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: 540Input: n = 225, m = 452, A = 10, B = 15
Output: 3389
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 <bits/stdc++.h>using namespace std;// Function to find the LCMint find_LCM(int x, int y){ return (x * y) / __gcd(x, y);}// Function to count the pairsint 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 codeint main(){ int n = 60, m = 90, A = 5, B = 10; cout << CountPairs(n, m, A, B); return 0;} |
Java
//Java implementation of above approachimport 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 methodfrom math import gcd# Function to find the LCMdef find_LCM(x, y) : return (x * y) // gcd(x, y)# Function to count the pairsdef 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 approachusing 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
<?php// PHP implementation of above approachfunction gcd($a, $b){ return $b == 0 ? $a : gcd($b, $a % $b);}// Function to find the LCMfunction find_LCM($x, $y){ return (int)(($x * $y) / gcd($x, $y));}// Function to count the pairsfunction CountPairs($n, $m, $A, $B){ $cnt = 0; $lcm = find_LCM($A, $B); for ($i = 1; $i <= $n; $i++) $cnt += (int)(($m + ($i % $lcm)) / $lcm); return $cnt;}// Driver code$n = 60; $m = 90; $A = 5; $B = 10;echo CountPairs($n, $m, $A, $B);// This code is contributed// by Akanksha Rai?> |
Javascript
<script>//Javascript implementation of above approach function gcd(a,b) { return b==0 ? a :gcd(b,a%b); } //Function to find the LCM function find_LCM(x,y) { return Math.floor((x * y) / gcd(x, y)); } //Function to count the pairs function CountPairs(n,m,A,B) { let cnt = 0; let lcm = find_LCM(A, B); for (let i = 1; i <= n; i++) cnt += Math.floor((m + (i % lcm)) / lcm); return cnt; } //Driver code let n = 60, m = 90, A = 5, B = 10; document.write(CountPairs(n, m, A, B)); // This code is contributed by rag2127</script> |
Output:
540
Time Complexity: O(n) for iterating from 1 till n.
Auxiliary Space: O(1) as no extra space is used.
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