标签：x1 double PTA Blog x2 && y1 第二次 y2
一、前言：
这次的期中考试相对于来说难度还是比较简单的，但是对于我来说容器的知识记得不是很牢固，所以导致没有满分，但是过后再去写的时候又没有那么难，pta包括期中考试整体方面的题量和难度来说是比较大的，期中考试的题目如果说你知识点不牢固，你有可能做不完，pta的题目难度很大，主要难度我觉得集中在图形的判断上，因为四边形和五边形的分割与判断，他的情况实在是太多了，比如输入10个点，如果前五个点形成三角形，后五个点也形成三角形，那就有100种情况，如果说这一百种情况纷纷列出来，那将是非常庞大的代码量，这么复杂的代码实在是太让人绝望了。五边形的分割也很让人头疼，比如五边形可以分割成三角形+六边形，三角形加四边形。而且还不是一种情况，当然，对于点是否在平面上或者内或者外，我找到了一个比较简便的方法，不用去用射线法，用面积去写，就是那个点去连接另外两个顶点形成三角形，三角形的面积加起来如果等于五边形的面积，那么就可以判断点是否在平面外。，我觉得这个想法是比较好的。
涉及到的知识点有很多的，有父类，容器，分类，类的创建，类的管理，正则表达式等等。
二、设计与分析：
1、四边形
四边形主题要运用了类的分类和正则表达式。
类图：
主要代码：
1 package 四边形;
2
3 import java.text.DecimalFormat;
4 import java.util.Scanner;
5
6 public class Main{
7
8 public static void main(String[] args) {
9 // TODO 自动生成的方法存根
10 Scanner in = new Scanner(System.in);
11 String number = in.nextLine();
12 String[] tokens = null;
13 for(int i=0;i<number.length();i++)
14 {
15 tokens=number.split("[:, ]");
16 }
17 Judge1.judge1(number,tokens);
18 }
19
20 }
21
22 class Judge1{
23 private static String df= "[1-5]:([+-]?\\d+(\\.\\d+)?,[+-]?\\d+(\\.\\d+)?\\s?)+";
24
25 public static void judge1(String number,String[] tokens) {
26 Double n = Double .parseDouble(tokens[0]);
27 if(tokens.length==9&&number.matches(df)&&(n==1||n==2||n==3))
28 Qua.quadrangle(tokens);
29 else if(tokens.length!=9&&number.matches(df)&&(n==1||n==2||n==3))
30 System.out.println("wrong number of points");
31 else if(tokens.length==13&&number.matches(df)&&n==4)
32 sixpoints.six(tokens);
33 else if(tokens.length!=13&&number.matches(df)&&n==4)
34 System.out.println("wrong number of points");
35 else if(tokens.length==11&&number.matches(df)&&n==5)
36 fivepoints.five(tokens);
37 else if(tokens.length!=11&&number.matches(df)&&n==5)
38 System.out.println("wrong number of points");
39 else if (!number.matches("[+-]?([1-9]\\d*|0)(\\.\\d+)?,[+-]?([1-9]\\d*|0)(\\.\\d+)?"))
40 System.out.println("Wrong Format");
41 else if (!number.matches("[1-5]:.+"))
42 System.out.println("Wrong Format");
43 }
44 }
45
46 class Judge2{//判断是否为四边形
47 public static boolean judge2(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4){
48 double[] core;
49 core=nodical.nod(x1, y1, x4, y4, x3, y3, x2, y2);
50 if ((y4 - y3) * (x4 - x2) == (y4 - y2) * (x4- x3))
51 return false;
52 else if ((y4- y3) * (x4 - x1) == (y4 - y1) * (x4 - x3))
53 return false;
54 else if ((y4 - y2) * (x4 - x1) == (y4 - y1) * (x4 - x2))
55 return false;
56 else if ((y3 - y2) * (x3 - x1) == (y3 - y1) * (x3 - x2))
57 return false; //任意三个顶点成直线，非四边形
58 else if((y1-y2)*(x3-x4)-(y3-y4)*(x1-x2)==0&&(y2-y3)*(x3-x4)-(y3-y4)*(x2-x3)==0)
59 return false;
60 if(( ((core[0]>x1&&core[0]<x2)||(core[0]>x2&&core[0]<x1)) &&((core[1]>y1&&core[1]<y2)||(core[1]>y2&&core[1]<y1)))&&( ((core[0]>x3&&core[0]<x4)||(core[0]>x4&&core[0]<x3)) &&((core[1]>y3&&core[1]<y4)||(core[1]>y4&&core[1]<y3))))
