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