一 Java博客作业( 四 ) _直线方程

< 4)System.out.print("wrong number of points");String[] point3 = dj[2].split(",");String[] point4 = dj[3].split(",");if (!dj[0].equals(dj[1]) && !dj[2].equals(dj[3])) {if (point1.length == 2 && point2.length == 2 && point3.length == 2 && point4.length == 2) {if (legal(point1[0]) && legal(point1[1]) && legal(point2[0]) && legal(point2[1]) && legal(point3[0])&& legal(point3[1]) && legal(point4[0]) && legal(point4[1])) {double x0 = Double.parseDouble(point1[0]);double y0 = Double.parseDouble(point1[1]);double x1 = Double.parseDouble(point2[0]);double y1 = Double.parseDouble(point2[1]);double x2 = Double.parseDouble(point3[0]);double y2 = Double.parseDouble(point3[1]);double x3 = Double.parseDouble(point4[0]);double y3 = Double.parseDouble(point4[1]);if (dj.length == 4 && (y3 - y2) / (x3 - x2) == (y1 - y0) / (x1 - x0))System.out.print("true");if ((y3 - y2) / (x3 - x2) != (y1 - y0) / (x1 - x0))System.out.print("false");if (dj.length > 4)//点数量的判断System.out.print("wrong number of points");} elseSystem.out.print("Wrong Format");} elseSystem.out.print("Wrong Format");} elseSystem.out.print("points coincide");}void Five(String x) {String[] dj = x.split(" ");String[] point1 = dj[0].split(",");String[] point2 = dj[1].split(",");if (dj.length < 4)System.out.print("wrong number of points");String[] point3 = dj[2].split(",");String[] point4 = dj[3].split(",");if (!dj[0].equals(dj[1]) && !dj[2].equals(dj[3])) {if (point1.length == 2 && point2.length == 2 && point3.length == 2 && point4.length == 2) {if (legal(point1[0]) && legal(point1[1]) && legal(point2[0]) && legal(point2[1]) && legal(point3[0])&& legal(point3[1]) && legal(point4[0]) && legal(point4[1])) {double x0 = Double.parseDouble(point1[0]);double y0 = Double.parseDouble(point1[1]);double x1 = Double.parseDouble(point2[0]);double y1 = Double.parseDouble(point2[1]);double x2 = Double.parseDouble(point3[0]);double y2 = Double.parseDouble(point3[1]);double x3 = Double.parseDouble(point4[0]);double y3 = Double.parseDouble(point4[1]);if ((y3 - y2) / (x3 - x2) == (y1 - y0) / (x1 - x0))//线平行的情况System.out.print("is parallel lines,have no intersection point");else {double yz = ((y0 - y1) * (y3 - y2) * x0 + (y3 - y2) * (x1 - x0) * y0//xz，yz为交点坐标+ (y1 - y0) * (y3 - y2) * x2 + (x2 - x3) * (y1 - y0) * y2)/ ((x1 - x0) * (y3 - y2) + (y0 - y1) * (x3 - x2));double xz = x2 + (x3 - x2) * (yz - y2) / (y3 - y2);if (dj.length == 4) {System.out.print(xz + "," + yz + " ");if ((y1 - y0) / (xz - x0) == (yz - y0) / (x1 - x0)，|| (y3 - y2) / (xz - x2) == (yz - y2) / (x3 - x2))System.out.print("true");elseSystem.out.print("false");}if (dj.length > 4)System.out.print("wrong number of points");}} elseSystem.out.print("Wrong Format");} elseSystem.out.print("Wrong Format");} elseSystem.out.print("points coincide");}}
第三次作业第三题
7-3 点线形系列3-三角形的计算 (48 分)
用户输入一组选项和数据，进行与三角形有关的计算。选项包括：
1：输入三个点坐标，判断是否是等腰三角形、等边三角形，判断结果输出true/false，两个结果之间以一个英文空格符分隔。
2：输入三个点坐标，输出周长、面积、重心坐标，三个参数之间以一个英文空格分隔，坐标之间以英文","分隔。
3：输入三个点坐标，输出是钝角、直角还是锐角三角形，依次输出三个判断结果（true/false），以一个英文空格分隔，
4：输入五个点坐标，输出前两个点所在的直线与三个点所构成的三角形相交的交点数量，如果交点有两个，则按面积大小依次输出三角形被直线分割成两部分的面积。若直线与三角形一条线重合，输出"The point is on the edge of the "
5：输入四个点坐标，输出第一个是否在后三个点所构成的三角形的内部（输出in the /outof ）。
必须使用射线法，原理：由第一个点往任一方向做一射线，射线与三角形的边的交点（不含点本身）数量如果为1，则在三角形内部。如果交点有两个或0个，则在三角形之外。若点在三角形的某条边上，输出"on the "