cc++ 实现控制台下字符画直线和画圆.doc

c/c+ 实现控制台下字符画直线，画圆DDA画直线法，DDA算法原理：=1/max(|x|,|y|) 中点BH画圆法八分法首先解决八分之一圆弧算法步骤：1.输入圆的半径R。2.计算初始值d=1.25-R、x=0、y=R。3.绘制点(x,y)及其在八分圆中的另外七个对称点。4.判断d的符号。若d0，则先将d更新为d+2x+3，再将(x,y)更新为(x+1,y)；否则先将d更新为d+2(x-y)+5，再将(x,y)更新为(x+1,y-1)。5.当xy时，重复步骤3和4。否则结束。改进：用d-0.25代替d算法步骤：1.输入圆的半径R。2.计算初始值d=1-R、x=0、y=R。3.绘制点(x,y)及其在八分圆中的另外七个对称点。4.判断d的符号。若d0，则先将d更新为d+2x+3，再将(x,y)更新为(x+1,y)；否则先将d更新为d+2(x-y)+5，再将(x,y)更新为(x+1,y-1)。5.当xy时，重复步骤3和4。否则结束。源代码：DDA画直线：#includeiostream#includecmathusing namespace std;int arr100100;void drawLine(int x0, int y0, int x1, int y1)int ymax = y1;int ymin = y0;if (y0 y1)ymax = y0;ymin = y1;int xmax = x0;if (x1 x0)xmax = x1;for (int j = ymin; j ymax; j+)for (int i = 0; i xmax ; i+)if (arrij = 1)cout *;else cout ;cout endl;cout endl;void DDAline(int x0, int y0, int x1, int y1)int dx, dy, epsl, k; float x, y, xIncre, yIncre;dx = x1 - x0; dy = y1 - y0; x = x0; y = y0;if (abs(dx)abs(dy) epsl = abs(dx);else epsl = abs(dy);xIncre = (float)(dx) / epsl;yIncre = (float)(dy) / epsl;for (k = 0; k = epsl; k+)int tx = (int)(x + 0.5);int ty = (int)(y + 0.5);arrtxty = 1;/putpixel(int)(x + 0.5), (int)(y + 0.5);x += xIncre; y += yIncre;drawLine(x0, y0, x1, y1);int main()int x0, y0,x1,y1;while (true)cout 请输入第一个点的x，y值：;cin x0 y0;cout 请输入第二个点的x，y值：;cin x1 y1;if (y0 y1)/交换值使得y1在y0下面int temp = y1;y1 = y0;y0 = temp;temp = x1;x1 = x0;x0 = temp;drawLine(x0, y0, x1, y1);return 0;中点BH画圆法#includestdio.hint arr100;/存储由算法找到的第一象限y=x 区间的所有的圆点坐标int arrXY100100;/存储最终的圆点矩阵/int r: r为要画的圆半径void Circle(int r,int numOfarr)int x, y;/找到其他划分区间的圆点坐标，在存储矩阵上标记for (int i = 0; i numOfarr; i+)/-x,-yif (i = 0)x = r;else x = -i + r;y = -arri + r; arrXYxy = 1;/-y,-xif (i = 0)y = r;else y = -i + r;x = -arri + r;arrXYxy = 1;/y,-xif (i = 0)y = r;else y = -i + r;x = arri + r;arrXYxy = 1;/x,-yif (i = 0)x = r;else x = i + r;y = -arri + r;arrXYxy = 1;/x,yif (i = 0)x = r;else x = i + r;y = arri + r;arrXYxy = 1;/y,xif (i = 0)y = r;else y = i + r;x = arri + r;arrXYxy = 1;/-y,xif (i = 0)y = r;else y = i + r;x = -arri + r;arrXYxy = 1;/-x,yif (i = 0)x = r;else x = -i + r;y = arri + r;arrXYxy = 1;for (int i = 0;i=2*r; i+)for (int j = 0; j = 2*r; j+) if (arrXYji = 0)printf( );else if(arrXYji=1)printf(*);else printf(+);printf(n);void Draw(int r)int x=0, y=r, d=1-r;while (x=y)arrx = y;/存储坐标，if (d 0)d += 2 * x + 3;elsed +