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 :Â
C++
#include <bits/stdc++.h>using namespace std;Â
// 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;Â
    // Return the multiplied value    return {x, y};}Â
Â
// 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;}Â
// Driver codeint main(){Â
    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.*;Â
class GFG {static class pair { Â Â Â Â int first, second; Â Â Â Â public pair(int first, int second) Â Â Â Â { Â Â Â Â Â Â Â Â this.first = first; Â Â Â Â Â Â Â Â this.second = second; Â Â Â Â } } Â
// 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;Â
    // Return the multiplied value    return new pair(x, y);}Â
Â
// 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;}Â
// 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); }}Â
// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the approachÂ
# 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) %MÂ
    # Return the multiplied value    return [x, y]Â
# 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 resÂ
# 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;Â Â Â Â Â class GFG {public class pair { Â Â Â Â public int first, second; Â Â Â Â public pair(int first, int second) Â Â Â Â { Â Â Â Â Â Â Â Â this.first = first; Â Â Â Â Â Â Â Â this.second = second; Â Â Â Â } } Â
// 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;Â
    // Return the multiplied value    return new pair(x, y);}Â
Â
// 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;}Â
// 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); }}Â
// This code is contributed by 29AjayKumar |
Javascript
function pair(first, second) {Â Â Â Â this.first = first;Â Â Â Â this.second = second;}Â
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);}Â
function compPow(complex, k, M) {Â Â Â Â let res = new pair(1, 0);Â Â Â Â while (k > 0) {Â Â Â Â Â Â Â Â if (k % 2 === 1) {Â Â Â Â Â Â Â Â Â Â Â Â res = multiply(res, complex, M);Â Â Â Â Â Â Â Â }Â Â Â Â Â Â Â Â complex = multiply(complex, complex, M);Â Â Â Â Â Â Â Â k = k >> 1;Â Â Â Â }Â Â Â Â return res;}Â
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);Â
// This code is contributed by abn95knd1. |
Output:
25 + i29
Time complexity: O(log k).
Auxiliary Space: O(1)
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