使用三个乘法运算的复数乘积

给定四个表示形式(a + bi)和(c + di)的两个复数的整数a，b，c和d ，任务是仅使用三个乘法运算来找到给定复数的乘积。

例子：

Input: a = 2, b = 3, c = 4 and d = 5
Output: -7 + 22i
Explanation:
Product is given by:
(2 + 3i)*(4 + 5i) = 2*4 + 4*3i + 2*5i + 3*5*(-1)
= 8 – 15 + (12 + 10)i
= -7 + 22i

Input: a = 3, b = 7, c = 6 and d = 2
Output: 4 + 48i

天真的方法：天真的方法是将给定的两个复数直接相乘为：

=> (a + bi)*(c + di)
=> a(c + di) + b*i(c + di)
=> a*c + ad*i + b*c*i + b*d*i*i
=> (a*c – b*d) + (a*d + b*c)*i

上述操作将需要四次乘法才能找到两个复数的乘积。

高效方法：上述方法需要四次乘法才能找到乘积。它可以简化为三个乘法：

两个复数的相乘如下：

(a + bi)*(c + di) = a*c – b*d + (a*d + b*c)i

简化实部：

real part = a*c – b*d
Let prod1 = a*c and prod2 = b*d.
Thus, real part = prod1 – prod2

简化虚构部分，如下所示：

imaginary part = a*d + b*c

Adding and subtracting a*c and b*d in the above imaginar part we have,

imaginary part = a*c – a*c + a*d + b*c + b*d – b*d,
On rearranging the terms we get,
=> a*b + b*c + a*d + b*d – a*c – b*d
=> (a + b)*c + (a + b)*d – a*c – b*d
=> (a + b)*(c + d) – a*c – b*d

Let prod3 = (a + b)*(c + d)
Then the imaginary part is given by prod3 – (prod1 + prod2).

因此，我们需要找到prod1 = a * c ， prod2 = b * d和prod3 =(a + b)*(c + d)的值。

因此，我们的最终答案将是：

Real Part = prod1 – prod2
Imaginary Part = prod3 – (prod1 + prod2)

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
  
// Function to multiply Complex
// Numbers with just three
// multiplications
void print_product(int a, int b,
                   int c, int d)
{
    // Find value of prod1, prod2 and prod3
    int prod1 = a * c;
    int prod2 = b * d;
    int prod3 = (a + b) * (c + d);
  
    // Real Part
    int real = prod1 - prod2;
  
    // Imaginary Part
    int imag = prod3 - (prod1 + prod2);
  
    // Print the result
    cout << real << " + " << imag << "i";
}
  
// Driver Code
int main()
{
    int a, b, c, d;
  
    // Given four Numbers
    a = 2;
    b = 3;
    c = 4;
    d = 5;
  
    // Function Call
    print_product(a, b, c, d);
    return 0;
}

Java

// Java program for the above approach
class GFG{
  
// Function to multiply Complex
// Numbers with just three
// multiplications
static void print_product(int a, int b,
                          int c, int d)
{
      
    // Find value of prod1, prod2 and prod3
    int prod1 = a * c;
    int prod2 = b * d;
    int prod3 = (a + b) * (c + d);
  
    // Real Part
    int real = prod1 - prod2;
  
    // Imaginary Part
    int imag = prod3 - (prod1 + prod2);
  
    // Print the result
    System.out.println(real + " + " +
                       imag + "i");
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given four numbers
    int a = 2;
    int b = 3;
    int c = 4;
    int d = 5;
      
    // Function call
    print_product(a, b, c, d);
}
}
  
// This code is contributed by Pratima Pandey

Python3

# Python3 program for the above approach 
  
# Function to multiply Complex 
# Numbers with just three 
# multiplications 
def print_product(a, b, c, d):
      
    # Find value of prod1, prod2 
    # and prod3
    prod1 = a * c
    prod2 = b * d
    prod3 = (a + b) * (c + d)
  
    # Real part
    real = prod1 - prod2
  
    # Imaginary part
    imag = prod3 - (prod1 + prod2)
  
    # Print the result
    print(real, " + ", imag, "i") 
  
# Driver code
  
# Given four numbers
a = 2
b = 3
c = 4
d = 5
  
# Function call
print_product(a, b, c, d) 
  
# This code is contributed by Vishal Maurya.

C#

// C# program for the above approach
using System;
class GFG{
  
// Function to multiply Complex
// Numbers with just three
// multiplications
static void print_product(int a, int b,
                          int c, int d)
{
    // Find value of prod1, prod2 and prod3
    int prod1 = a * c;
    int prod2 = b * d;
    int prod3 = (a + b) * (c + d);
  
    // Real Part
    int real = prod1 - prod2;
  
    // Imaginary Part
    int imag = prod3 - (prod1 + prod2);
  
    // Print the result
    Console.Write(real + " + " + imag + "i");
}
  
// Driver Code
public static void Main()
{
    int a, b, c, d;
  
    // Given four Numbers
    a = 2;
    b = 3;
    c = 4;
    d = 5;
  
    // Function Call
    print_product(a, b, c, d);
}
}
  
// This code is contributed by Code_Mech

输出：

-7 + 22i

时间复杂度： O(1)
辅助空间： O(1)