Check if there is any pair in a given range with GCD is divisible by k

Given a range, we need to check if is their any pairs in the segment whose GCD is divisible by k.

Examples:

Input : l=4, r=6, k=2
Output : YES
There are two numbers 4 and 6 whose GCD is 2 which is divisible by 2.

Input : l=3 r=5 k=4
Output : NO
Their is no such pair whose gcd is divisible by 5.

Basically we need to count such numbers in range l to r such that they are divisible by k. Because if we choose any such two numbers then their gcd is also a multiple of k. Now if count of such numbers is greater than one then we can form pair, otherwise it’s impossible to form a pair (x, y) such that gcd(x, y) is divisible by k.



Below is the implementation of above approach:

C++

#include <bits/stdc++.h>
using namespace std;
  
// function to count such possible numbers
bool Check_is_possible(int l, int r, int k)
{
    int count = 0;
  
    for (int i = l; i <= r; i++) {
  
        // if i is divisible by k
        if (i % k == 0)
            count++;
    }
  
    // if count of such numbers
    // is greater than one
    return (count > 1);
}
  
// Driver code
int main()
{
    int l = 4, r = 12;
    int k = 5;
  
    if (Check_is_possible(l, r, k))
        cout << "YES\n";
    else
        cout << "NO\n";
    return 0;
}

C#

using System;

// function to count such
// possible numbers
class GFG
{
public bool Check_is_possible(int l, int r,
int k)
{
int count = 0;

for (int i = l; i <= r; i++) { // if i is divisible by k if (i % k == 0) count++; } // if count of such numbers // is greater than one return (count > 1);
}



// Driver code
public static void Main()
{
GFG g = new GFG();
int l = 4, r = 12;
int k = 5;

if (g.Check_is_possible(l, r, k))
Console.WriteLine(“YES\n”);
else
Console.WriteLine(“NO\n”);
}
}

// This code is contributed
// by Soumik

Output:

YES

Time Complexity: O(r – l + 1)

An efficient solution is based on the efficient approach discussed here.

// Program to count the numbers divisible
// by k in a given range
#include <bits/stdc++.h>
using namespace std;
  
// Returns count of numbers in [l r] that
// are divisible by k.
int Check_is_possible(int l, int r, int k)
{
    int div_count = (r / k) - (l / k);
  
    // Add 1 explicitly as l is divisible by k
    if (l % k == 0)
        div_count++;
  
    // l is not divisible by k
    return (div_count > 1);
}
  
// Driver Code
int main()
{
    int l = 30, r = 70, k = 10;
  
    if (Check_is_possible(l, r, k))
        cout << "YES\n";
    else
        cout << "NO\n";
    return 0;
}

Output:

YES


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Improved By : SoumikMondal



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