Dado un número entero N , la tarea es imprimir los primeros N términos cuya suma de dígitos sea un múltiplo de 10 . Los primeros términos de la serie son 19, 28, 37, 46, 55, …
Ejemplos:
Entrada: N = 5
Salida: 19 28 37 46 55
Entrada: N = 10
Salida: 19 28 37 46 55 64 73 82 91 109
Planteamiento: Se puede observar que para obtener el N- ésimo término de la serie requerida, encuentre la suma de los dígitos de N. Si la suma ya es un múltiplo de 10 , agregue el dígito 0 al final de N ; de lo contrario, agregue el dígito mínimo posible al final, de modo que la nueva suma de dígitos sea un múltiplo de 10 .
Por ejemplo, para obtener el término 19 , dado que la suma de los dígitos ya es un múltiplo de 10 , agregue 0 y 190 es el término 19 de la serie.
Para N = 5 , el dígito mínimo que se puede agregar para hacer la suma de dígitos como un múltiplo de 10 es 5 y 55 es el quinto término de la serie.
A continuación se muestra la implementación del enfoque anterior:
C++
#include <bits/stdc++.h>
using namespace std;
const int TEN = 10;
// Function to return the
// sum of digits of n
int digitSum(int n)
{
int sum = 0;
while (n > 0) {
// Add last digit to the sum
sum += n % TEN;
// Remove last digit
n /= TEN;
}
return sum;
}
// Function to return the nth term
// of the required series
int getNthTerm(int n)
{
int sum = digitSum(n);
// If sum of digit is already
// a multiple of 10 then append 0
if (sum % TEN == 0)
return (n * TEN);
// To store the minimum digit
// that must be appended
int extra = TEN - (sum % TEN);
// Return n after appending
// the required digit
return ((n * TEN) + extra);
}
// Function to print the first n terms
// of the required series
void firstNTerms(int n)
{
for (int i = 1; i <= n; i++)
cout << getNthTerm(i) << " ";
}
// Driver code
int main()
{
int n = 10;
firstNTerms(n);
return 0;
}
Java
// Java implementation of the above approach
class GFG
{
final static int TEN = 10;
// Function to return the
// sum of digits of n
static int digitSum(int n)
{
int sum = 0;
while (n > 0)
{
// Add last digit to the sum
sum += n % TEN;
// Remove last digit
n /= TEN;
}
return sum;
}
// Function to return the nth term
// of the required series
static int getNthTerm(int n)
{
int sum = digitSum(n);
// If sum of digit is already
// a multiple of 10 then append 0
if (sum % TEN == 0)
return (n * TEN);
// To store the minimum digit
// that must be appended
int extra = TEN - (sum % TEN);
// Return n after appending
// the required digit
return ((n * TEN) + extra);
}
// Function to print the first n terms
// of the required series
static void firstNTerms(int n)
{
for (int i = 1; i <= n; i++)
System.out.print(getNthTerm(i) + " ");
}
// Driver code
public static void main (String[] args)
{
int n = 10;
firstNTerms(n);
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 code for above implementation TEN = 10 # Function to return the # sum of digits of n def digitSum(n): sum = 0 while (n > 0): # Add last digit to the sum sum += n % TEN # Remove last digit n //= TEN return sum # Function to return the nth term # of the required series def getNthTerm(n): sum = digitSum(n) # If sum of digit is already # a multiple of 10 then append 0 if (sum % TEN == 0): return (n * TEN) # To store the minimum digit # that must be appended extra = TEN - (sum % TEN) # Return n after appending # the required digit return ((n * TEN) + extra) # Function to print the first n terms # of the required series def firstNTerms(n): for i in range(1, n + 1): print(getNthTerm(i), end = " ") # Driver code n = 10 firstNTerms(n) # This code is contributed by Mohit Kumar
C#
// C# Program to Find the Unique elements
// in linked lists
using System;
class GFG
{
readonly static int TEN = 10;
// Function to return the
// sum of digits of n
static int digitSum(int n)
{
int sum = 0;
while (n > 0)
{
// Add last digit to the sum
sum += n % TEN;
// Remove last digit
n /= TEN;
}
return sum;
}
// Function to return the nth term
// of the required series
static int getNthTerm(int n)
{
int sum = digitSum(n);
// If sum of digit is already
// a multiple of 10 then append 0
if (sum % TEN == 0)
return (n * TEN);
// To store the minimum digit
// that must be appended
int extra = TEN - (sum % TEN);
// Return n after appending
// the required digit
return ((n * TEN) + extra);
}
// Function to print the first n terms
// of the required series
static void firstNTerms(int n)
{
for (int i = 1; i <= n; i++)
Console.Write(getNthTerm(i) + " ");
}
// Driver code
public static void Main (String[] args)
{
int n = 10;
firstNTerms(n);
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
const TEN = 10;
// Function to return the
// sum of digits of n
function digitSum(n)
{
let sum = 0;
while (n > 0) {
// Add last digit to the sum
sum += n % TEN;
// Remove last digit
n = Math.floor(n / TEN);
}
return sum;
}
// Function to return the nth term
// of the required series
function getNthTerm(n)
{
let sum = digitSum(n);
// If sum of digit is already
// a multiple of 10 then append 0
if (sum % TEN == 0)
return (n * TEN);
// To store the minimum digit
// that must be appended
let extra = TEN - (sum % TEN);
// Return n after appending
// the required digit
return ((n * TEN) + extra);
}
// Function to print the first n terms
// of the required series
function firstNTerms(n)
{
for (let i = 1; i <= n; i++)
document.write(getNthTerm(i) + " ");
}
// Driver code
let n = 10;
firstNTerms(n);
// This code is contributed by Surbhi Tyagi.
</script>
Producción:
19 28 37 46 55 64 73 82 91 109