# Count total divisors of A or B in a given range

• Difficulty Level : Easy
• Last Updated : 21 May, 2021

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.

1. Divide n by a to obtain total count of all numbers(1 to n) divisible by ‘a’.

2. Divide m-1 by a to obtain total count of all numbers(1 to m-1) divisible by ‘a’.

3. 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’.

1. Find LCM of ‘a’ and ‘b’.

2. Divide n by LCM to obtain the count of numbers(1 to n) divisible by both ‘a’ and ‘b’.

3. Divide m-1 by LCM to obtain the count of numbers(1 to m-1) divisible by both ‘a’ and ‘b’.

4. 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 ``using` `namespace` `std;` `// Utility function to find LCM of two numbers``int` `FindLCM(``int` `a, ``int` `b)``{``    ``return` `(a * b) / __gcd(a, b);``}` `// Function to calculate all divisors in given range``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 code``int` `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 range` `import` `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``# numbers``def` `FindLCM(a, b):``    ``return` `(a ``*` `b) ``/` `__gcd(a, b);`  `# Function to calculate all divisors in``# given range``def` `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 code``m ``=` `3``n ``=` `11``a ``=` `2``b ``=` `3``;``print``(rangeDivisor(m, n, a, b))``m ``=` `11``n ``=` `1000000``a ``=` `6``b ``=` `35``print``(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

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## Javascript

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Output:

```6
190475```

Time complexity: O(log(MAX(a, b))
Auxiliary space: O(1)
This article is contributed by Shubham Bansal. If you like GeeksforGeeks 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 GeeksforGeeks main page and help other Geeks.

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