将“ a”中的第K位提高为“ b”次幂

由于三个数a，b和K，发现第k个位数从右侧A ^B
例子：

Input : a = 3, b = 3, 
        k = 1
Output : 7
Explanation
3^3 = 27 for k = 1. First digit is 7 in 27

Input : a = 5, b = 2, 
        k = 2
Output : 2
Explanation
5^2 = 25 for k = 2. First digit is 2 in 25

方法
1)计算a ^ b
2)反复删除最后一位直到第k位不符合

C++

// CPP program for finding k-th digit in a^b
#include 
using namespace std;
 
// To compute k-th digit in a^b
int kthdigit(int a, int b, int k)
{
    // computing a^b
    int p = pow(a, b);
 
    int count = 0;
    while (p > 0 && count < k) {
 
        // getting last digit
        int rem = p % 10;
 
        // increasing count by 1
        count++;
 
        // if current number is required digit
        if (count == k)
            return rem;
 
        // remove last digit
        p = p / 10;
    }
 
    return 0;
}
 
// Driver code
int main()
{
    int a = 5, b = 2;
    int k = 1;
    cout << kthdigit(a, b, k);
    return 0;
}

Java

// Java program for finding k-th digit in a^b
import java.util.*;
import java.lang.*;
 
public class GfG {
    // To compute k-th digit in a^b
    public static int kthdigit(int a, int b, int k)
    {
        // Computing a^b
        int p = (int)Math.pow(a, b);
 
        int count = 0;
        while (p > 0 && count < k) {
 
            // Getting last digit
            int rem = p % 10;
 
            // Increasing count by 1
            count++;
 
            // If current number is required digit
            if (count == k)
                return rem;
 
            // Remove last digit
            p = p / 10;
        }
 
        return 0;
    }
     
    // Driver Code
    public static void main(String argc[]) {
        int a = 5, b = 2;
        int k = 1;
        System.out.println(kthdigit(a, b, k));
    }
     
}
 
// This code is contributed by Sagar Shukla.

Python3

# Python3 code to compute k-th
# digit in a^b
def kthdigit(a, b, k):
     
    # computin a^b in python
    p = a ** b
    count = 0
     
    while (p > 0 and count < k):
         
        # getting last digit
        rem = p % 10
 
        # increasing count by 1
        count = count + 1
 
        # if current number is
        # required digit
        if (count == k):
            return rem
 
        # remove last digit
        p = p / 10;
     
# driver code   
a = 5
b = 2
k = 1
ans = kthdigit(a, b, k)
print (ans)
 
# This code is contributed by Saloni Gupta

C#

// C# program for finding k-th digit in a^b
using System;
 
public class GfG {
     
    // To compute k-th digit in a^b
    public static int kthdigit(int a, int b, int k)
    {
        // Computing a^b
        int p = (int)Math.Pow(a, b);
 
        int count = 0;
        while (p > 0 && count < k) {
 
            // Getting last digit
            int rem = p % 10;
 
            // Increasing count by 1
            count++;
 
            // If current number is required digit
            if (count == k)
                return rem;
 
            // Remove last digit
            p = p / 10;
        }
 
        return 0;
    }
     
    // Driver Code
    public static void Main() {
        int a = 5, b = 2;
        int k = 1;
        Console.WriteLine(kthdigit(a, b, k));
    }
     
}
 
// This code is contributed by vt_m.

PHP

 0 and $count < $k)
    {
 
        // getting last digit
        $rem = $p % 10;
 
        // increasing count by 1
        $count++;
 
        // if current number is
        // required digit
        if ($count == $k)
            return $rem;
 
        // remove last digit
        $p = $p / 10;
    }
 
    return 0;
}
 
    // Driver Code
    $a = 5;
    $b = 2;
    $k = 1;
    echo kthdigit($a, $b, $k);
 
// This code is contributed by anuj_67.
?>

Javascript

 
// JavaScript program for finding k-th digit in a^b
 
// To compute k-th digit in a^b
    function kthdigit(a, b, k)
    {
        // Computing a^b
        let p = Math.pow(a, b);
   
        let count = 0;
        while (p > 0 && count < k) {
   
            // Getting last digit
            let rem = p % 10;
   
            // Increasing count by 1
            count++;
   
            // If current number is required digit
            if (count == k)
                return rem;
   
            // Remove last digit
            p = p / 10;
        }
   
        return 0;
    }
       
 
// Driver code
         
        let a = 5, b = 2;
        let k = 1;
        document.write(kthdigit(a, b, k));  
 

输出：

5

如何避免溢出?
我们可以在模10sup> k下找到幂，以避免溢出。在求模的幂后，我们需要返回幂的第一位。