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 ith 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)