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.
Input
An integer n is given.
Output
Print 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.
Constraints
0 ≤ n ≤ 6
Sample Input 1
1
Sample Output 1
0.00000000 0.00000000
33.33333333 0.00000000
50.00000000 28.86751346
66.66666667 0.00000000
100.00000000 0.00000000
Sample Input 2
2
Sample Output 2
0.00000000 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
/*
^....0
^ .1 ^1^
.. 01
1.^ 1.0
^ 1 ^ ^0.1
1 ^ ^..^
0. ^ 0^
.0 1 .^
.1 ^0 .........001^
.1 1. .111100....01^
00 ^ 11^ ^1. .1^
1.^ ^0 0^
.^ ^0..1
.1 1..^
1 .0 ^ ^
00. ^^0.^
^ 0 ^^110.^
0 0 ^ ^^^10.01
^^ 10 1 1 ^^^1110.1
01 10 1.1 ^^^1111110
010 01 ^^ ^^^1111^1.^ ^^^
10 10^ 0^ 1 ^^111^^^0.1^ 1....^
11 0 ^^11^^^ 0.. ....1^ ^ ^
1. 0^ ^11^^^ ^ 1 111^ ^ 0.
10 00 11 ^^^^^ 1 0 1.
0^ ^0 ^0 ^^^^ 0 0.
0^ 1.0 .^ ^^^^ 1 1 .0
^.^ ^^ 0^ ^1 ^^^^ 0. ^.1
1 ^ 11 1. ^^^ ^ ^ ..^
^..^ ^1 ^.^ ^^^ .0 ^.0
0..^ ^0 01 ^^^ .. 0..^
1 .. .1 ^.^ ^^^ 1 ^ ^0001
^ 1. 00 0. ^^^ ^.0 ^.1
. 0^. ^.^ ^.^ ^^^ ..0.0
1 .^^. .^ 1001 ^^ ^^^ . 1^
. ^ ^. 11 0. 1 ^ ^^ 0.
0 ^. 0 ^0 1 ^^^ 0.
0.^ 1. 0^ 0 .1 ^^^ ..
.1 1. 00 . .1 ^^^ ..
1 1. ^. 0 .^ ^^ ..
0. 1. .^ . 0 .
.1 1. 01 . . ^ 0
^.^ 00 ^0 1. ^ 1 1
.0 00 . ^^^^^^ .
.^ 00 01 ..
1. 00 10 1 ^
^.1 00 ^. ^^^ .1
.. 00 .1 1..01 ..
1.1 00 1. ..^ 10
^ 1^ 00 ^.1 0 1 1
.1 00 00 ^ 1 ^
. 00 ^.^ 10^ ^^
1.1 00 00 10^
..^ 1. ^. 1.
0 1 ^. 00 00 .^
^ ^. ^ 1 00 ^0000^ ^ 01
1 0 ^. 00.0^ ^00000 1.00.1 11
. 1 0 1^^0.01 ^^^ 01
.^ ^ 1 1^^ ^.^
1 1 0.
.. 1 ^
1 1
^ ^ .0
1 ^ 1
.. 1.1 ^0.0
^ 0 1..01^^100000..0^
1 1 ^ 1 ^^1111^ ^^
0 ^ ^ 1 1000^
.1 ^.^ . 00
.. 1.1 0. 0
1. . 1. .^
1. 1 1. ^0
^ . ^.1 00 01
^.0 001. .^
*/
// 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 <iostream>
#include <cstdio>
#include <cmath>
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
", s.x, s.y);
koch(n - 1, s, u);
printf("%.8f %.8f
", u.x, u.y);
koch(n - 1, u, t);
printf("%.8f %.8f
", 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
", a.x, a.y);
koch(n, a, b);
printf("%.8f %.8f
", b.x, b.y);
return 0;
}