Find power of power under mod of a prime
Given four numbers A, B, C and M, where M is prime number. Our task is to find ABC (mod M).
Example:
Input : A = 2, B = 4, C = 3, M = 23 Output : 6 43 = 64 so, 2^64(mod 23) = 6
A Naive Approach is to calculate res = BC and then calculate Ares % M by modular exponential. The problem of this approach is that we can’t apply directly mod M on BC, so we have to calculate this value without mod M. But if we solve it directly then we will come up with the large value of exponent of A which will definitely overflow in final answer.
An Efficient approach is to reduce the BC to a smaller value by using the Fermat’s Little Theorem, and then apply Modular exponential.
According the Fermat's little a(M - 1) = 1 (mod M) if M is a prime. So if we rewrite BC as x*(M-1) + y, then the task of computing ABC becomes Ax*(M-1) + y which can be written as Ax*(M-1)*Ay. From Fermat's little theorem, we know Ax*(M-1) = 1. So task of computing ABC reduces to computing Ay What is the value of y? From BC = x * (M - 1) + y, y can be written as BC % (M-1) We can easily use the above theorem such that we can get A ^ (B ^ C) % M = (A ^ y ) % M Now we only need to find two things as:- 1. y = (B ^ C) % (M - 1) 2. Ans = (A ^ y) % M
C++
// C++ program to find (a^b) mod m for a large 'a'#include<bits/stdc++.h>using namespace std;// Iterative Function to calculate (x^y)%p in O(log y)unsigned int power(unsigned int x, unsigned int y, unsigned int p){ unsigned int res = 1; // Initialize result x = x % p; // Update x if it is more than or // equal to p while (y > 0) { // If y is odd, multiply x with result if (y & 1) res = (res*x) % p; // y must be even now y = y>>1; // y = y/2 x = (x*x) % p; } return res;}unsigned int Calculate(unsigned int A, unsigned int B, unsigned int C, unsigned int M){ unsigned int res, ans; // Calculate B ^ C (mod M - 1) res = power(B, C, M-1); // Calculate A ^ res ( mod M ) ans = power(A, res, M); return ans;}// Driver program to run the caseint main(){ // M must be be a Prime Number unsigned int A = 3, B = 9, C = 4, M = 19; cout << Calculate(A, B, C, M); return 0;} |
Java
// Java program to find (a^b)// mod m for a large 'a'import java.util.*;class GFG { // Iterative Function to// calculate (x^y)%p in O(log y)static int power(int x, int y, int p) { // Initialize result int res = 1; // Update x if it is more // than or equal to p x = x % p; while (y > 0) { // If y is odd, multiply x with result if (y % 2 != 0) res = (res * x) % p; // y must be even now y = y >> 1; // y = y/2 x = (x * x) % p; } return res;}static int Calculate(int A, int B, int C, int M) { int res, ans; // Calculate B ^ C (mod M - 1) res = power(B, C, M - 1); // Calculate A ^ res ( mod M ) ans = power(A, res, M); return ans;}// Driver codepublic static void main(String[] args) { // M must be be a Prime Number int A = 3, B = 9, C = 4, M = 19; System.out.print(Calculate(A, B, C, M));}}// This code is contributed by Anant Agarwal. |
Python
# Python program to calculate the ansdef calculate(A, B, C, M): # Calculate B ^ C (mod M - 1) res = pow(B, C, M-1) # Calculate A ^ res ( mod M ) ans = pow(A, res, M) return ans# Driver program to run the caseA = 3B = 9C = 4# M must be Prime NumberM = 19print( calculate(A, B, C, M) ) |
C#
// C# program to find (a^b) mod m for// a large 'a'using System;class GFG { // Iterative Function to calculate // (x^y)%p in O(log y) static int power(int x, int y, int p) { // Initialize result int res = 1; // Update x if it is more // than or equal to p x = x % p; while (y > 0) { // If y is odd, multiply x // with result if (y % 2 != 0) res = (res * x) % p; // y must be even now y = y >> 1; // y = y/2 x = (x * x) % p; } return res; } static int Calculate(int A, int B, int C, int M) { int res, ans; // Calculate B ^ C (mod M - 1) res = power(B, C, M - 1); // Calculate A ^ res ( mod M ) ans = power(A, res, M); return ans; } // Driver code public static void Main() { // M must be be a Prime Number int A = 3, B = 9, C = 4, M = 19; Console.Write(Calculate(A, B, C, M)); }}// This code is contributed by nitin mittal. |
PHP
<?php// PHP program to find// (a^b) mod m for a// large 'a'// Iterative Function// to calculate (x^y)%p// in O(log y)function power($x, $y, $p){ $res = 1; // Initialize result $x = $x % $p; // Update x if it // is more than or // equal to p while ($y > 0) { // If y is odd, multiply // x with result if ($y & 1) $res = ($res * $x) % $p; // y must be even now $y = $y >> 1; // y = y/2 $x = ($x * $x) % $p; } return $res;}function Calculate($A, $B, $C, $M){ $res; $ans; // Calculate B ^ C // (mod M - 1) $res = power($B, $C, $M - 1); // Calculate A ^ // res ( mod M ) $ans = power($A, $res, $M); return $ans;}// Driver Code// M must be be// a Prime Number$A = 3; $B = 9;$C = 4; $M = 19;echo Calculate($A, $B, $C, $M);// This code is contributed// by ajit?> |
Javascript
<script>// Javascript program to find (a^b)// mod m for a large 'a' // Iterative Function to // calculate (x^y)%p in O(log y)function power(x, y, p) { // Initialize result let res = 1; // Update x if it is more // than or equal to p x = x % p; while (y > 0) { // If y is odd, multiply x with result if (y % 2 != 0) res = (res * x) % p; // y must be even now y = y >> 1; // y = y/2 x = (x * x) % p; } return res;} function Calculate(A, B, C, M) { let res, ans; // Calculate B ^ C (mod M - 1) res = power(B, C, M - 1); // Calculate A ^ res ( mod M ) ans = power(A, res, M); return ans;}// Driver Code // M must be be a Prime Number let A = 3, B = 9, C = 4, M = 19; document.write(Calculate(A, B, C, M)); </script> |
Output:
18
Time Complexity: O(log(B) + log(C))
Auxiliary space: O(1)
This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
