return the count of numbers static int countNumbers(int arr[], int n, int l, int r) { // Function call to find the // LCM of N numbers int lcm = findLCM(arr, n); // Return the count of numbers int count = (r / lcm) - ((l - 1) / lcm); return count; } // Driver Code public static void main(String args[]) { int arr[] = { 1, 4, 2 }; int n = arr.length; int l = 1, r = 10; System.out.println(countNumbers(arr, n, l, r)); } } // This code is contributed by Mukul Singh |
chevron_right
filter_none
Python3
# Python program to count numbers in # a range that are divisible by all # array elements import math # Function to find the lcm of array def findLCM(arr, n): lcm = arr[0]; # Iterate in the array for i in range(1, n - 1): # Find lcm lcm = (lcm * arr[i]) / math.gcd(arr[i], lcm); return lcm; # Function to return the count of numbers def countNumbers(arr, n, l, r): # Function call to find the # LCM of N numbers lcm = int(findLCM(arr, n)); # Return the count of numbers count = (r / lcm) - ((l - 1) / lcm); print(int(count)); # Driver Code arr = [1, 4, 2]; n = len(arr); l = 1; r = 10; countNumbers(arr, n, l, r); # This code is contributed # by Shivi_Aggarwal |
chevron_right
filter_none
C#
// C# program to count numbers in a range // that are divisible by all array elements using System; class GFG { // Function to calculate gcd static int __gcd(int a, int b) { // Everything divides 0 if (a == 0 || b == 0) return 0; // base case if (a == b) return a; // a is greater if (a > b) return __gcd(a - b, b); return __gcd(a, b - a); } // Function to find the lcm of array static int findLCM(int []arr, int n) { int lcm = arr[0]; // Iterate in the array for (int i = 1; i < n; i++) { // Find lcm lcm = (lcm * arr[i]) / __gcd(arr[i], lcm); } return lcm; } // Function to return the count of numbers static int countNumbers(int []arr, int n, int l, int r) { // Function call to find the // LCM of N numbers int lcm = findLCM(arr, n); // Return the count of numbers int count = (r / lcm) - ((l - 1) / lcm); return count; } // Driver Code public static void Main() { int []arr = { 1, 4, 2 }; int n = arr.Length; int l = 1, r = 10; Console.WriteLine(countNumbers(arr, n, l, r)); } } // This code is contributed by Ryuga |
chevron_right
filter_none
PHP
<?php // PHP program to count numbers in a range // that are divisible by all array elements // Function to calculate gcd function __gcd($a, $b) { // Everything divides 0 if ($a == 0 || $b == 0) return 0; // base case if ($a == $b) return $a; // a is greater if ($a > $b) return __gcd($a - $b, $b); return __gcd($a, $b - $a); } // Function to find the lcm of array function findLCM($arr, $n) { $lcm = $arr[0]; // Iterate in the array for ($i = 1; $i < $n; $i++) { // Find lcm $lcm = ($lcm * $arr[$i]) / __gcd($arr[$i], $lcm); } return $lcm; } // Function to return the count of numbers function countNumbers($arr, $n, $l, $r) { // Function call to find the // LCM of N numbers $lcm = findLCM($arr, $n); // Return the count of numbers $count = (int)($r / $lcm) - (int)(($l - 1) / $lcm); return $count; } // Driver Code $arr = array(1, 4, 2); $n = sizeof($arr); $l = 1; $r = 10; echo countNumbers($arr, $n, $l, $r); // This code is contributed // by Akanksha Rai ?> |
chevron_right
filter_none
Output:
2
Recommended Posts:
- Minimum elements to be added in a range so that count of elements is divisible by K
- Count numbers in range 1 to N which are divisible by X but not by Y
- Count of numbers in a range that does not contain the digit M and which is divisible by M.
- Count of Numbers in a Range divisible by m and having digit d in even positions
- Count numbers divisible by K in a range with Fibonacci digit sum for Q queries
- Count elements which divide all numbers in range L-R
- Count of elements not divisible by any other elements of Array
- Count the number of elements in an array which are divisible by k
- Count of pairs of Array elements which are divisible by K when concatenated
- Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M
- Count numbers in a range with digit sum divisible by K having first and last digit different
- Queries for GCD of all numbers of an array except elements in a given range
- Queries for count of array elements with values in given range with updates
- Array Range Queries to count Powerful numbers with updates
- Array range queries to count the number of Fibonacci numbers with updates
- Count of distinct numbers in an Array in a range for Online Queries using Merge Sort Tree
- No of pairs (a[j] >= a[i]) with k numbers in range (a[i], a[j]) that are divisible by x
- Numbers that are not divisible by any number in the range [2, 10]
- Range Queries to count elements lying in a given Range : MO's Algorithm
- Count integers in a range which are divisible by their euler totient value
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
