复数的模幂

给定四个整数A ， B ， K ， M 。任务是找到(A + iB) ^K ％M ，这也是一个复数。 A + iB代表复数。

例子：

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

先决条件：模幂

方法：
一种有效的方法类似于单个数字的模幂。在这里，我们有两个数字A，B，而不是一个数字。因此，将一对整数作为参数传递给函数，而不是一个数字。

下面是上述方法的实现：

C++

#include 
using namespace std;
  
// Function to multiply two complex numbers modulo M
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 {x, y};
}
  
  
// Function to calculate the complex modular exponentiation
pair compPow(pair complex, int k, int M)
{
    // Here, res is initialised to (1 + i0)
    pair 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 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

输出：

25 + i29

时间复杂度： O(log k)。