使用三次乘法运算的复数乘积
给定四个整数 a、b、c 和 d ，它们表示形式为 (a + bi) 和 (c + di) 的两个复数，任务是仅使用三次乘法运算来找到给定复数的乘积。 示例:
输入: a = 2，b = 3，c = 4，d = 5 输出: -7 + 22i 说明: 产品给出为: (2+3i)(4+5i)= 2 * 4+4 * 3i+2 * 5i+3 * 5 (1) = 8–15+(12+10)I =-7+22i【T10
天真方法:天真方法是将给定的两个复数直接相乘，如下所示:
= >(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
上述运算需要四次乘法才能求出两个复数的乘积。 高效方法:上述方法需要四次乘法才能找到乘积。可以简化为三次乘法如下: 两个复数的乘法如下:
(a+bi)*(c+di)= a * c–b * d+(a * d+b * c)I
简化实数部分:
实部= a * c–b * d 让 prod1 = ac，prod 2 = b * d 因此，实部= prod 1–prod 2*
将虚部简化如下:
虚数部分= ad + bc 在上面的想象部分中加减 ac 和 bd 我们有， 虚数部分= a * c–a * c+a * d+b * c+b * d–b * d， 关于重新排列我们得到的项， =>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 让 prod3 = (a + b)(c + d) 然后虚部由prod 3 –( prod 1+prod)给出
因此，我们需要找到 prod1 = a * c 、 prod2 = b * d 、 prod3 = ( a + b ) * ( c + d ) 的值。 那么，我们的最终答案将是:
实部= prod 1–prod 2 虚部= prod 3 –( prod 1+prod 2)
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<script>
// Javascript program for the above approach
// Function to multiply Complex
// Numbers with just three
// multiplications
function print_product( a, b, c, d)
{
// Find value of prod1, prod2 and prod3
let prod1 = a * c;
let prod2 = b * d;
let prod3 = (a + b) * (c + d);
// Real Part
let real = prod1 - prod2;
// Imaginary Part
let imag = prod3 - (prod1 + prod2);
// Print the result
document.write(real + " + " + imag + "i");
}
// Driver Code
let a, b, c, d;
// Given four Numbers
a = 2;
b = 3;
c = 4;
d = 5;
// Function Call
print_product(a, b, c, d);
</script>
Output:
-7 + 22i
时间复杂度:O(1) T5】辅助空间: O(1)
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