Count number of pairs (A <= N, B <= N) such that gcd (A , B) is B
Given a number n. We need find the number of ordered pairs of a and b such gcd(a, b) is b itself
Examples :
Input : n = 2 Output : 3 (1, 1) (2, 2) and (2, 1) Input : n = 3 Output : 5 (1, 1) (2, 2) (3, 3) (2, 1) and (3, 1)
Naive approach : gcd(a, b) = b means b is a factor of a. So total number of pairs will be equal to sum of divisors for each a = 1 to n. Please refer find all divisors of a natural number for implementation.
Efficient approach : gcd(a, b) = b means that a is a multiple of b. So total number of pairs will be sum of number of multiples of each b (where b varies from 1 to n) which are less than or equal to n.
For a number i, number of multiples of i is less than or equal to floor(n/i). So what we need to do is just sum the floor(n/i) for each i = 1 to n and print it. But more optimizations can be done. floor(n/i) can have atmost 2*sqrt(n) values for i >= sqrt(n). floor(n/i) can vary from 1 to sqrt(n) and similarly for i = 1 to sqrt(n) floor(n/i) can have values from 1 to sqrt(n). So total of 2*sqrt(n) distinct values
let floor(n/i) = k k <= n/i < k + 1 n/k+1 < i <= n/k floor(n/k+1) < i <= floor(n/k) Thus for given k the largest value of i for which the floor(n/i) = k is floor(n/k) and all the set of i for which the floor(n/i) = k are consecutive
CPP
// C++ implementation of counting pairs// such that gcd (a, b) = b#include <bits/stdc++.h>using namespace std;// returns number of valid pairsint CountPairs(int n){ // initialize k int k = n; // loop till imin <= n int imin = 1; // Initialize result int ans = 0; while (imin <= n) { // max i with given k floor(n/k) int imax = n / k; // adding k*(number of i with // floor(n/i) = k to ans ans += k * (imax - imin + 1); // set imin = imax + 1 and k = n/imin imin = imax + 1; k = n / imin; } return ans;}// Driver functionint main(){ cout << CountPairs(1) << endl; cout << CountPairs(2) << endl; cout << CountPairs(3) << endl; return 0;} |
Java
// Java implementation of counting pairs// such that gcd (a, b) = bclass GFG { // returns number of valid pairs static int CountPairs(int n) { // initialize k int k = n; // loop till imin <= n int imin = 1; // Initialize result int ans = 0; while (imin <= n) { // max i with given k floor(n/k) int imax = n / k; // adding k*(number of i with // floor(n/i) = k to ans ans += k * (imax - imin + 1); // set imin = imax + 1 // and k = n/imin imin = imax + 1; k = n / imin; } return ans; } // Driver code public static void main(String[] args) { System.out.println(CountPairs(1)); System.out.println(CountPairs(2)); System.out.println(CountPairs(3)); }}// This code is contributed by Anant Agarwal. |
Python3
# Python implementation of counting# pairs such that gcd (a, b) = b# returns number of valid pairsdef CountPairs(n): # initialize k k = n # loop till imin <= n imin = 1 # Initialize result ans = 0 while(imin <= n): # max i with given k floor(n / k) imax = n / k # adding k*(number of i with # floor(n / i) = k to ans ans += k * (imax - imin + 1) # set imin = imax + 1 and # k = n / imin imin = imax + 1 k = n / imin return ans # Driver codeprint(CountPairs(1))print(CountPairs(2))print(CountPairs(3))# This code is contributed by Anant Agarwal. |
C#
// C# implementation of counting// pairs such that gcd (a, b) = busing System;class GFG { // returns number of valid pairs static int CountPairs(int n) { // initialize k int k = n; // loop till imin <= n int imin = 1; // Initialize result int ans = 0; while (imin <= n) { // max i with given // k floor(n / k) int imax = n / k; // adding k * (number of i // with floor(n / i) = k // to ans ans += k * (imax - imin + 1); // set imin = imax + 1 // and k = n / imin imin = imax + 1; k = n / imin; } return ans; } // Driver code public static void Main(String []args) { Console.WriteLine(CountPairs(1)); Console.WriteLine(CountPairs(2)); Console.WriteLine(CountPairs(3)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP implementation of counting// pairs such that gcd (a, b) = b// returns number of valid pairsfunction CountPairs($n){ // initialize k $k = $n; // loop till imin <= n $imin = 1; // Initialize result $ans = 0; while ($imin <= $n) { // max i with given k floor(n/k) $imax = $n / $k; // adding k*(number of i with // floor(n/i) = k to ans $ans += $k * ($imax - $imin + 1); // set imin = imax + 1 // and k = n/imin $imin = $imax + 1; $k = (int)($n / $imin); } return $ans;}// Driver Codeecho(CountPairs(1) . "\n");echo(CountPairs(2) . "\n");echo(CountPairs(3) . "\n");// This code is contributed by Ajit.?> |
Javascript
<script>// Javascript implementation of counting pairs// such that gcd (a, b) = b// returns number of valid pairsfunction CountPairs(n){ // initialize k let k = n; // loop till imin <= n let imin = 1; // Initialize result let ans = 0; while (imin <= n) { // max i with given k floor(n/k) let imax = Math.floor(n / k); // adding k*(number of i with // floor(n/i) = k to ans ans += k * (imax - imin + 1); // set imin = imax + 1 and k = n/imin imin = imax + 1; k = Math.floor(n / imin); } return ans;}// Driver function document.write(CountPairs(1) + "<br>"); document.write(CountPairs(2) + "<br>"); document.write(CountPairs(3) + "<br>");// This is code is contributed by Mayank Tyagi</script> |
Output :
1 3 5
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