📜  查找不能被A整除的第N个自然数

📅  最后修改于: 2021-06-27 02:44:15             🧑  作者: Mango

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:

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。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


Javascript


输出:
7