Dado un número entero N , la tarea es imprimir los primeros K múltiplos de N utilizando operadores bit a bit .
Ejemplos:
Entrada: N = 16, K = 7
Salida:
16 * 1 = 16
16 * 2 = 32
16 * 3 = 48
16 * 4 = 64
16 * 5 = 80
16 * 6 = 96
16 * 7 = 112
Entrada: N = 7, K = 10
Salida:
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
Acercarse:
Siga los pasos a continuación para resolver el problema:
- Iterar hasta K.
- Para cada iteración, imprima el valor actual de N .
- Luego, calcule la suma de 2 i para cada i- ésimo conjunto de bits de N. Agregue esta suma a N y repita desde el paso anterior.
A continuación se muestra la implementación del enfoque anterior:
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>
Producción:
16 * 1 = 16 16 * 2 = 32 16 * 3 = 48 16 * 4 = 64 16 * 5 = 80 16 * 6 = 96 16 * 7 = 112
Complejidad temporal: O(Klog 2 N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por PranjalKumar4 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA