Find Last Digit of a^b for Large Numbers

You are given two integer numbers, the base a (number of digits d, such that 1 <= d <= 1000) and the index b (0 <= b <= 922*10^15). You have to find the last digit of a^b.
Examples: 

Input  : 3 10
Output : 9

Input  : 6 2
Output : 6

Input  : 150 53
Output : 0

After taking few examples, we can notice below pattern.

Number  |  Last digits that repeat in cycle
  1     |     1
  2     |  4, 8, 6, 2
  3     |  9, 7, 1, 3
  4     |  6, 4
  5     |  5
  6     |  6
  7     |  9, 3, 1, 7
  8     |  4, 2, 6, 8
  9     |  1, 9

In the given table, we can see that maximum length for cycle repetition is 4. 
Example: 2*2 = 4*2 = 8*2 = 16*2 = 32 last digit in 32 is 2 that means after multiplying 4 times digit repeat itself. So the algorithm is very simple .
Source : Brilliants.org
Algorithm :

1. Since number are very large we store them as a string.
2. Take last digit in base a.
3. Now calculate b%4. Here b is very large. 
   • If b%4==0 that means b is completely divisible by 4, so our exponent now will be exp = 4 because by multiplying number 4 times, we get the last digit according to cycle table in above diagram.
   • If b%4!=0 that means b is not completely divisible by 4, so our exponent now will be exp=b%4 because by multiplying number exponent times, we get the last digit according to cycle table in above diagram.
   • Now calculate ldigit = pow( last_digit_in_base, exp ).
   • Last digit of a^b will be ldigit%10.

Below is the implementation of above algorithm. 

C++

// C++ code to find last digit of a^b
#include <bits/stdc++.h>
using namespace std;

// Function to find b % a
int Modulo(int a, char b[])
{
    // Initialize result
    int mod = 0;

    // calculating mod of b with a to make
    // b like 0 <= b < a
    for (int i = 0; i < strlen(b); i++)
        mod = (mod * 10 + b[i] - '0') % a;

    return mod; // return modulo
}

// function to find last digit of a^b
int LastDigit(char a[], char b[])
{
    int len_a = strlen(a), len_b = strlen(b);

    // if a and b both are 0
    if (len_a == 1 && len_b == 1 && b[0] == '0' && a[0] == '0')
        return 1;

    // if exponent is 0
    if (len_b == 1 && b[0] == '0')
        return 1;

    // if base is 0
    if (len_a == 1 && a[0] == '0')
        return 0;

    // if exponent is divisible by 4 that means last
    // digit will be pow(a, 4) % 10.
    // if exponent is not divisible by 4 that means last
    // digit will be pow(a, b%4) % 10
    int exp = (Modulo(4, b) == 0) ? 4 : Modulo(4, b);

    // Find last digit in 'a' and compute its exponent
    int res = pow(a[len_a - 1] - '0', exp);

    // Return last digit of result
    return res % 10;
}

// Driver program to run test case
int main()
{
    char a[] = "117", b[] = "3";
    cout << LastDigit(a, b);
    return 0;
}

Java

// Java code to find last digit of a^b
import java.io.*;
import java.math.*;

class GFG {

    // Function to find b % a
    static int Modulo(int a, char b[])
    {
        // Initialize result
        int mod = 0;

        // calculating mod of b with a to make
        // b like 0 <= b < a
        for (int i = 0; i < b.length; i++)
            mod = (mod * 10 + b[i] - '0') % a;

        return mod; // return modulo
    }

    // Function to find last digit of a^b
    static int LastDigit(char a[], char b[])
    {
        int len_a = a.length, len_b = b.length;

        // if a and b both are 0
        if (len_a == 1 && len_b == 1 && b[0] == '0' && a[0] == '0')
            return 1;

        // if exponent is 0
        if (len_b == 1 && b[0] == '0')
            return 1;

        // if base is 0
        if (len_a == 1 && a[0] == '0')
            return 0;

        // if exponent is divisible by 4 that means last
        // digit will be pow(a, 4) % 10.
        // if exponent is not divisible by 4 that means last
        // digit will be pow(a, b%4) % 10
        int exp = (Modulo(4, b) == 0) ? 4 : Modulo(4, b);

        // Find last digit in 'a' and compute its exponent
        int res = (int)(Math.pow(a[len_a - 1] - '0', exp));

        // Return last digit of result
        return res % 10;
    }

    // Driver program to run test case
    public static void main(String args[]) throws IOException
    {
        char a[] = "117".toCharArray(), b[] = { '3' };
        System.out.println(LastDigit(a, b));
    }
}

// This code is contributed by Nikita Tiwari.

