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++
// 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; } |
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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. |
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Python3
# 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. |
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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. |
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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. ?> |
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Output :
3
This article is contributed by Shashank Mishra ( Gullu ). This article is reviewed by team geeksforgeeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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Improved By : nitin mittal
