# Count pairs (i,j) such that (i+j) is divisible by A and B both

Given n, m, A and B. The task is to count the number of pairs of integers (x, y) such that 1 x n and 1 y m and (x+y) mod A and (x+y) mod B both equals to 0.

Examples:

Input: n = 60, m = 90, A = 5, B = 10
Output: 540

Input: n = 225, m = 452, A = 10, B = 15
Output: 3389


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: If (x+y) is divisible by both A and B then basically LCM of A and B is the smallest divisor of (x+y). So we calculate all numbers that is less than or equal to m and divisible by LCM of them and when iterating with the loop then we check if the present number is divisible by LCM of A and B.

Below is the implementation of the above approach:

## C++

 // C++ implementation of above approach  #include  using namespace std;     // Function to find the LCM  int find_LCM(int x, int y)  {      return (x * y) / __gcd(x, y);  }     // Function to count the pairs  int CountPairs(int n, int m, int A, int B)  {      int cnt = 0;      int lcm = find_LCM(A, B);         for (int i = 1; i <= n; i++)          cnt += (m + (i % lcm)) / lcm;         return cnt;  }     // Driver code  int main()  {      int n = 60, m = 90, A = 5, B = 10;         cout << CountPairs(n, m, A, B);         return 0;  }

## Java

 //Java implementation of above approach  import java.util.*;  public class ACE {         static int gcd(int a,int b)      {          return b==0 ? a :gcd(b,a%b);      }             //Function to find the LCM      static int find_LCM(int x, int y)      {       return (x * y) / gcd(x, y);      }         //Function to count the pairs      static int CountPairs(int n, int m, int A, int B)      {       int cnt = 0;       int lcm = find_LCM(A, B);          for (int i = 1; i <= n; i++)           cnt += (m + (i % lcm)) / lcm;          return cnt;      }         //Driver code      public static void main(String[] args) {                     int n = 60, m = 90, A = 5, B = 10;             System.out.println(CountPairs(n, m, A, B));         }     }

## Python 3

 # Python3 implementation of  # above approach     # from math lib import gcd method  from math import gcd     # Function to find the LCM   def find_LCM(x, y) :         return (x * y) // gcd(x, y)     # Function to count the pairs   def CountPairs(n, m, A, B) :         cnt = 0     lcm = find_LCM(A, B)         for i in range(1, n + 1) :          cnt += (m + (i % lcm)) // lcm         return cnt     # Driver code       if __name__ == "__main__" :         n, m, A, B = 60, 90, 5, 10        print(CountPairs(n, m, A, B))     # This code is contributed  # by ANKITRAI1

## C#

 // C# implementation of above approach  using System;     class GFG  {      static int gcd(int a,int b)      {          return b == 0 ? a : gcd(b, a % b);      }             // Function to find the LCM      static int find_LCM(int x, int y)      {      return (x * y) / gcd(x, y);      }         //Function to count the pairs      static int CountPairs(int n, int m,                            int A, int B)      {          int cnt = 0;          int lcm = find_LCM(A, B);                 for (int i = 1; i <= n; i++)              cnt += (m + (i % lcm)) / lcm;                 return cnt;      }         // Driver code      public static void Main()       {          int n = 60, m = 90, A = 5, B = 10;             Console.WriteLine(CountPairs(n, m, A, B));      }  }     // This Code is contributed by mits

## PHP

 

Output:

540


Time Complexity: O(n)

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