Write a program which reads an integer n and draws a Koch curve based on recursive calles of depth n.
The Koch curve is well known as a kind of fractals.
You can draw a Koch curve in the following algorithm:
Divide a given segment (p1, p2) into three equal segments.
Replace the middle segment by the two sides of an equilateral triangle (s, u, t) of the same length as the segment.
Repeat this procedure recursively for new segments (p1, s), (s, u), (u, t), (t, p2).
You should start (0, 0), (100, 0) as the first segment.
An integer n is given.
OutputPrint each point (x, y) of the Koch curve. Print a point in a line. You should start the point(0, 0), which is the endpoint of the first segment and end with the point (100, 0), the other endpoint so that you can draw the Koch curve as an unbroken line. Each solution should be given as a decimal with an arbitrary number of fractional digits, and with an absolute error of at most 10-4.
Constraints0 ≤ n ≤ 6
Sample Input 11
Sample Output 10.00000000 0.00000000 33.33333333 0.00000000 50.00000000 28.86751346 66.66666667 0.00000000 100.00000000 0.00000000
Sample Input 22
Sample Output 20.00000000 0.00000000 11.11111111 0.00000000 16.66666667 9.62250449 22.22222222 0.00000000 33.33333333 0.00000000 38.88888889 9.62250449 33.33333333 19.24500897 44.44444444 19.24500897 50.00000000 28.86751346 55.55555556 19.24500897 66.66666667 19.24500897 61.11111111 9.62250449 66.66666667 0.00000000 77.77777778 0.00000000 83.33333333 9.62250449 88.88888889 0.00000000 100.00000000 0.00000000
思路koch函数包含三个参数，分别是递归深度n，以及线段的两个端点a,b。
递归函数首先会求出线段a，b的三等分点s，t,然后求能够使得线段as，st，tb组成正三角形的点u。
接下来，函数会顺次进行以下处理： 1.对线段as递归调用koch，输出s的坐标。 2.对线段su递归调用koch，输出u的坐标。 3.对线段ut递归调用koch，输出t的坐标。 4.对线段tb递归调用koch。
s与t的坐标可通过矢量运算求得，以点s为起点，采用旋转矩阵将t逆时针旋转60°，即可得到点u。
code/*
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*/ // Virtual_Judge —— Koch Curve Aizu - ALDS1_5_C .cpp created by VB_KoKing on 2019-05-04:16. /* Procedural objectives：
Variables required by the program:
Procedural thinking：
Functions required by the program:
*/ /* My dear Max said：
"I like you,
So the first bunch of sunshine I saw in the morning is you,
The first gentle breeze that passed through my ear is you,
The first star I see is also you.
The world I see is all your shadow."
FIGHTING FOR OUR FUTURE！！！
*/ #include #include #include using namespace std; struct point { double x, y; }; void koch(int n, point a, point b) { if (n == 0) return; point s, t, u; double th = M_PI * 60.0 / 180.0; s.x = (2.0 * a.x + 1.0 * b.x) / 3.0; s.y = (2.0 * a.y + 1.0 * b.y) / 3.0; t.x = (1.0 * a.x + 2.0 * b.x) / 3.0; t.y = (1.0 * a.y + 2.0 * b.y) / 3.0; u.x = (t.x - s.x) * cos(th) - (t.y - s.y) * sin(th) + s.x; u.y = (t.x - s.x) * sin(th) + (t.y - s.y) * cos(th) + s.y; koch(n - 1, a, s); printf("%.8f %.8f\n", s.x, s.y); koch(n - 1, s, u); printf("%.8f %.8f\n", u.x, u.y); koch(n - 1, u, t); printf("%.8f %.8f\n", t.x, t.y); koch(n - 1, t, b); } int main() { int n; point a, b; cin >> n; a.x = 0; a.y = 0; b.x = 100; b.y = 0; printf("%.8f %.8f\n", a.x, a.y); koch(n, a, b); printf("%.8f %.8f\n", b.x, b.y); return 0; }
