[C] Print language "heart-shaped" pattern in C

In your world, it is not that a little romantic program apes do not understand? Well, not exactly, program ape world is very romantic drops.

Fool version

int main()
{
    printf("");
    return 0;
}

Advanced Edition

// Version a: single character heart-shaped pattern 
#include <stdio.h> int main ( void ) 
{ a float A, X, Y;
     for (Y = 1.5F ; Y> - 1.5F ; Y- = 0.1f ) 
    { for (X = - 1.5F ; X < 1.5F ; = X + 0.05f ) 
        { 
            a = X * Y * X + Y- . 1 ;
             // where the @ sign is the heart-shaped pattern printed symbols, Flexible char CH = AX * A * A * X * Y * Y * Y <= 0.0f ? ' @ ' : ' ' 
            the putchar (CH);  
 

    
        
              ;
            //或者putchar(a*a*a-x*x*y*y*y<=0.0f?'*':' ');
        }
        printf("\n");
    }
    
    return 0;
}

 

 

 

 

//空心版
#include <stdio.h>
#include <math.h>

// The first heart shape in hollow
// (x^2+y^2-1)^3 - x^2*y^3 = 0
// y ~ (-1.1 , 1.3 )
// x ~ (-1.2 , 1.2 )
int main() {

    for (float y = 1.3; y >= -1.1; y -= 0.06)
    {
        for (float x = -1.2; x <= 1.2; x += 0.025)
        {
            if (pow((x * x + y * y - 1.0), 3) - x * x * y * y * y <= 0.0)
                printf(" ");
            else
                printf("*");
        }
        printf("\n");
    }

    return 0;
}

 

 

 

//版本二:可用多个字符组成打印
#include <stdio.h>

int main(void)
{
    float f, x, y, z;
    for (y = 1.5f; y > -1.5f; y -= 0.1f)
    {
        for (x = -1.5f; x < 1.5f; x += 0.05f)
        {
            z = x * x + y * y - 1;
            f = z * z * z - x * x * y * y * y;
            putchar(f <= 0.0f ? "lihuan"[(int)(f * -8.0f)] : ' ');
            //这里的lihuan字符可更改成符号或其他字母,汉字不行,图案会变形
        }
        printf("\n");
    }

    return 0;
}

 

 

 

//加颜色版
#include <stdio.h> int main(int argc, char* argv[]) { float f, x, y, z; for (y = 1.5f; y > -1.5f; y -= 0.1f) { for (x = -1.5f; x < 1.5f; x += 0.05f) { z = x * x + y * y - 1; f = z * z * z - x * x * y * y * y; // =================================== printf("\033[47m"); char buf[100] = { "lihuan" };
//这里的“lihuan”可以改为其他字符,注意,中文不行哦!
if (f <= 0.0f) { printf("\033[1;31m%c\033[0m", buf[(int)(f * -8.0f)]); } else { printf("%c", ' '); } printf("\033[0m"); // ===================================== } printf("\n"); } return 0; }

 

 

 加强版:

#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(int argc, char *argv[]) 
{
    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');
    }
}

 

 变形版:

#include <stdio.h>
 
int main()
{
    int  i, j, k, l, m;    
    char c = '*';    //ASCII码里面 3 就是一个字符小爱心 
    
    for (i=1; i<=5; i++)                printf("\n");    //开头空出5行 
    
    for (i=1; i<=3; i++) 
    {    //前3行中间有空隙分开来写 
        for (j=1; j<=32-2*i; j++)        
            printf(" ");    //左边的空格,每下一行左边的空格比上一行少2个 //8*n-2*i 
        for (k=1; k<=4*i+1; k++)        
            printf("%c", c);//输出左半部分字符小爱心 
        for (l=1; l<=13-4*i; l++)        
            printf(" ");    //中间的空格,每下一行的空格比上一行少4个 
        for (m=1; m<=4*i+1; m++)        
            printf("%c", c);//输出右半部分字符小爱心
        printf("\n");        //每一行输出完毕换行 
    }
    
    for (i=1; i<=3; i++) 
    {    //下3行中间没有空格 
        for (j=1; j<=24+1; j++)            
            printf(" ");    //左边的空格 //8*(n-1)+1
        for (k=1; k<=29; k++)            
            printf("%c", c);//输出字符小爱心
        printf("\n");        //每一行输出完毕换行
    }
    
    for (i=7; i>=1; i--) 
    {    //下7行 
        for (j=1; j<=40-2*i; j++)        
            printf(" ");    //左边的空格,每下一行左边的空格比上一行少2个//8*(n+1)-2*i
        for (k=1; k<=4*i-1; k++)        
            printf("%c", c);//每下一行的字符小爱心比上一行少4个(这个循环是i--) 
        printf("\n");        //每一行输出完毕换行
    }
    
    for (i=1; i<=39; i++)                
        printf(" ");    //最后一行左边的空格 
    
    printf("%c\n", c);        //最后一个字符小爱心 
    
    for (i=1; i<=5; i++)                
        printf("\n");    //最后空出5行 
    
    return 0;
}

 

#include <stdio.h>
#include <math.h>

#define A 3

void print(void)
{
 double x,y;

 for(y=-4;y<=0;y+=0.3)
 {
     for(x=-4;x<=4;x+=0.2)
         if(fabs(sqrt(x*x+y*y)-A*sin(2*atan(y/x)))<=1 ||fabs(sqrt(x*x+y*y)-A*sin(2*atan(-y/x)))<=1)
         printf("*");
     else    printf(" ");
     printf("\n");
 }
 for(y=-1;y<=0;y+=0.2)
 {
     for(x=-4;x<=4;x+=0.2)
        if(fabs(y)-0.65*x*x>=0.2)
         printf("*");
     else     printf(" ");
     printf("\n");
 }
}

int main(void)
{
 print();
 getchar();
 return 0;
}

 

 

 

算法链接:http://mathworld.wolfram.com/HeartSurface.html

代码主要参考网址:https://www.zhihu.com/question/20187195

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Origin www.cnblogs.com/HGNET/p/12053561.html