# 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 pairs` `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 function` `int` `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) = b` `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) {` ` ` `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 pairs` `def` `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 code` `print` `(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) = b` `using` `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 pairs` `function` `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 Code` `echo` `(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 pairs` `function` `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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