Given four integers A, B, K, M. The task is to find (A + iB)K % M which is a complex number too. A + iB represents a complex number. Examples:
Input : A = 2, B = 3, K = 4, M = 5 Output: 1 + i*0 Input : A = 7, B = 3, K = 10, M = 97 Output: 25 + i*29
Prerequisite: Modular Exponentiation Approach: An efficient approach is similar to the modular exponentiation of a single number. Here, instead of a single we have two number A, B. So, pass a pair of integers as a parameter to the function instead of a single number. Below is the implementation of the above approach :
C++
#include <bits/stdc++.h>using namespace std;// Function to multiply two complex numbers modulo Mpair<int, int> Multiply (pair<int, int> p, pair<int, int> q, int M){ // Multiplication of two complex numbers is // (a + ib)(c + id) = (ac - bd) + i(ad + bc) int x = ((p.first * q.first) % M - (p.second * q.second) % M + M) % M; int y = ((p.first * q.second) % M + (p.second * q.first) % M) %M; // Return the multiplied value return {x, y};}// Function to calculate the complex modular exponentiationpair<int, int> compPow(pair<int, int> complex, int k, int M){ // Here, res is initialised to (1 + i0) pair<int, int> res = { 1, 0 }; while (k > 0) { // If k is odd if (k & 1) { // Multiply 'complex' with 'res' res = Multiply(res, complex, M); } // Make complex as complex*complex complex = Multiply(complex, complex, M); // Make k as k/2 k = k >> 1; } //Return the required answer return res;}// Driver codeint main(){ int A = 7, B = 3, k = 10, M = 97; // Function call pair<int, int> ans = compPow({A, B}, k, M); cout << ans.first << " + i" << ans.second; return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {static class pair { int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } // Function to multiply two complex numbers modulo Mstatic pair Multiply (pair p, pair q, int M){ // Multiplication of two complex numbers is // (a + ib)(c + id) = (ac - bd) + i(ad + bc) int x = ((p.first * q.first) % M - (p.second * q.second) % M + M) % M; int y = ((p.first * q.second) % M + (p.second * q.first) % M) % M; // Return the multiplied value return new pair(x, y);}// Function to calculate the // complex modular exponentiationstatic pair compPow(pair complex, int k, int M){ // Here, res is initialised to (1 + i0) pair res = new pair(1, 0 ); while (k > 0) { // If k is odd if (k % 2 == 1) { // Multiply 'complex' with 'res' res = Multiply(res, complex, M); } // Make complex as complex*complex complex = Multiply(complex, complex, M); // Make k as k/2 k = k >> 1; } // Return the required answer return res;}// Driver codepublic static void main(String[] args){ int A = 7, B = 3, k = 10, M = 97; // Function call pair ans = compPow(new pair(A, B), k, M); System.out.println(ans.first + " + i" + ans.second); }}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the approach# Function to multiply two complex numbers modulo Mdef Multiply (p, q, M): # Multiplication of two complex numbers is # (a + ib)(c + id) = (ac - bd) + i(ad + bc) x = ((p[0] * q[0]) % M - \ (p[1] * q[1]) % M + M) % M y = ((p[0] * q[1]) % M + \ (p[1] * q[0]) % M) %M # Return the multiplied value return [x, y]# Function to calculate the# complex modular exponentiationdef compPow(complex, k, M): # Here, res is initialised to (1 + i0) res = [1, 0] while (k > 0): # If k is odd if (k & 1): # Multiply 'complex' with 'res' res = Multiply(res, complex, M) # Make complex as complex*complex complex = Multiply(complex, complex, M) # Make k as k/2 k = k >> 1 # Return the required answer return res# Driver codeif __name__ == '__main__': A = 7 B = 3 k = 10 M = 97 # Function call ans = compPow([A, B], k, M) print(ans[0], "+ i", end = "") print(ans[1]) # This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approachusing System; class GFG {public class pair { public int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } // Function to multiply two complex numbers modulo Mstatic pair Multiply (pair p, pair q, int M){ // Multiplication of two complex numbers is // (a + ib)(c + id) = (ac - bd) + i(ad + bc) int x = ((p.first * q.first) % M - (p.second * q.second) % M + M) % M; int y = ((p.first * q.second) % M + (p.second * q.first) % M) % M; // Return the multiplied value return new pair(x, y);}// Function to calculate the // complex modular exponentiationstatic pair compPow(pair complex, int k, int M){ // Here, res is initialised to (1 + i0) pair res = new pair(1, 0 ); while (k > 0) { // If k is odd if (k % 2 == 1) { // Multiply 'complex' with 'res' res = Multiply(res, complex, M); } // Make complex as complex*complex complex = Multiply(complex, complex, M); // Make k as k/2 k = k >> 1; } // Return the required answer return res;}// Driver codepublic static void Main(String[] args){ int A = 7, B = 3, k = 10, M = 97; // Function call pair ans = compPow(new pair(A, B), k, M); Console.WriteLine(ans.first + " + i" + ans.second); }}// This code is contributed by 29AjayKumar |
Javascript
function pair(first, second) { this.first = first; this.second = second;}function multiply(p, q, M) { // Multiplication of two complex numbers is // (a + ib)(c + id) = (ac - bd) + i(ad + bc) let x = ((p.first * q.first) % M - (p.second * q.second) % M + M) % M; let y = ((p.first * q.second) % M + (p.second * q.first) % M) % M; return new pair(x, y);}function compPow(complex, k, M) { let res = new pair(1, 0); while (k > 0) { if (k % 2 === 1) { res = multiply(res, complex, M); } complex = multiply(complex, complex, M); k = k >> 1; } return res;}let A = 7, B = 3, k = 10, M = 97;let ans = compPow(new pair(A, B), k, M);console.log(ans.first + " + i" + ans.second);// This code is contributed by abn95knd1. |
Output:
25 + i29
Time complexity: O(log k).
Auxiliary Space: O(1)
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