Dado un número N , la tarea es imprimir los primeros N números de modo que cada número y el reverso del número sea divisible por su suma de dígitos .
Ejemplo:
Entrada: N = 4
Salida: 1 2 3 4
Explicación:
El reverso de cada número de un solo dígito es el mismo número. Y, todo número es divisible por sí mismo.
Entrada: N = 12
Salida: 1 2 3 4 5 6 7 8 9 10 12 18
Enfoque: La idea es iterar a través de cada número desde el 1 y calcular la suma de los dígitos. Para cada uno de esos números, verifica si el número y el reverso del número son divisibles por la suma o no. Por lo tanto, se siguen los siguientes pasos para calcular la respuesta:
- Inicialice el contador a uno e itere todos los números uno por uno.
- Para cada número, encuentra el reverso del número .
- Mientras encuentra el reverso del número, calcule la suma de los dígitos del número .
- Ahora, verifica si el número y el reverso del número son divisibles por la suma de sus dígitos.
- Si es así, incremente el contador. Repita los pasos anteriores hasta que este contador sea igual a N .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the sum of digits
int digit_sum(int n)
{
int sum = 0, m;
// Loop to iterate through every
// digit of the number
while (n > 0) {
m = n % 10;
sum = sum + m;
n = n / 10;
}
// Returning the sum of digits
return (sum);
}
// Function to calculate the reverse
// of a number
int reverse(int n)
{
int r = 0;
// Loop to calculate the reverse
// of the number
while (n != 0) {
r = r * 10;
r = r + n % 10;
n = n / 10;
}
// Return the reverse of the
// number
return (r);
}
// Function to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
void operation(int n)
{
int i = 1, a, count = 0, r;
// Loop to continuously check and
// generate number until there
// are n outputs
while (count < n) {
// Variable to hold the sum of
// the digit of the number
a = digit_sum(i);
// Computing the reverse of the
// number
r = reverse(i);
// Checking if the condition satisfies.
// Increment the count and print the
// number if it satisfies.
if (i % a == 0 && r % a == 0) {
cout << i << " ";
count++;
i++;
}
else
i++;
}
}
// Driver code
int main()
{
int n = 10;
operation(n);
}
Java
// Java program to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
import java.util.*;
class GFG{
// Function to calculate the sum of digits
static int digit_sum(int n)
{
int sum = 0, m;
// Loop to iterate through
// every digit of the number
while (n > 0)
{
m = n % 10;
sum = sum + m;
n = n / 10;
}
// Returning the sum of digits
return (sum);
}
// Function to calculate the
// reverse of a number
static int reverse(int n)
{
int r = 0;
// Loop to calculate the
// reverse of the number
while (n != 0)
{
r = r * 10;
r = r + n % 10;
n = n / 10;
}
// Return the reverse
// of the number
return (r);
}
// Function to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
static void operation(int n)
{
int i = 1, a, count = 0, r;
// Loop to continuously check and
// generate number until there
// are n outputs
while (count < n)
{
// Variable to hold the sum
// of the digit of the number
a = digit_sum(i);
// Computing the reverse of the
// number
r = reverse(i);
// Checking if the condition satisfies.
// Increment the count and print the
// number if it satisfies.
if (i % a == 0 && r % a == 0)
{
System.out.print(i + " ");
count++;
i++;
}
else
i++;
}
}
// Driver code
public static void main(String args[])
{
int n = 10;
operation(n);
}
}
// This code is contributed by shivanisinghss2110
Python3
# Python3 program to print the first N numbers # such that every number and the reverse # of the number is divisible by its # sum of digits # Function to calculate the sum of digits def digit_sum(n): sum = 0 # Loop to iterate through every # digit of the number while (n > 0): m = n % 10; sum = sum + m; n = n //10 # Returning the sum of digits return (sum) # Function to calculate the reverse # of a number def reverse(n): r = 0 # Loop to calculate the reverse # of the number while (n != 0): r = r * 10 r = r + n % 10 n = n // 10 # Return the reverse of the # number return (r) # Function to print the first N numbers # such that every number and the reverse # of the number is divisible by its # sum of digits def operation(n): i = 1 count = 0 # Loop to continuously check and # generate number until there # are n outputs while (count < n): # Variable to hold the sum of # the digit of the number a = digit_sum(i) # Computing the reverse of the # number r = reverse(i) # Checking if the condition satisfies. # Increment the count and print the # number if it satisfies. if (i % a == 0 and r % a == 0): print(i, end = " ") count += 1 i += 1 else: i += 1 # Driver code if __name__ == '__main__': n = 10 operation(n) # This code is contributed by Samarth
C#
// C# program to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
using System;
class GFG{
// Function to calculate the sum of digits
static int digit_sum(int n)
{
int sum = 0, m;
// Loop to iterate through
// every digit of the number
while (n > 0)
{
m = n % 10;
sum = sum + m;
n = n / 10;
}
// Returning the sum of digits
return (sum);
}
// Function to calculate the
// reverse of a number
static int reverse(int n)
{
int r = 0;
// Loop to calculate the
// reverse of the number
while (n != 0)
{
r = r * 10;
r = r + n % 10;
n = n / 10;
}
// Return the reverse
// of the number
return (r);
}
// Function to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
static void operation(int n)
{
int i = 1, a, count = 0, r;
// Loop to continuously check and
// generate number until there
// are n outputs
while (count < n)
{
// Variable to hold the sum
// of the digit of the number
a = digit_sum(i);
// Computing the reverse of the
// number
r = reverse(i);
// Checking if the condition satisfies.
// Increment the count and print the
// number if it satisfies.
if (i % a == 0 && r % a == 0)
{
Console.Write(i + " ");
count++;
i++;
}
else
i++;
}
}
// Driver code
public static void Main()
{
int n = 10;
operation(n);
}
}
// This code is contributed by Code_Mech
Javascript
<script>
// Javascript program to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
// Function to calculate the sum of digits
function digit_sum(n)
{
let sum = 0, m;
// Loop to iterate through every
// digit of the number
while (n > 0) {
m = n % 10;
sum = sum + m;
n = parseInt(n / 10, 10);
}
// Returning the sum of digits
return (sum);
}
// Function to calculate the reverse
// of a number
function reverse(n)
{
let r = 0;
// Loop to calculate the reverse
// of the number
while (n != 0) {
r = r * 10;
r = r + n % 10;
n = parseInt(n / 10, 10);
}
// Return the reverse of the
// number
return (r);
}
// Function to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
function operation(n)
{
let i = 1, a, count = 0, r;
// Loop to continuously check and
// generate number until there
// are n outputs
while (count < n) {
// Variable to hold the sum of
// the digit of the number
a = digit_sum(i);
// Computing the reverse of the
// number
r = reverse(i);
// Checking if the condition satisfies.
// Increment the count and print the
// number if it satisfies.
if (i % a == 0 && r % a == 0) {
document.write(i + " ");
count++;
i++;
}
else
i++;
}
}
let n = 10;
operation(n);
</script>
Producción:
1 2 3 4 5 6 7 8 9 10
Complejidad temporal: O(n * log 10 n)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por srutichatterjee2024 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA