查找不能被A整除的第N个自然数

给定两个整数A和N ，我们的任务是找到不能被A整除的第N个自然数。

例子：

Input: A = 4, N = 12
Output: 15
Explanation:
The series starting from 1 excluding the multiples of A would be 1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17 and the 12th term which is not divisible by 4 is 15.

Input: A = 3, N = 20
Output: 29
Explanation:
The series starting from 1 excluding the multiples of A would be 1, 2, 4, 5, 7, 8, 10, 11 and so on and the Nth number which is not divisible by 3 is 29.

方法：

为了解决上述问题，我们必须观察到，由于跳过了A的倍数，所以每个(A – 1)整数之后都有一个间隙。为了找到数字N，我们将N除以(A – 1)，然后将其存储在变量中，我们说商。现在将A与该变量相乘，然后将余数相加，我们将得到结果答案。

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

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

但是，如果余数为0，则表示它是集合中的最后一个元素。让我们通过一个例子来理解它。

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

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

下面是上述方法的实现：

C++

// C++ code to Find the Nth number which
// is not divisible by A from the series
#include
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

// 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

# 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#

// 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

输出：

7