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 :
- Since number are very large we store them as a string.
- Take last digit in base a.
- 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++ 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 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. |
def last_digit(a, b):
a = int (a)
b = int (b)
# if a and b both are 0
if a = = 0 and b = = 0 :
return 1
# if exponent is 0
if b = = 0 :
return 1
# if base is 0
if a = = 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 b % 4 = = 0 :
res = 4
else :
res = b % 4
# Find last digit in 'a' and compute its exponent
num = pow (a, res)
# Return last digit of num
return num % 10
a = "117"
b = "3"
print (last_digit(a,b))
# This code is contributed by Naimish14. |
// 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 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. ?> |
<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 reviewed by team geeksforgeeks.