在十进制基数中从给定数字的基数B的右数中找到第N个数字

给定一个以十进制为底的数字A ，任务是从底数B的最后一个数字中找到第N^个数字
例子：

Input: A = 100, N = 3, B = 4 
Output: 2 
Explanation: 
(100)₄ = 1210 
3^rd digit from last is 2 
Input: A = 50, N = 3, B = 5 
Output: 2 
Explanation: 
(50)₅ = 200 
3^rd digit from last is 2 
 

方法：想法是跳过基数B中给定数字的(N-1)个数字，方法是将数字除以B(N – 1)次，然后将当前数字的模数除以B，得到第N^个右边的数字。
下面是上述方法的实现：

C++

// C++ Implementation to find Nth digit
// from right in base B
 
#include 
using namespace std;
 
// Function to compute Nth digit
// from right in base B
int nthDigit(int a, int n, int b)
{
 
    // Skip N-1 Digits in Base B
    for (int i = 1; i < n; i++)
        a = a / b;
 
    // Nth Digit from right in Base B
    return a % b;
}
 
// Driver Code
int main()
{
    int a = 100;
    int n = 3;
    int b = 4;
    cout << nthDigit(a, n, b);
    return 0;
}

Java

// Java Implementation to find Nth digit
// from right in base B
import java.util.*;
 
class GFG
{
 
// Function to compute Nth digit
// from right in base B
static int nthDigit(int a, int n, int b)
{
 
    // Skip N-1 Digits in Base B
    for (int i = 1; i < n; i++)
        a = a / b;
 
    // Nth Digit from right in Base B
    return a % b;
}
 
// Driver Code
public static void main(String[] args)
{
    int a = 100;
    int n = 3;
    int b = 4;
    System.out.print(nthDigit(a, n, b));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Ptyhon3 Implementation to find Nth digit
# from right in base B
 
# Function to compute Nth digit
# from right in base B
def nthDigit(a, n, b):
 
    # Skip N-1 Digits in Base B
    for i in range(1, n):
        a = a // b
 
    # Nth Digit from right in Base B
    return a % b
 
# Driver Code
a = 100
n = 3
b = 4
print(nthDigit(a, n, b))
 
# This code is contributed by ApurvaRaj

C#

// C# Implementation to find Nth digit
// from right in base B
using System;
 
class GFG
{
 
    // Function to compute Nth digit
    // from right in base B
    static int nthDigit(int a, int n, int b)
    {
     
        // Skip N-1 Digits in Base B
        for (int i = 1; i < n; i++)
            a = a / b;
     
        // Nth Digit from right in Base B
        return a % b;
    }
     
    // Driver Code
    public static void Main()
    {
        int a = 100;
        int n = 3;
        int b = 4;
        Console.Write(nthDigit(a, n, b));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

 
// Javascript Implementation to find Nth digit
// from right in base B
 
// Function to compute Nth digit
// from right in base B
function nthDigit(a, n, b)
{
 
    // Skip N-1 Digits in Base B
    for (var i = 1; i < n; i++)
        a = parseInt(a / b);
 
    // Nth Digit from right in Base B
    return a % b;
}
 
// Driver Code
var a = 100;
var n = 3;
var b = 4;
document.write(nthDigit(a, n, b));
 
// This code is contributed by rutvik_56.

输出：

2