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
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
output:
-7 22i
时间复杂度:o(1) t5】辅助空间: o(1)
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