c++中使用复数的几何|集合 2
原文:https://www . geesforgeks . org/geometry-using-complex-numbers-c-set-2/
在浏览了之前的帖子后，我们知道了什么是复数，以及如何使用它们来模拟笛卡尔平面中的点。现在，我们将了解如何在 C++中使用来自 STL 的复杂类。 要使用来自 STL 的复杂类，我们使用#include <复杂>
定义点类 我们可以通过 typedef 复数<双>点来定义我们的点类；在节目开始的时候。点的 X 和 Y 坐标分别是复数的实部和虚部。要访问我们的 X 和 Y 坐标，我们可以使用#定义如下来宏实()和 imag()函数:
# include <complex>
typedef complex<double> point;
# define x real()
# define y imag()
缺点:由于 x 和 y 已经作为宏使用，这些不能作为变量使用。然而，这个缺点并不代表它的许多优点。
卡片打印处理机（Card Print Processor 的缩写）
// CPP program to illustrate
// the definition of point class
#include <iostream>
#include <complex>
using namespace std;
typedef complex<double> point;
// X-coordinate is equivalent to the real part
// Y-coordinate is equivalent to the imaginary part
#define x real()
#define y imag()
int main()
{
point P(2.0, 3.0);
cout << "The X-coordinate of point P is: " << P.x << endl;
cout << "The Y-coordinate of point P is: " << P.y << endl;
return 0;
}
输出:
The X-coordinate of point P is: 2
The Y-coordinate of point P is: 3
关于平面内 P 个单点 P 的属性实现:
- P 的 X 坐标: P.x
- P 的 Y 坐标: P.y
- P 距原点(0，0)的距离: abs(P)
- OP 与 X 轴的夹角，其中 O 是原点: arg(z)
- P 绕原点的旋转: P *极坐标(r，θ)
卡片打印处理机（Card Print Processor 的缩写）
// CPP program to illustrate
// the implementation of single point attributes
#include <iostream>
#include <complex>
using namespace std;
typedef complex<double> point;
#define x real()
#define y imag()
// The constant PI for providing angles in radians
#define PI 3.1415926535897932384626
// Function used to display X and Y coordinates of a point
void displayPoint(point P)
{
cout << "(" << P.x << ", " << P.y << ")" << endl;
}
int main()
{
point P(4.0, 3.0);
// X-Coordinate and Y-coordinate
cout << "The X-coordinate of point P is: " << P.x << endl;
cout << "The Y-coordinate of point P is: " << P.y << endl;
// Distances of P from origin
cout << "The distance of point P from orgin is: " << abs(P) <<endl;
cout << "The squared distance of point P from orgin is: " << norm(P) <<endl;
// Tangent Angle made by OP with the X-Axis
cout << "The angle made by OP with the X-Axis is: "
<< arg(P) << " radians" << endl;
cout << "The angle made by OP with the X-Axis is: "
<< arg(P)*(180/PI) << " degrees" << endl;
// Rotation of P about origin
// The angle of rotation = 90 degrees
point P_rotated = P * polar(1.0, PI/2);
cout<<"The point P on rotating 90 degrees anti-clockwise becomes: P_rotated";
displayPoint(P_rotated);
return 0;
}
输出:
The X-coordinate of point P is: 4
The Y-coordinate of point P is: 3
The distance of point P from orgin is: 5
The squared distance of point P from orgin is: 25
The angle made by OP with the X-Axis is: 0.643501 radians
The angle made by OP with the X-Axis is: 36.8699 degrees
The point P on rotating 90 degrees anti-clockwise becomes: P_rotated(-3, 4)
- 向量加法: P + Q
- 向量减法:P–Q
- 欧氏距离:ABS(P–Q)
- PQ 线斜率:tan(arg(Q–P))
point A = conj(P) * Q
- 点积: A.x
- 叉积大小:绝对值(A.y)
卡片打印处理机（Card Print Processor 的缩写）
// CPP program to illustrate
// the implementation of two point attributes
#include <iostream>
#include <complex>
using namespace std;
typedef complex<double> point;
#define x real()
#define y imag()
// Constant PI for providing angles in radians
#define PI 3.1415926535897932384626
// Function used to display X and Y coordinates of a point
void displayPoint(point P)
{
cout << "(" << P.x << ", " << P.y << ")" << endl;
}
int main()
{
point P(2.0, 3.0);
point Q(3.0, 4.0);
// Addition and Subtraction
cout << "Addition of P and Q is: P+Q"; displayPoint(P+Q);
cout << "Subtraction of P and Q is: P-Q"; displayPoint(P-Q);
// Distances between points P and Q
cout << "The distance between point P ans Q is: " << abs(P-Q) <<endl;
cout << "The squared distance between point P ans Q is: " << norm(P-Q) <<endl;
// Slope of line PQ
cout << "The angle of elevation for line PQ is: "
<< arg(Q-P)*(180/PI) << " degrees" << endl;
cout << "The slope of line PQ is: " << tan(arg(Q-P)) <<endl;
// Construction of point A
point A = conj(P)*Q;
// Dot Product and Cross Product
cout << "The dot product P.Q is: " << A.x << endl;
cout << "The magnitude of cross product PxQ is: " << abs(A.y) << endl;
return 0;
}
输出:
Addition of P and Q is: P+Q(5, 7)
Subtraction of P and Q is: P-Q(-1, -1)
The distance between point P ans Q is: 1.41421
The squared distance between point P ans Q is: 2
The angle of elevation for line PQ is: 45 degrees
The slope of line PQ is: 1
The dot product P.Q is: 18
The magnitude of cross product PxQ is: 1
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