Encuentra el N-ésimo número natural que no es divisible por A

Dados dos enteros A y N , nuestra tarea es encontrar el N-ésimo número natural que no sea divisible por A.

Ejemplos: 

Entrada: A = 4, N = 12 
Salida: 15 
Explicación: 
La serie a partir de 1 excluyendo los múltiplos de A sería 1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17 y el término 12 que no es divisible por 4 es 15.

Entrada: A = 3, N = 20 
Salida: 29 
Explicación: 
La serie que comienza en 1 excluyendo los múltiplos de A sería 1, 2, 4, 5, 7, 8, 10, 11 y así sucesivamente y el N-ésimo número que es no divisible por 3 es 29. 
 

Acercarse:

Para resolver el problema mencionado anteriormente, debemos observar que hay un espacio después de cada número entero (A – 1) a medida que se omiten los múltiplos de A. Para encontrar que el número N se encuentra en qué conjunto, dividimos N por (A – 1) y lo almacenamos en una variable, digamos cociente . Ahora, multiplicando A con esa variable y sumando el resto obtendremos la respuesta resultante. 

If A = 4, N = 7:
quotient = 7 / 3 = 2 
remainder = 7 % 3 = 1

So the answer is 
(A * quotient) + remainder 
= 4 * 2 + 1 = 9

Pero si el resto es 0 significa que es el último elemento del conjunto. Entendámoslo con un ejemplo. 
 

If A = 4, N = 6:
quotient = 6 / 3 = 2 
remainder = 6 % 3 = 0

So the answer is 
(A * quotient) - 1 
= 4 * 2 - 1 = 7

A continuación se muestra la implementación del enfoque anterior:

// C++ code to Find the Nth number which
// is not divisible by A from the series
#include<bits/stdc++.h>
using namespace std;
 
void findNum(int n, int k)
{
     
    // Find the quotient and the remainder
    // when k is divided by n-1
    int q = k / (n - 1);
    int r = k % (n - 1);
    int a;
 
    // If the remainder is not 0
    // multiply n by q and subtract 1
    // if remainder is 0 then
    // multiply n by q
    // and add the remainder
    if(r != 0)
       a = (n * q) + r;
    else
       a = (n * q) - 1;
 
    // Print the answer
    cout << a;
}
 
// Driver code
int main()
{
    int A = 4, N = 6;
     
    findNum(A, N);
    return 0;
}
 
// This code is contributed by PratikBasu

// Java code to Find the Nth number which
// is not divisible by A from the series
class GFG {
     
static void findNum(int n, int k)
{
 
    // Find the quotient and the remainder
    // when k is divided by n-1
    int q = k / (n - 1);
    int r = k % (n - 1);
    int a = 0;
     
    // If the remainder is not 0
    // multiply n by q and subtract 1
    // if remainder is 0 then
    // multiply n by q
    // and add the remainder
    if (r != 0)
        a = (n * q) + r;
    else
        a = (n * q) - 1;
 
    // Print the answer
    System.out.println(a);
}
 
// Driver code
public static void main(String[] args)
{
    int A = 4;
    int N = 6;
 
    findNum(A, N);
}
}
 
// This code is contributed by 29AjayKumar

# Python3 code to Find the Nth number which
# is not divisible by A from the series
 
def findNum(n, k):
     
    # Find the quotient and the remainder
    # when k is divided by n-1
    q = k//(n-1)
    r = k % (n-1)
     
    # if the remainder is not 0
    # multiply n by q and subtract 1
     
    # if remainder is 0 then
    # multiply n by q
    # and add the remainder
    if(r != 0):
        a = (n * q)+r
    else:
        a = (n * q)-1
         
    # print the answer
    print(a)
 
# driver code
A = 4
N = 6
findNum(A, N)

// C# code to find the Nth number which
// is not divisible by A from the series
using System;
 
class GFG{
     
static void findNum(int n, int k)
{
 
    // Find the quotient and the remainder
    // when k is divided by n-1
    int q = k / (n - 1);
    int r = k % (n - 1);
    int a = 0;
     
    // If the remainder is not 0
    // multiply n by q and subtract 1
    // if remainder is 0 then
    // multiply n by q
    // and add the remainder
    if (r != 0)
        a = (n * q) + r;
    else
        a = (n * q) - 1;
 
    // Print the answer
    Console.WriteLine(a);
}
 
// Driver code
public static void Main(String[] args)
{
    int A = 4;
    int N = 6;
 
    findNum(A, N);
}
}
 
// This code is contributed by amal kumar choubey

<script>
 
// Javascript code to Find the Nth number which
// is not divisible by A from the series
function findNum(n, k)
{
     
    // Find the quotient and the remainder
    // when k is divided by n-1
    let q = parseInt(k / (n - 1));
    let r = k % (n - 1);
    let a;
 
    // If the remainder is not 0
    // multiply n by q and subtract 1
    // if remainder is 0 then
    // multiply n by q
    // and add the remainder
    if (r != 0)
        a = (n * q) + r;
    else
        a = (n * q) - 1;
 
    // Print the answer
    document.write(a);
}
 
// Driver code
let A = 4, N = 6;
 
findNum(A, N);
 
// This code is contributed by rishavmahato348
 
</script>

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