圣诞节来了，教你用各类编程语言手写一棵圣诞树|python|圣诞树|方块|源代码|编程语言

时间已到深冬，今晚就是平安夜了，你会给你的男/女朋友送上一颗平安果吗？

作为一名科技宅，很抱歉不能给每一位小伙伴都送上苹果，那就送给大家一个技能吧，用各种编程语言手写一棵圣诞树，祈福各位关注我的小伙伴都平平安安！

C/C++

如何用 C 语言画一个“圣诞树”，就不介绍太简单的方法了（for循环打印几个' * '）。直接看网上大神写的圣诞树，有没有很不错的感觉！

下方是C语言源代码，可以自行编译运行。

#include

#define PI 3.14159265359float sx, sy;float sdCircle(float px, float py, float r) { float dx = px - sx, dy = py - sy; return sqrtf(dx * dx + dy * dy) - r;}float opUnion(float d1, float d2) { return d1 < d2 ? d1 : d2;}#define T px + scale * r * cosf(theta), py + scale * r * sin(theta)float f(float px, float py, float theta, float scale, int n) { float d = 0.0f; for (float r = 0.0f; r < 0.8f; r += 0.02f) d = opUnion(d, sdCircle(T, 0.05f * scale * (0.95f - r))); if (n > 0) for (int t = -1; t <= 1; t += 2) { float tt = theta + t * 1.8f; float ss = scale * 0.9f; for (float r = 0.2f; r < 0.8f; r += 0.1f) { d = opUnion(d, f(T, tt, ss * 0.5f, n - 1)); ss *= 0.8f; } } return d;}int main(int argc, char* argv[]) { int n = argc > 1 ? atoi(argv[1]) : 3; for (sy = 0.8f; sy > 0.0f; sy -= 0.02f, putchar('\n')) for (sx = -0.35f; sx < 0.35f; sx += 0.01f) putchar(f(0, 0, PI * 0.5f, 1.0f, n) < 0 ? '*' : ' ');}

由于原来圣诞树的叶子和枝干都是由' * '来代替，风格太过单一。经过作者的优化，成功升级为3D效果，还是蛮炫酷的。由于较过多的浮点运算，使得运行时间较长，运行下面的程序花了近1060s ，动手能力强的小伙伴可以尝试优化一下。

下方是源代码：

#include

#define PI 3.14159265359ffloat sx, sy;typedef float Mat[4][4];typedef float Vec[4];void scale(Mat* m, float s) { Mat temp = { {s,0,0,0}, {0,s,0,0 }, { 0,0,s,0 }, { 0,0,0,1 } }; memcpy(m, &temp, sizeof(Mat));}void rotateY(Mat* m, float t) { float c = cosf(t), s = sinf(t); Mat temp = { {c,0,s,0}, {0,1,0,0}, {-s,0,c,0}, {0,0,0,1} }; memcpy(m, &temp, sizeof(Mat));}void rotateZ(Mat* m, float t) { float c = cosf(t), s = sinf(t); Mat temp = { {c,-s,0,0}, {s,c,0,0}, {0,0,1,0}, {0,0,0,1} }; memcpy(m, &temp, sizeof(Mat));}void translate(Mat* m, float x, float y, float z) { Mat temp = { {1,0,0,x}, {0,1,0,y}, {0,0,1,z}, {0,0,0,1} }; memcpy(m, &temp, sizeof(Mat));}void mul(Mat* m, Mat a, Mat b) { Mat temp; for (int j = 0; j < 4; j++) for (int i = 0; i < 4; i++) { temp[j][i] = 0.0f; for (int k = 0; k < 4; k++) temp[j][i] += a[j][k] * b[k][i]; } memcpy(m, &temp, sizeof(Mat)); }void transformPosition(Vec* r, Mat m, Vec v) { Vec temp = { 0, 0, 0, 0 }; for (int j = 0; j < 4; j++) for (int i = 0; i < 4; i++) temp[j] += m[j][i] * v[i]; memcpy(r, &temp, sizeof(Vec)); }float transformLength(Mat m, float r) { return sqrtf(m[0][0] * m[0][0] + m[0][1] * m[0][1] + m[0][2] * m[0][2]) * r;}float sphere(Vec c, float r) { float dx = c[0] - sx, dy = c[1] - sy; float a = dx * dx + dy * dy; return a < r * r ? sqrtf(r * r - a) + c[2] : -1.0f;}float opUnion(float z1, float z2) { return z1 > z2 ? z1 : z2;}float f(Mat m, int n) { float z = -1.0f; for (float r = 0.0f; r < 0.8f; r += 0.02f) { Vec v = { 0.0f, r, 0.0f, 1.0f }; transformPosition(&v, m, v); z = opUnion(z, sphere(v, transformLength(m, 0.05f * (0.95f - r)))); } if (n > 0) { Mat ry, rz, s, t, m2, m3; rotateZ(&rz, 1.8f); for (int p = 0; p < 6; p++) { rotateY(&ry, p * (2 * PI / 6)); mul(&m2, ry, rz); float ss = 0.45f; for (float r = 0.2f; r < 0.8f; r += 0.1f) { scale(&s, ss); translate(&t, 0.0f, r, 0.0f); mul(&m3, s, m2); mul(&m3, t, m3); mul(&m3, m, m3); z = opUnion(z, f(m3, n - 1)); ss *= 0.8f; } } } return z;}float f0(float x, float y, int n) { sx = x; sy = y; Mat m; scale(&m, 1.0f); return f(m, n);}int main(int argc, char* argv[]) { int n = argc > 1 ? atoi(argv[1]) : 3; float zoom = argc > 2 ? atof(argv[2]) : 1.0f; for (float y = 0.8f; y > -0.0f; y -= 0.02f / zoom, putchar('\n')) for (float x = -0.35f; x < 0.35f; x += 0.01f / zoom) { float z = f0(x, y, n); if (z > -1.0f) { float nz = 0.001f; float nx = f0(x + nz, y, n) - z; float ny = f0(x, y + nz, n) - z; float nd = sqrtf(nx * nx + ny * ny + nz * nz); float d = (nx - ny + nz) / sqrtf(3) / nd; d = d > 0.0f ? d : 0.0f; // d = d < 1.0f ? d : 1.0f; putchar(".-:=+*#%@@"[(int)(d * 9.0f)]); } else putchar(' '); }}

