Generate first K multiples of N using Bitwise operators

Given an integer N, the task is to print the first K multiples of N using Bitwise Operators.

Examples:

Input: N = 16, K = 7 
Output:
16 * 1 = 16 
16 * 2 = 32 
16 * 3 = 48 
16 * 4 = 64 
16 * 5 = 80 
16 * 6 = 96 
16 * 7 = 112
Input: N = 7, K = 10 
Output:
7 * 1 = 7 
7 * 2 = 14 
7 * 3 = 21 
7 * 4 = 28 
7 * 5 = 35 
7 * 6 = 42 
7 * 7 = 49 
7 * 8 = 56 
7 * 9 = 63 
7 * 10 = 70 

Approach:

Follow the steps below to solve the problem: 

• Iterate up to K.
• For each iteration, print current value of N.
• Then, calculate the sum of 2ⁱ for every i^th set bit of N. Add this sum to N and repeat from the step above.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to print the first K
// multiples of N
void Kmultiples(int n, int k)
{
    int a = n;

    for (int i = 1; i <= k; i++) {

        // Print the value of N*i
        cout << n << " * " << i << " = "
             << a << endl;
        int j = 0;

        // Iterate each bit of N and add
        // pow(2, pos), where pos is the
        // index of each set bit
        while (n >= (1 << j)) {

            // Check if current bit at
            // pos j is fixed or not
            a += n & (1 << j);

            // For next set bit
            j++;
        }
    }
}

// Driver Code
int main()
{
    int N = 16, K = 7;

    Kmultiples(N, K);

    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG{

// Function to print the first K
// multiples of N
static void Kmultiples(int n, int k)
{
    int a = n;

    for (int i = 1; i <= k; i++)
    {

        // Print the value of N*i
        System.out.print(n+ " * " +  i+ " = "
             + a +"\n");
        int j = 0;

        // Iterate each bit of N and add
        // Math.pow(2, pos), where pos is the
        // index of each set bit
        while (n >= (1 << j))
        {

            // Check if current bit at
            // pos j is fixed or not
            a += n & (1 << j);

            // For next set bit
            j++;
        }
    }
}

// Driver Code
public static void main(String[] args)
{
    int N = 16, K = 7;

    Kmultiples(N, K);
}
}

// This code is contributed by Rohit_ranjan

Python3

# Python3 program to implement
# the above approach

# Function to print the first K
# multiples of N
def Kmultiples(n, k):

    a = n

    for i in range(1, k + 1):

        # Print the value of N*i
        print("{} * {} = {}".format(n, i, a))
        j = 0

        # Iterate each bit of N and add
        # pow(2, pos), where pos is the
        # index of each set bit
        while(n >= (1 << j)):

            # Check if current bit at
            # pos j is fixed or not
            a += n & (1 << j)

            # For next set bit
            j += 1

# Driver Code
N = 16
K = 7

Kmultiples(N, K)

# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to print the first K
// multiples of N
static void Kmultiples(int n, int k)
{
    int a = n;

    for(int i = 1; i <= k; i++)
    {

        // Print the value of N*i
        Console.Write(n + " * " + i +
                          " = " + a + "\n");
        int j = 0;

        // Iterate each bit of N and add
        // Math.Pow(2, pos), where pos is
        //  the index of each set bit
        while (n >= (1 << j))
        {

            // Check if current bit at
            // pos j is fixed or not
            a += n & (1 << j);

            // For next set bit
            j++;
        }
    }
}

// Driver Code
public static void Main(String[] args)
{
    int N = 16, K = 7;

    Kmultiples(N, K);
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>
// javascript program to implement
// the above approach
// Function to print the first K
// multiples of N
function Kmultiples(n , k)
{
    var a = n;

    for (i = 1; i <= k; i++)
    {

        // Print the value of N*i
        document.write(n+ " * " +  i+ " = "
             + a +"<br>");
        var j = 0;

        // Iterate each bit of N and add
        // Math.pow(2, pos), where pos is the
        // index of each set bit
        while (n >= (1 << j))
        {

            // Check if current bit at
            // pos j is fixed or not
            a += n & (1 << j);

            // For next set bit
            j++;
        }
    }
}

// Driver Code
var N = 16, K = 7;

Kmultiples(N, K);

// This code contributed by Princi Singh
</script>

Output:

16 * 1 = 16
16 * 2 = 32
16 * 3 = 48
16 * 4 = 64
16 * 5 = 80
16 * 6 = 96
16 * 7 = 112

Time Complexity: O(Klog₂N) 
Auxiliary Space: O(1)