将作为字符串给出的 N 个复数相乘
原文:https://www . geesforgeks . org/multiply-n-complex-numbers-以字符串形式给出/
给定字符串形式的 N 复数，任务是打印这些 N 复数的乘法。
示例:
输入: N = 3，V = { 3 + 1i，2 + 1i，5 + -7i } 输出: 10+-60i 说明: 首先我们将(3 + 1i)和(2+1i)相乘得到 5+5i。下一步，我们将 5+5i 和-5+-7i 相乘，得到最终结果 10+-60i 。
输入: N = 3，V = {“7+4i”、“-12+1i”、“-16+-7i”、“12+18i”} 输出: -9444+35442i
接近
- 首先，从头开始迭代，取前两个字符串，并删除它们。
- 接下来，将字符串转换为带有适当符号的数字。将字符串的实部和虚部存储在单独的变量中。取 a 为第一弦的实部， b 为第一弦的虚部。取 c 为第二弦的实部， d 为第二弦的虚部。
- 接下来，我们将通过计算 a 和 c 的乘积，并将其与 b 的乘积相减，再与 b 和 d 的乘积相减，来计算实数的合成值。对于虚部，我们将计算 a 和 d 的乘积之和，以及 b 和 c 的乘积。
- 然后，我们将生成一个临时字符串来获取之前计算的实值和虚值之和。
- 我们将把结果字符串推进向量。重复以上步骤，直到向量中只剩下一个元素。
- 返回向量中最后一个剩余的元素，这就是想要的答案。
下面是我们上述方法的实现:
C++
// C++ Program to multiply
// N complex Numbers
#include <bits/stdc++.h>
using namespace std;
#define ll long long
// Function which returns the
// string in digit format
vector<long long int> findnum(string s1)
{
vector<long long int> v;
// a : real
// b : imaginary
int a = 0, b = 0;
int sa = 0, sb = 0, i = 0;
// sa : sign of a
// sb : sign of b
if (s1[0] == '-') {
sa = 1;
i = 1;
}
// Extract the real number
while (isdigit(s1[i])) {
a = a * 10 + (int(s1[i]) - 48);
i++;
}
if (s1[i] == '+') {
sb = 0;
i += 1;
}
if (s1[i] == '-') {
sb = 1;
i += 1;
}
// Extract the imaginary part
while (i < s1.length() && isdigit(s1[i])) {
b = b * 10 + (int(s1[i]) - 48);
i++;
}
if (sa)
a *= -1;
if (sb)
b *= -1;
v.push_back(a);
v.push_back(b);
return v;
}
string complexNumberMultiply(vector<string> v)
{
// if size==1 means we reached at result
while (v.size() != 1) {
// Extract the first two elements
vector<ll> v1 = findnum(v[0]);
vector<ll> v2 = findnum(v[1]);
// Remove them
v.erase(v.begin());
v.erase(v.begin());
// Calculate and store the real part
ll r = (v1[0] * v2[0] - v1[1] * v2[1]);
// Calculate and store the imaginary part
ll img = v1[0] * v2[1] + v1[1] * v2[0];
string res = "";
// Append the real part
res += to_string(r);
res += '+';
// Append the imaginary part
res += to_string(img) + 'i';
// Insert into vector
v.insert(v.begin(), res);
}
return v[0];
}
// Driver Function
int main()
{
int n = 3;
vector<string> v = { "3+1i",
"2+1i", "-5+-7i" };
cout << complexNumberMultiply(v) << "\n";
return 0;
}
Python 3
# Python3 program to multiply
# N complex Numbers
# Function which returns the
# in digit format
def findnum(s1):
v = []
# a : real
# b : imaginary
a = 0
b = 0
sa = 0
sb = 0
i = 0
# sa : sign of a
# sb : sign of b
if (s1[0] == '-'):
sa = 1
i = 1
# Extract the real number
while (s1[i].isdigit()):
a = a * 10 + (int(s1[i]))
i += 1
if (s1[i] == '+'):
sb = 0
i += 1
if (s1[i] == '-'):
sb = 1
i += 1
# Extract the imaginary part
while (i < len(s1) and s1[i].isdigit()):
b = b * 10 + (int(s1[i]))
i += 1
if (sa):
a *= -1
if (sb):
b *= -1
v.append(a)
v.append(b)
return v
def complexNumberMultiply(v):
# If size==1 means we reached at result
while (len(v) != 1):
# Extract the first two elements
v1 = findnum(v[0])
v2 = findnum(v[1])
# Remove them
del v[0]
del v[0]
# Calculate and store the real part
r = (v1[0] * v2[0] - v1[1] * v2[1])
# Calculate and store the imaginary part
img = v1[0] * v2[1] + v1[1] * v2[0]
res = ""
# Append the real part
res += str(r)
res += '+'
# Append the imaginary part
res += str(img) + 'i'
# Insert into vector
v.insert(0, res)
return v[0]
# Driver code
if __name__ == '__main__':
n = 3
v = [ "3+1i", "2+1i", "-5+-7i" ]
print(complexNumberMultiply(v))
# This code is contributed by mohit kumar 29
Output:
10+-60i
时间复杂度:O(N) T5】辅助空间: O(N)
版权属于：月萌API www.moonapi.com，转载请注明出处
