所以我制作了这个程序，你可以在其中输入圆或线的参数，它将通过在显示器上绘制数组来显示所述对象。
它的工作原理是将坐标系"投影"到阵列上。(该程序还要求您给出数组的分辨率，列和行的数量相同。(然后，对于数组的每个单元格，它会检查圆/线是否与单元格相交。如果是，或者在给定范围内，则单元格将获得值1。如果超出范围，则为0。当所有单元格都已给定值时，程序将显示数组。所以最后你会看到一个由1组成的圆或线，其余的数组将显示为零。
问题是打印数组需要相对较长的时间(7到10秒(，而实际计算几乎不需要时间。
我的问题是，正如标题中所说，阵列的显示过程能否以某种方式加快？还是我做错了什么？我正在使用Code:：Blocks作为我的编译器。
我知道我的代码可能优化得很差，但我一周前才开始编程。因此，如果代码很难理解，请原谅我。
提前谢谢！
#include <stdio.h>
#include <stdlib.h>
int main()
{
float x = 0, y = 0, ypos= 0 , xpos = 0, radius = 0, rsqrd = 0, rcheck = 0, thick = 0, grad = 0, offs = 0, lcheck = 0;
int matsize = 0, i, j, branch = 0;
char filled;
printf("n0 - circlen1 - linenDo you want to draw a circle or a line? (0/1) ");
scanf("%d", &branch);
if(branch == 0)
{
printf("Value of radius: ");
scanf("%f", &radius);
printf("Position of circle on the x axis: ");
scanf("%f", &xpos);
printf("Position of circle on the y axis: ");
scanf("%f", &ypos);
printf("Is the circle filled? (y/n) ");
scanf(" %c", &filled);
if(filled == 'n')
{
printf("The thickness of circle: ");
scanf("%f", &thick);
}
if(filled == 'y' || filled == 'n')
{
printf("Resolution: ");
scanf("%d" , &matsize);
printf("n");
}
rsqrd = radius*radius; //rsqrd is equal to radius squared.
x = -1*(matsize/2); //with this I make sure that the x and y values start from the top right corner of the matrix, so that each x, y value corresponds to the correct cell position (i, j)
y = matsize/2;
int mat[matsize][matsize];
if(filled == 'n')
{
for(i = 0; i < matsize; i++)
{
for(j = 0; j < matsize; j++)
{
rcheck = ((y - ypos)*(y - ypos)) + ((x - xpos)*(x - xpos)); // calculating the equation of the circle with the x and y values taking the offset into account
if(abs(rcheck-rsqrd) <= (thick*thick))
{
mat[i][j] = 1;
}
else
{
mat[i][j] = 0;
}
x = x+1; //stepping the values of x and y so they stay with the corresponding cell
}
x = -1*(matsize/2);
y = y-1;
}
}
if(filled =='y')
{
for(i = 0; i < matsize; i++)
{
for(j = 0; j < matsize; j++)
{
rcheck = ((y - ypos)*(y - ypos)) + ((x - xpos)*(x - xpos)); // calculating the equation of the circle with the x and y values taking the offset into account
if(rcheck <= rsqrd)
{
mat[i][j] = 1;
}
else
{
mat[i][j] = 0;
}
x = x+1; //stepping the values of x and y so they stay with the corresponding cell
}
x = -1*(matsize/2);
y = y-1;
}
}
if(filled == 'y' || filled == 'n')
{
for(i = 0; i < matsize; i++) // displaying the matrix
{ //
for(j = 0; j < matsize; j++) //
{ //
printf("%d ",mat[i][j]); //
} //
printf("n"); //
} //
}
}
if(branch == 1)
{
printf("Value of gradient: ");
scanf("%f", &grad);
printf("Value of offset: ");
scanf("%f", &offs);
printf("Thickness of line: ");
scanf("%f", &thick);
printf("Resoultion: ");
scanf("%d", &matsize);
x = -1*(matsize/2); //with this I make sure that the x and y values start from the top right corner of the matrix, so that each x, y value corresponds to the correct cell position (i, j)
y = matsize/2;
int mat[matsize][matsize];
for(i = 0; i < matsize; i++)
{
for(j = 0; j < matsize; j++)
{
lcheck = y - (x * grad); // calculating the equation of the circle with the x and y values taking the offset into account
if(abs(lcheck-offs) <= thick)
{
mat[i][j] = 1;
}
else
{
mat[i][j] = 0;
}
x = x+1; //stepping the values of x and y so they stay with the corresponding cell
}
x = -1*(matsize/2);
y = y-1;
}
if(branch == 1)
{
for(i = 0; i < matsize; i++) // displaying the matrix
{ //
for(j = 0; j < matsize; j++)//
{ //
printf("%d ",mat[i][j]);//
} //
printf("n"); //
} //
}
}
return 0;
}
正如我在评论中所说，也许这与这个堆栈溢出问答有关
在读了一点之后，您还可以尝试缓冲stdout，以使其更快。