Given four integers m, n, a, b. Find how many integers from range m to n are divisible by a or b.
Examples :
Input: 3 11 2 3
Output: 6
Explanation:
m = 3, n = 11, a = 2, b = 3
There are total 6 numbers from 3 to
11 which are divisible by 2 or 3 i.e,
3, 4, 6, 8, 9, 10
Input: arr[] = {11, 1000000, 6, 35}
Output: 190475
A Naive approach is to run a loop from m to n and count all numbers which are divisible by either a or b. Time complexity of this approach will be O(m – n) which will definitely time out for large values of m.
An efficient approach is to use simple LCM and division method.
- Divide n by a to obtain total count of all numbers(1 to n) divisible by ‘a’.
- Divide m-1 by a to obtain total count of all numbers(1 to m-1) divisible by ‘a’.
- Subtract the count of step 1 and 2 to obtain total divisors in range m to n.
Now we have a total number of divisors of ‘a’ in given range. Repeat the above to count total divisors of ‘b’.
Add these to obtain total count of divisors ‘a’ and ‘b’.
But the number divisible by both a and b counted twice. Therefore, to remove this ambiguity we can use LCM of a and b to count total number divisible by both ‘a’ and ‘b’.
- Find LCM of ‘a’ and ‘b’.
- Divide n by LCM to obtain the count of numbers(1 to n) divisible by both ‘a’ and ‘b’.
- Divide m-1 by LCM to obtain the count of numbers(1 to m-1) divisible by both ‘a’ and ‘b’.
- Subtract the count of steps 2 and 3 to obtain total divisors of both ‘a’ and ‘b’.
Now subtract this result from the previous calculated result to find total count of all unique divisors of ‘a’ or ‘b’.
C++
// C++ program to count total divisors of 'a'// or 'b' in a given range #include <bits/stdc++.h>using namespace std;// Utility function to find LCM of two numbersint FindLCM(int a, int b){ return (a * b) / __gcd(a, b);}// Function to calculate all divisors in given rangeint rangeDivisor(int m, int n, int a, int b){ // Find LCM of a and b int lcm = FindLCM(a, b); int a_divisor = n / a - (m - 1) / a; int b_divisor = n / b - (m - 1) / b; // Find common divisor by using LCM int common_divisor = n / lcm - (m - 1) / lcm; int ans = a_divisor + b_divisor - common_divisor; return ans;}// Driver codeint main(){ int m = 3, n = 11, a = 2, b = 3; cout << rangeDivisor(m, n, a, b) << endl; m = 11, n = 1000000, a = 6, b = 35; cout << rangeDivisor(m, n, a, b); return 0;} |
Java
// Java program to count total divisors of 'a'// or 'b' in a given rangeimport java.math.BigInteger;class Test{ // Utility method to find LCM of two numbers static int FindLCM(int a, int b) { return (a * b) / new BigInteger(a+"").gcd(new BigInteger(b+"")).intValue(); } // method to calculate all divisors in given range static int rangeDivisor(int m, int n, int a, int b) { // Find LCM of a and b int lcm = FindLCM(a, b); int a_divisor = n / a - (m - 1) / a; int b_divisor = n / b - (m - 1) / b; // Find common divisor by using LCM int common_divisor = n / lcm - (m - 1) / lcm; int ans = a_divisor + b_divisor - common_divisor; return ans; } // Driver method public static void main(String args[]) { int m = 3, n = 11, a = 2, b = 3; System.out.println(rangeDivisor(m, n, a, b)); m = 11; n = 1000000 ; a = 6; b = 35; System.out.println(rangeDivisor(m, n, a, b)); }} |
Python3
