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 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 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. |
Python3
# 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)
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