# Python 3 code to find last digit of a ^ b

import math

# Function to find b % a
def Modulo(a, b) :
    # Initialize result
    mod = 0

    # calculating mod of b with a to make
    # b like 0 <= b < a
    for i in range(0, len(b)) :
        mod = (mod * 10 + (int)(b[i])) % a

    return mod # return modulo


# function to find last digit of a ^ b
def LastDigit(a, b) :
    len_a = len(a)
    len_b = len(b)

    # if a and b both are 0
    if (len_a == 1 and len_b == 1 and b[0] == '0' and a[0] == '0') :
        return 1

    # if exponent is 0
    if (len_b == 1 and b[0]=='0') :
        return 1

    # if base is 0
    if (len_a == 1 and a[0] == '0') :
        return 0

    # if exponent is divisible by 4 that means last
    # digit will be pow(a, 4) % 10.
    # if exponent is not divisible by 4 that means last
    # digit will be pow(a, b % 4) % 10
    if((Modulo(4, b) == 0)) :
        exp = 4
    else : 
        exp = Modulo(4, b)

    # Find last digit in 'a' and compute its exponent
    res = math.pow((int)(a[len_a - 1]), exp)

    # Return last digit of result
    return res % 10
    

# Driver program to run test case
a = ['1', '1', '7']
b = ['3']
print(LastDigit(a, b))

# This code is contributed to Nikita Tiwari.

C#

// C# code to find last digit of a^b.
using System;

class GFG {

    // Function to find b % a
    static int Modulo(int a, char[] b)
    {

        // Initialize result
        int mod = 0;

        // calculating mod of b with a
        // to make b like 0 <= b < a
        for (int i = 0; i < b.Length; i++)
            mod = (mod * 10 + b[i] - '0') % a;

        // return modulo
        return mod;
    }

    // Function to find last digit of a^b
    static int LastDigit(char[] a, char[] b)
    {
        int len_a = a.Length, len_b = b.Length;

        // if a and b both are 0
        if (len_a == 1 && len_b == 1 &&
                   b[0] == '0' && a[0] == '0')
            return 1;

        // if exponent is 0
        if (len_b == 1 && b[0] == '0')
            return 1;

        // if base is 0
        if (len_a == 1 && a[0] == '0')
            return 0;

        // if exponent is divisible by 4
        // that means last digit will be
        // pow(a, 4) % 10. if exponent is
        //not divisible by 4 that means last
        // digit will be pow(a, b%4) % 10
        int exp = (Modulo(4, b) == 0) ? 4
                            : Modulo(4, b);

        // Find last digit in 'a' and
        // compute its exponent
        int res = (int)(Math.Pow(a[len_a - 1]
                                - '0', exp));

        // Return last digit of result
        return res % 10;
    }

    // Driver program to run test case
    public static void Main()
    {

        char[] a = "117".ToCharArray(),
        b = { '3' };

        Console.Write(LastDigit(a, b));
    }
}

// This code is contributed by nitin mittal.

PHP

<?php
// php code to find last digit of a^b

// Function to find b % a
function Modulo($a, $b)
{

    // Initialize result
    $mod = 0;

    // calculating mod of b with a to make
    // b like 0 <= b < a
    for ($i = 0; $i < strlen($b); $i++)
        $mod = ($mod * 10 + $b[$i] - '0') % $a;

    return $mod; // return modulo
}

// function to find last digit of a^b
function LastDigit($a, $b)
{
    $len_a = strlen($a); $len_b = strlen($b);

    // if a and b both are 0
    if ($len_a == 1 && $len_b == 1 &&
                $b[0] == '0' && $a[0] == '0')
        return 1;

    // if exponent is 0
    if ($len_b == 1 && $b[0] == '0')
        return 1;

    // if base is 0
    if ($len_a == 1 && $a[0] == '0')
        return 0;

    // if exponent is divisible by 4 that
    // means last digit will be pow(a, 4)
    // % 10. if exponent is not divisible
    // by 4 that means last digit will be
    // pow(a, b%4) % 10
    $exp = (Modulo(4, $b) == 0) ? 4 :
                              Modulo(4, $b);

    // Find last digit in 'a' and compute
    // its exponent
    $res = pow($a[$len_a - 1] - '0', $exp);

    // Return last digit of result
    return $res % 10;
}

// Driver program to run test case
$a = "117";
$b = "3";
echo LastDigit($a, $b);

// This code is contributed by nitin mittal.
?>

Javascript

<script>

// Javascript code to find last digit of a^b

// Function to find b % a
function Modulo(a, b)
{

    // Initialize result
    let mod = 0;

    // calculating mod of b with a to make
    // b like 0 <= b < a
    for (let i = 0; i < b.length; i++)
        mod = (mod * 10 + b[i] - '0') % a;

    return mod; // return modulo
}

// function to find last digit of a^b
function LastDigit(a, b)
{
    let len_a = a.length;
    let len_b = b.length;

    // if a and b both are 0
    if (len_a == 1 && len_b == 1 &&
                b[0] == '0' && a[0] == '0')
        return 1;

    // if exponent is 0
    if (len_b == 1 && b[0] == '0')
        return 1;

    // if base is 0
    if (len_a == 1 && a[0] == '0')
        return 0;

    // if exponent is divisible by 4 that
    // means last digit will be pow(a, 4)
    // % 10. if exponent is not divisible
    // by 4 that means last digit will be
    // pow(a, b%4) % 10
    exp = (Modulo(4, b) == 0) ? 4 :
                            Modulo(4, b);

    // Find last digit in 'a' and compute
    // its exponent
    res = Math.pow(a[len_a - 1] - '0', exp);

    // Return last digit of result
    return res % 10;
}

// Driver program to run test case
let a = "117";
let b = "3";
document.write(LastDigit(a, b));

// This code is contributed by _saurabh_jaiswal

</script>

Output : 

3

This article is contributed by Shashank Mishra ( Gullu ). This article is reviewed by team geeksforgeeks. 
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