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Program to find Nth term divisible by a or b
  • Difficulty Level : Hard
  • Last Updated : 02 Jun, 2021

Given two integers a   and b   . The task is to find the Nth term which is divisible by either of a   or b   .
Examples : 
 

Input : a = 2, b = 5, N = 10
Output : 16

Input : a = 3, b = 7, N = 25
Output : 57

 

Naive Approach: A simple approach is to traverse over all the terms starting from 1 until we find the desired Nth term which is divisible by either of a   or b   . This solution has time complexity of O(N).
Efficient Approach: The idea is to use Binary search. Here we can calculate how many numbers from 1 to num   are divisible by either a or b by using formula:
termCount= num/a + num/b - num/lcm(a, b)
All the multiples of lcm(a, b) will be divisible by both a   and b   so we need to remove these terms. Now if the number of divisible terms is less than N we will increase the low position of binary search otherwise decrease high until number of divisible terms is equal to N.
Below is the implementation of the above idea : 
 

C++




// C++ program to find nth term
// divisible by a or b
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return
// gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
 
    return gcd(b % a, a);
}
 
// Function to calculate how many numbers
// from 1 to num are divisible by a or b
int divTermCount(int a, int b, int lcm, int num)
{
    // calculate number of terms divisible by a and
    // by b then, remove the terms which is are
    // divisible by both a and b
    return num / a + num / b - num / lcm;
}
 
// Binary search to find the nth term
// divisible by a or b
int findNthTerm(int a, int b, int n)
{
    // set low to 1 and high to max(a, b)*n, here
    // we have taken high as 10^18
    int low = 1, high = INT_MAX, mid;
    int lcm = (a * b) / gcd(a, b);
 
    while (low < high) {
        mid = low + (high - low) / 2;
 
        // if the current term is less than
        // n then we need to increase low
        // to mid + 1
        if (divTermCount(a, b, lcm, mid) < n)
            low = mid + 1;
 
        // if current term is greater than equal to
        // n then high = mid
        else
            high = mid;
    }
 
    return low;
}
 
// Driver code
int main()
{
    int a = 2, b = 5, n = 10;
    cout << findNthTerm(a, b, n) << endl;
 
    return 0;
}

Java




// Java program to find nth term
// divisible by a or b
class GFG
{
// Function to return
// gcd of a and b
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
 
    return gcd(b % a, a);
}
 
// Function to calculate how many numbers
// from 1 to num are divisible by a or b
static int divTermCount(int a, int b,
                        int lcm, int num)
{
    // calculate number of terms
    // divisible by a and by b then,
    // remove the terms which is are
    // divisible by both a and b
    return num / a + num / b - num / lcm;
}
 
// Binary search to find the
// nth term divisible by a or b
static int findNthTerm(int a, int b, int n)
{
    // set low to 1 and high to max(a, b)*n, 
    // here we have taken high as 10^18
    int low = 1, high = Integer.MAX_VALUE, mid;
    int lcm = (a * b) / gcd(a, b);
 
    while (low < high)
    {
        mid = low + (high - low) / 2;
 
        // if the current term is less
        // than n then we need to increase
        // low to mid + 1
        if (divTermCount(a, b, lcm, mid) < n)
            low = mid + 1;
 
        // if current term is greater
        // than equal to n then high = mid
        else
            high = mid;
    }
 
    return low;
}
 
// Driver code
public static void main (String[] args)
{
    int a = 2, b = 5, n = 10;
    System.out.println(findNthTerm(a, b, n));
}
}
 
// This code is contributed by Smitha

Python3




# Python 3 program to find nth term
# divisible by a or b
import sys
 
# Function to return gcd of a and b
def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)
 
# Function to calculate how many numbers
# from 1 to num are divisible by a or b
def divTermCount(a, b, lcm, num):
 
    # calculate number of terms divisible
    # by a and by b then, remove the terms
    # which are divisible by both a and b
    return num // a + num // b - num // lcm
 
# Binary search to find the nth term
# divisible by a or b
def findNthTerm(a, b, n):
 
    # set low to 1 and high to max(a, b)*n,
    # here we have taken high as 10^18
    low = 1; high = sys.maxsize
    lcm = (a * b) // gcd(a, b)
    while low < high:
        mid = low + (high - low) // 2
 
        # if the current term is less
        # than n then we need to increase 
        # low to mid + 1
        if divTermCount(a, b, lcm, mid) < n:
            low = mid + 1
 
        # if current term is greater
        # than equal to n then high = mid
        else:
            high = mid
    return low
 
# Driver code
a = 2; b = 5; n = 10
print(findNthTerm(a, b, n))
 
# This code is contributed by Shrikant13

C#




// C# program to find nth term
// divisible by a or b
using System;
 
class GFG
{
// Function to return gcd of a and b
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
 
    return gcd(b % a, a);
}
 
// Function to calculate how many numbers
// from 1 to num are divisible by a or b
static int divTermCount(int a, int b,
                        int lcm, int num)
{
    // calculate number of terms
    // divisible by a and by b then,
    // remove the terms which is are
    // divisible by both a and b
    return num / a + num / b - num / lcm;
}
 
// Binary search to find the
// nth term divisible by a or b
static int findNthTerm(int a, int b, int n)
{
    // set low to 1 and high to max(a, b)*n,
    // here we have taken high as 10^18
    int low = 1, high = int.MaxValue, mid;
    int lcm = (a * b) / gcd(a, b);
 
    while (low < high)
    {
        mid = low + (high - low) / 2;
 
        // if the current term is less
        // than n then we need to increase
        // low to mid + 1
        if (divTermCount(a, b, lcm, mid) < n)
            low = mid + 1;
 
        // if current term is greater
        // than equal to n then high = mid
        else
            high = mid;
    }
 
    return low;
}
 
// Driver code
static public void Main ()
{
    int a = 2, b = 5, n = 10;
    Console.WriteLine(findNthTerm(a, b, n));
}
}
 
// This code is contributed by Sach_Code

Javascript




<script>
 
// JavaScript program to find nth term
// divisible by a or b
 
// Function to return
// gcd of a and b
function gcd(a , b)
{
    if (a == 0)
        return b;
 
    return gcd(b % a, a);
}
 
// Function to calculate how many numbers
// from 1 to num are divisible by a or b
function divTermCount(a , b, lcm , num)
{
    // calculate number of terms
    // divisible by a and by b then,
    // remove the terms which is are
    // divisible by both a and b
    return parseInt(num / a) +
    parseInt(num / b) - parseInt(num / lcm);
}
 
// Binary search to find the
// nth term divisible by a or b
function findNthTerm(a , b , n)
{
    // set low to 1 and high to max(a, b)*n, 
    // here we have taken high as 10^18
    var low = 1, high = Number.MAX_VALUE, mid;
    var lcm = parseInt((a * b) / gcd(a, b));
 
    while (low < high)
    {
        mid = low + parseInt((high - low) / 2);
 
        // if the current term is less
        // than n then we need to increase
        // low to mid + 1
        if (divTermCount(a, b, lcm, mid) < n)
            low = mid + 1;
 
        // if current term is greater
        // than equal to n then high = mid
        else
            high = mid;
    }
 
    return low;
}
 
// Driver code
var a = 2, b = 5, n = 10;
document.write(findNthTerm(a, b, n));
 
 
// This code is contributed by Amit Katiyar
 
</script>
Output: 
16

 

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