python

提供一种最简单的python实现方法，哈哈！不喜勿喷！

源代码如下：

height = 5stars = 1for i in range(height): print((' ' * (height - i)) + ('*' * stars)) stars += 2print((' ' * height) + '|')

也可以使用简单的图形库Turtle来实现，就像下面这样子。原理也很简单，参照前面最简单的实现方法，加载图形库后把原有的*使用绿色的方块代替，在各个角和顶部采用红色圆形代替，树干用棕色方块代替即可，最后在适当润色一下。

源代码：

import turtlescreen = turtle.Screen()screen.setup(800,600)circle = turtle.Turtle()circle.shape('circle')circle.color('red')circle.speed('fastest')circle.up()square = turtle.Turtle()square.shape('square')square.color('green')square.speed('fastest')square.up()circle.goto(0,280)circle.stamp()k = 0for i in range(1, 17): y = 30*i for j in range(i-k): x = 30*j square.goto(x,-y+280) square.stamp() square.goto(-x,-y+280) square.stamp() if i % 4 == 0: x = 30*(j+1) circle.color('red') circle.goto(-x,-y+280) circle.stamp() circle.goto(x,-y+280) circle.stamp() k += 2 if i % 4 == 3: x = 30*(j+1) circle.color('yellow') circle.goto(-x,-y+280) circle.stamp() circle.goto(x,-y+280) circle.stamp()square.color('brown')for i in range(17,20): y = 30*i for j in range(3): x = 30*j square.goto(x,-y+280) square.stamp() square.goto(-x,-y+280) square.stamp()turtle.exitonclick()

当然还有另外一种实现方式，是一种动态的效果。

下面是源代码：

from turtle import *import randomimport timen = 80.0speed("fastest")screensize(bg='seashell')left(90)forward(3*n)color("orange", "yellow")begin_fill()left(126)for i in range(5): forward(n/5) right(144) forward(n/5) left(72)end_fill()right(126)color("dark green")backward(n*4.8)def tree(d, s): if d <= 0: return forward(s) tree(d-1, s*.8) right(120) tree(d-3, s*.5) right(120) tree(d-3, s*.5) right(120) backward(s)tree(15, n)backward(n/2)for i in range(200): a = 200 - 400 * random.random() b = 10 - 20 * random.random() up() forward(b) left(90) forward(a) down() if random.randint(0, 1) == 0: color('tomato') else: color('wheat') circle(2) up() backward(a) right(90) backward(b)time.sleep(60)