如何用C语言画一个高逼格的"心形"？

本文转自知乎：http://www.zhihu.com/question/20187195                Milo Yip大神的回答

#include <stdio.h>



 




int main() {



    for (float y = 1.5f; y > -1.5f; y -= 0.1f) {



        for (float x = -1.5f; x < 1.5f; x += 0.05f) {



            float a = x * x + y * y - 1;



            putchar(a * a * a - x * x * y * y * y <= 0.0f ? '*' : ' ');



        }



        putchar('\n');



    }



}

 再来个有花纹的。（这其实是该函数的Level set）

#include <stdio.h>




 



int main() {



    for (float y = 1.5f; y > -1.5f; y -= 0.1f) {



        for (float x = -1.5f; x < 1.5f; x += 0.05f) {




            float z = x * x + y * y - 1;



            float f = z * z * z - x * x * y * y * y;



            putchar(f <= 0.0f ? ".:-=+*#%@"[(int)(f * -8.0f)] : ' ');



        }



        putchar('\n');



    }



}

简单使用迭代法求解，用Finite difference求法矢量，用wrapped diffuse着色。

#include <stdio.h>



#include <math.h>



 




float f(float x, float y, float z) {



    float a = x * x + 9.0f / 4.0f * y * y + z * z - 1;



    return a * a * a - x * x * z * z * z - 9.0f / 80.0f * y * y * z * z * z;



}



 




float h(float x, float z) {



    for (float y = 1.0f; y >= 0.0f; y -= 0.001f)



        if (f(x, y, z) <= 0.0f)



            return y;



    return 0.0f;



}



 




int main() {



    for (float z = 1.5f; z > -1.5f; z -= 0.05f) {



        for (float x = -1.5f; x < 1.5f; x += 0.025f) {



            float v = f(x, 0.0f, z);



            if (v <= 0.0f) {



                float y0 = h(x, z);



                float ny = 0.01f;



                float nx = h(x + ny, z) - y0;



                float nz = h(x, z + ny) - y0;



                float nd = 1.0f / sqrtf(nx * nx + ny * ny + nz * nz);



                float d = (nx + ny - nz) * nd * 0.5f + 0.5f;



                putchar(".:-=+*#%@"[(int)(d * 5.0f)]);



            }



            else




                putchar(' ');



        }



        putchar('\n');



    }



}

「3D」版，简单使用迭代法求解，用Finite difference求法矢量，用wrapped diffuse着色。

#include <stdio.h>



#include <math.h>



 




float f(float x, float y, float z) {



    float a = x * x + 9.0f / 4.0f * y * y + z * z - 1;



    return a * a * a - x * x * z * z * z - 9.0f / 80.0f * y * y * z * z * z;



}



 




float h(float x, float z) {



    for (float y = 1.0f; y >= 0.0f; y -= 0.001f)



        if (f(x, y, z) <= 0.0f)



            return y;



    return 0.0f;



}



 




int main() {



    for (float z = 1.5f; z > -1.5f; z -= 0.05f) {



        for (float x = -1.5f; x < 1.5f; x += 0.025f) {



            float v = f(x, 0.0f, z);



            if (v <= 0.0f) {



                float y0 = h(x, z);



                float ny = 0.01f;



                float nx = h(x + ny, z) - y0;



                float nz = h(x, z + ny) - y0;



                float nd = 1.0f / sqrtf(nx * nx + ny * ny + nz * nz);



                float d = (nx + ny - nz) * nd * 0.5f + 0.5f;



                putchar(".:-=+*#%@"[(int)(d * 5.0f)]);



            }



            else




                putchar(' ');



        }



        putchar('\n');



    }



}

把「3D版」输出至PPM文件，可以用Photoshop打开。另外降低了ny的值导致有超有趣的pattern，就保留下来吧。

#ifdef _MSC_VER



#define _CRT_SECURE_NO_WARNINGS



#endif



#include <stdio.h>



#include <math.h>



 




float f(float x, float y, float z) {



    float a = x * x + 9.0f / 4.0f * y * y + z * z - 1;



    return a * a * a - x * x * z * z * z - 9.0f / 80.0f * y * y * z * z * z;



}



 




float h(float x, float z) {



    for (float y = 1.0f; y >= 0.0f; y -= 0.001f)



        if (f(x, y, z) <= 0.0f)



            return y;



    return 0.0f;



}



 




int main() {



    FILE* fp = fopen("heart.ppm", "w");



    int sw = 512, sh = 512;



    fprintf(fp, "P3\n%d %d\n255\n", sw, sh);



    for (int sy = 0; sy < sh; sy++) {



        float z = 1.5f - sy * 3.0f / sh;



        for (int sx = 0; sx < sw; sx++) {



            float x = sx * 3.0f / sw - 1.5f;



            float v = f(x, 0.0f, z);



            int r = 0;



            if (v <= 0.0f) {



                float y0 = h(x, z);



                float ny = 0.001f;



                float nx = h(x + ny, z) - y0;



                float nz = h(x, z + ny) - y0;



                float nd = 1.0f / sqrtf(nx * nx + ny * ny + nz * nz);



                float d = (nx + ny - nz) / sqrtf(3) * nd * 0.5f + 0.5f;



                r = (int)(d * 255.0f);



            }



            fprintf(fp, "%d 0 0 ", r);



        }



        fputc('\n', fp);



    }



    fclose(fp);



}



 



 



原文出处：http://www.zhihu.com/question/20187195