解释像耳朵一样的甜甜圈(最后部分)
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时间:2024-12-18 13:27:34 441浏览 收藏
编程并不是一个机械性的工作,而是需要有思考,有创新的工作,语法是固定的,但解决问题的思路则是依靠人的思维,这就需要我们坚持学习和更新自己的知识。今天golang学习网就整理分享《解释像耳朵一样的甜甜圈(最后部分)》,文章讲解的知识点主要包括,如果你对文章方面的知识点感兴趣,就不要错过golang学习网,在这可以对大家的知识积累有所帮助,助力开发能力的提升。
c 的完整代码是
#include <math.h> #include <stdio.h> #include <string.h> #include <unistd.h> typedef struct { double a1; double a2; double a3; } singlerow; typedef struct { singlerow a1; singlerow a2; singlerow a3; } matrix; singlerow multiply(singlerow m1, matrix m2) { singlerow res; res.a1 = m1.a1 * m2.a1.a1 + m1.a2 * m2.a2.a1 + m1.a3 * m2.a3.a1; res.a2 = m1.a1 * m2.a1.a2 + m1.a2 * m2.a2.a2 + m1.a3 * m2.a3.a2; res.a3 = m1.a1 * m2.a1.a3 + m1.a2 * m2.a2.a3 + m1.a3 * m2.a3.a3; return res; } int main() { int screen_width = 80, height = 22; char buffer[1760]; float zbuffer[1760]; float a = 0, b = 0; int r2 = 2, r1 = 1; printf("\x1b[2j"); while (1) { memset(buffer, ' ', 1760); memset(zbuffer, 0, 7040); for (float theta = 0; theta < 6.28; theta += 0.07) { for (float phi = 0; phi < 6.28; phi += 0.02) { singlerow circle = {2 + cos(theta), sin(theta), 0}; // rotation on y-axis matrix ry = {{cos(phi), 0, sin(phi)}, {0, 1, 0}, {-sin(phi), 0, cos(phi)}}; // rotation on x-axis matrix rx = {{1, 0, 0}, {0, cos(a), sin(a)}, {0, -sin(a), cos(a)}}; // rotation on z-axis matrix rz = {{cos(b), sin(b), 0}, {-sin(b), cos(b), 0}, {0, 0, 1}}; singlerow donut = multiply(circle, ry); singlerow rotatex = multiply(donut, rx); // we will consider it as [nx, ny, nz] singlerow spinningdonut = multiply(rotatex, rz); float recinz = 1 / (spinningdonut.a3 + 5); int x = 40 + 30 * spinningdonut.a1 * recinz; int y = 12 + 15 * spinningdonut.a2 * recinz; // o is index of current buffer int o = x + screen_width * y; int l = 8 * (spinningdonut.a2 - spinningdonut.a3 + 2 * cos(b) * sin(a) * sin(phi) - 2 * cos(phi) * cos(theta) * sin(b) - 2 * cos(phi) * sin(b) + 2 * cos(a) * sin(phi) ); // donut luminicity will be seen by these characters // these 12 char charout[] = ".,-~:;=!*#$@"; if (x < screen_width && y < height && zbuffer[o] < recinz) { zbuffer[o] = recinz; // if l < 0, . will be buffer buffer[o] = charout[l > 0 ? l : 0]; } } } // clear screen printf("\x1b[h"); for (int i = 0; i <1761; i++) { // on every 80th character, new line will be printed // if there's a reminder then buffer printed putchar(i % 80 ? buffer[i] : 10); a += 0.00004; b += 0.00002; } // timer to slow down speed a bit usleep(10000); } return 0; }
java的完整代码是
