Dados cuatro enteros A , B , K , M . La tarea es encontrar (A + iB) K % M que también es un número complejo. A + iB representa un número complejo. Ejemplos:
Entrada: A = 2, B = 3, K = 4, M = 5 Salida: 1 + i*0 Entrada: A = 7, B = 3, K = 10, M = 97 Salida: 25 + i*29
Requisito previo : Enfoque de exponenciación modular : un enfoque eficiente es similar a la exponenciación modular de un solo número. Aquí, en lugar de uno solo, tenemos dos números A, B. Entonces, pase un par de enteros como parámetro a la función en lugar de un solo número. A continuación se muestra la implementación del enfoque anterior:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to multiply two complex numbers modulo M
pair<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 exponentiation
pair<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 code
int 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 approach
import 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 M
static 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 exponentiation
static 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 code
public 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 M def 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 exponentiation def 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 code if __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 approach
using 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 M
static 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 exponentiation
static 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 code
public 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
Producción:
25 + i29
Complejidad temporal: O(log k).
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por greenindia y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA