C语言圣诞树的实现示例
时间:2022-09-16 10:16:49|栏目:C代码|点击: 次
你们要的圣诞树它来啦!
快去送给心爱的人吧!
效果如下:
#define _CRT_SECURE_NO_WARNINGS 1 #include <math.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #define PI 3.14159265359f float 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[]) { //这个里可以加打印信息哟,为他添加专属的printf 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(' '); } }
上一篇:数据结构之堆的具体使用
栏 目:C代码
下一篇:c语言static关键字用法详解
本文标题:C语言圣诞树的实现示例
本文地址:http://www.codeinn.net/misctech/213847.html