import java.util.Arrays; class singleRow { public double a1; public double a2; public double a3; public singleRow(double a1, double a2, double a3) { this.a1 = a1; this.a2 = a2; this.a3 = a3; } } class Matrix { public singleRow a1; public singleRow a2; public singleRow a3; public Matrix(singleRow a1, singleRow a2, singleRow a3) { this.a1 = new singleRow(a1.a1, a1.a2, a1.a3); this.a2 = new singleRow(a2.a1, a2.a2, a2.a3); this.a3 = new singleRow(a3.a1, a3.a2, a3.a3); } public static singleRow multiply(singleRow m1, Matrix m2) { singleRow res = new singleRow(0, 0, 0); res.a1 = (m1.a1 * m2.a1.a1) + (m1.a2 * m2.a2.a1) + (m1.a3 * m2.a3.a1); res.a2 = (m1.a1 * m2.a1.a2) + (m1.a2 * m2.a2.a2) + (m1.a3 * m2.a3.a2); res.a3 = (m1.a1 * m2.a1.a3) + (m1.a2 * m2.a2.a3) + (m1.a3 * m2.a3.a3); return res; } } public class Main { public static void main() { int screen_width = 80, height = 22; char[] buffer = new char[1760]; double[] zBuffer = new double[1760]; double A = 0, B = 0; int R2 = 2, R1 = 1; System.out.print("\u001b[2J"); while (true) { Arrays.fill(buffer, 0, 1760, ' '); Arrays.fill(zBuffer, 0, 1760, 0); for (float theta = 0; theta < 6.28; theta += 0.07) { for (float phi = 0; phi < 6.28; phi += 0.02) { singleRow circle = {2 + Math.cos(theta), Math.sin(theta), 0}; // rotation on Y-axis Matrix Ry = new Matrix( new singleRow(Math.cos(phi), 0, Math.sin(phi)), new singleRow(0, 1, 0), new singleRow(-Math.sin(phi), 0, Math.cos(phi)) ); // rotation on X-axis Matrix Rx = new Matrix( new singleRow(1, 0, 0), new singleRow(0, Math.cos(A), Math.sin(A)), new singleRow(0, -Math.sin(A), Math.cos(A)) ); // rotation on Z-axis Matrix Rz = new Matrix( new singleRow(Math.cos(B), Math.sin(B), 0), new singleRow(-Math.sin(B), Math.cos(B), 0), new singleRow(0, 0, 1) ); singleRow donut = Matrix.multiply(circle, Ry); singleRow rotateX = Matrix.multiply(donut, Rx); // We will consider it as [Nx, Ny, Nz] singleRow spinningDonut = Matrix.multiply(rotateX, Rz); float reciNz = 1 / (spinningDonut.a3 + 5); int x = 40 + 30 * spinningDonut.a1 * reciNz; int y = 12 + 15 * spinningDonut.a2 * reciNz; // o is index of current buffer int o = x + screen_width * y; int L = 8 * (spinningDonut.a2 - spinningDonut.a3 + 2 * Math.cos(B) * Math.sin(A) * Math.sin(phi) - 2 * Math.cos(phi) * Math.cos(theta) * Math.sin(B) - 2 * Math.cos(phi) * Math.sin(B) + 2 * Math.cos(A) * Math.sin(phi) ); // donut luminicity will be seen by these characters // these 12 char[] charOpts = {'.', ',', '-', '~', ':', ';', '=', '!', '*', '#', '$', '@'}; if (x < screen_width && y < height && zBuffer[o] < reciNz) { zBuffer[o] = reciNz; // If L < 0, . will be buffer buffer[o] = charOut[L > 0 ? L : 0]; } } } // Clear screen System.out.print("\u001b[H"); for (int i = 0; i <1761; i++) { // On every 80th character, new line will be printed // If there's a reminder then buffer printed System.out.print(i % 80 ? buffer[i] : 10); A += 0.00004; B += 0.00002; } } } }
到这里,我们也就讲完了《解释像耳朵一样的甜甜圈(最后部分)》的内容了。个人认为,基础知识的学习和巩固,是为了更好的将其运用到项目中,欢迎关注golang学习网公众号,带你了解更多关于的知识点!
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