# python program to count total divisors# of 'a' or 'b' in a given range def __gcd(x, y): if x > y: small = y else: small = x for i in range(1, small+1): if((x % i == 0) and (y % i == 0)): gcd = i return gcd # Utility function to find LCM of two# numbersdef FindLCM(a, b): return (a * b) / __gcd(a, b);# Function to calculate all divisors in# given rangedef rangeDivisor(m, n, a, b): # Find LCM of a and b lcm = FindLCM(a, b) a_divisor = int( n / a - (m - 1) / a) b_divisor = int(n / b - (m - 1) / b) # Find common divisor by using LCM common_divisor =int( n / lcm - (m - 1) / lcm) ans = a_divisor + b_divisor - common_divisor return ans# Driver codem = 3n = 11a = 2b = 3;print(rangeDivisor(m, n, a, b))m = 11n = 1000000a = 6b = 35print(rangeDivisor(m, n, a, b))# This code is contributed by Sam007 |
C#
// C# program to count total divisors// of 'a' or 'b' in a given range using System;class GFG { static int GCD(int num1, int num2) { int Remainder; while (num2 != 0) { Remainder = num1 % num2; num1 = num2; num2 = Remainder; } return num1; } // Utility function to find LCM of // two numbers static int FindLCM(int a, int b) { return (a * b) / GCD(a, b); } // Function to calculate all divisors in given range static int rangeDivisor(int m, int n, int a, int b) { // Find LCM of a and b int lcm = FindLCM(a, b); int a_divisor = n / a - (m - 1) / a; int b_divisor = n / b - (m - 1) / b; // Find common divisor by using LCM int common_divisor = n / lcm - (m - 1) / lcm; int ans = a_divisor + b_divisor - common_divisor; return ans; } public static void Main () { int m = 3, n = 11, a = 2, b = 3; Console.WriteLine(rangeDivisor(m, n, a, b)); m = 11; n = 1000000; a = 6; b = 35; Console.WriteLine(rangeDivisor(m, n, a, b)); }}// This code is contributed by Sam007. |
PHP
<?php// PHP program to count total// divisors of 'a' or 'b' in // a given range function gcd($a, $b) { return ($a % $b) ? gcd($b, $a % $b) : $b;}// Utility function to // find LCM of two numbersfunction FindLCM($a, $b){ return ($a * $b) / gcd($a, $b);}// Function to calculate // all divisors in given rangefunction rangeDivisor($m, $n, $a, $b){ // Find LCM of a and b $lcm = FindLCM($a, $b); $a_divisor = $n / $a - ($m - 1) / $a; $b_divisor = $n / $b - ($m - 1) / $b; // Find common divisor by using LCM $common_divisor = $n / $lcm - ($m - 1) / $lcm; $ans = $a_divisor + $b_divisor - $common_divisor; return $ans;}// Driver Code$m = 3;$n = 11;$a = 2;$b = 3;print(ceil(rangeDivisor($m, $n, $a, $b)));echo "\n";$m = 11;$n = 1000000;$a = 6;$b = 35;print(ceil(rangeDivisor($m, $n,$a,$b)));// This code is contributed by Sam007?> |
Javascript
<script> // JavaScript program to count total divisors // of 'a' or 'b' in a given range function GCD(num1, num2) { let Remainder; while (num2 != 0) { Remainder = num1 % num2; num1 = num2; num2 = Remainder; } return num1; } // Utility function to find LCM of // two numbers function FindLCM(a, b) { return parseInt((a * b) / GCD(a, b), 10); } // Function to calculate all divisors in given range function rangeDivisor(m, n, a, b) { // Find LCM of a and b let lcm = FindLCM(a, b); let a_divisor = parseInt(n / a, 10) - parseInt((m - 1) / a, 10); let b_divisor = parseInt(n / b, 10) - parseInt((m - 1) / b, 10); // Find common divisor by using LCM let common_divisor = parseInt(n / lcm, 10) - parseInt((m - 1) / lcm, 10); let ans = a_divisor + b_divisor - common_divisor; return ans; } let m = 3, n = 11, a = 2, b = 3; document.write(rangeDivisor(m, n, a, b) + "</br>"); m = 11; n = 1000000; a = 6; b = 35; document.write(rangeDivisor(m, n, a, b)); </script> |
Output:
6 190475
Time complexity: O(log(MAX(a, b))
Auxiliary space: O(1)
This article is contributed by Shubham Bansal. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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