Program to find Nth term divisible by a or b
Given two integers 
and 
. The task is to find the Nth term which is divisible by either of or .
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 or . 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 
are divisible by either a or b by using formula:
All the multiples of lcm(a, b) will be divisible by both and 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; } |
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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 |
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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 = 10print(findNthTerm(a, b, n)) # This code is contributed by Shrikant13 |
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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 |
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Output:
16
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