61 return false;
62 else
63 return true;
64 }
65 }
66
67 class Judge3{//判断是否为三角形
68 public static int judge3(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4) {
69 if( ((y1-y2)*(x2-x4)-(y2-y4)*(x1-x2)==0&&(y3-y2)*(x2-x4)-(y2-y4)*(x3-x2)!=0&&( (y1>y2&&y1<y4)||(y1>y4&&y1<y2)||(x1>x2&&x1<x4)||(x1>x4&&x1<x2) )) || (x1-x2==0&&y1-y2==0) || (x1-x4==0&&y1-y4==0) )
70 return 1;//顶点2,3,4
71 else if( ((y2-y1)*(x1-x3)-(y1-y3)*(x2-x1)==0&&(y4-y1)*(x1-x3)-(y1-y3)*(x4-x1)!=0&&( (y2>y1&&y2<y3)||(y2>y3&&y2<y1)||(x2>x1&&x2<x3)||(x2>x3&&x2<x1) )) || (x2-x1==0&&y2-y1==0) || (x2-x3==0&&y2-y3==0) )
72 return 2;//顶点1,3,4
73 else if( ((y3-y2)*(x2-x4)-(y2-y4)*(x3-x2)==0&&(y1-y2)*(x2-x4)-(y2-y4)*(x1-x2)!=0&&( (y3>y2&&y3<y4)||(y3>y4&&y3<y2)||(x3>x2&&x3<x4)||(x3>x4&&x3<x2) )) || (x3-x2==0&&y3-y2==0) || (x3-x4==0&&y3-y4==0) )
74 return 3;//顶点1,2,4
75 else if( ((y4-y1)*(x1-x3)-(y1-y3)*(x4-x1)==0&&(y2-y1)*(x1-x3)-(y1-y3)*(x2-x1)!=0&&( (y4>y1&&y4<y3)||(y4>y3&&y4<y1)||(x4>x1&&x4<x3)||(x4>x3&&x4<x1) )) || (x4-x1==0&&y4-y1==0) || (x4-x3==0&&y4-y3==0) )
76 return 4;//顶点1,2,3
77 else
78 return 0;
79 }
80
81 }
82
83 class Judge4{//四边形判断点是否在四边形内
84 public static int judge4(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4,double x5,double y5) {
85 double s1,s2,s3,s4,area,area1;
86 double l1,l2,l3,l4,l5;
87 double n1,n2,n3,n4,n5;
88 l1=length.longth(x2, y2, x5, y5);
89 l2=length.longth(x2, y2, x3, y3);
90 l3=length.longth(x3, y3, x4, y4);
91 l4=length.longth(x4, y4, x5, y5);
92 l5=length.longth(x2, y2, x4, y4);
93 n1=length.longth(x1, y1, x2, y2);
94 n2=length.longth(x1, y1, x3, y3);
95 n3=length.longth(x1, y1, x4, y4);
96 n4=length.longth(x1, y1, x5, y5);
97 area=0.5*l1*l4*triangle.sin(l5, l1, l4)+0.5*l2*l3*triangle.sin(l5, l2, l3);
98 s1=0.5*n1*n2*triangle.sin(l2, n1, n2);
99 s2=0.5*n2*n3*triangle.sin(l3, n3, n2);
100 s3=0.5*n3*n4*triangle.sin(l4, n3, n4);
101 s4=0.5*n1*n4*triangle.sin(l1, n1, n4);
102 n5=(float)(s1+s2+s3+s4);
103 area1=(float)area;
104 if(area1-n5!=0)
105 return 0;
106 else
107 {
108 if(n1+n2-l2==0||n2+n3-l3==0||n3+n4-l4==0||n4+n1-l1==0)//判断点是否在线上
109 return 2;
110 else
111 return 1;
112 }
113
114 }
115 }
116
117 class Judge5{//三角形
118 public static int judge5(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4) {
119 double s1,s2,s3,area;
120 double l1,l2,l3,l4,l5,l6;
121 float s,area1;
122 l1=length.longth(x2, y2, x3, y3);
123 l2=length.longth(x3, y3, x4, y4);
124 l3=length.longth(x4, y4, x2, y2);
125 l4=length.longth(x1, y1, x2, y2);
126 l5=length.longth(x1, y1, x3, y3);
127 l6=length.longth(x1, y1, x4, y4);
128 area=0.5*l1*l3*triangle.sin(l2, l1, l3);
129 s1=0.5*l4*l5*triangle.sin(l1, l4, l5);
130 s2=0.5*l5*l6*triangle.sin(l2, l5, l6);
131 s3=0.5*l6*l4*triangle.sin(l3, l4, l6);
132 s=(float)(s1+s2+s3);
133 area1=(float)(area);
134 if(area1-s!=0)
135 return 0;
136 else
137 {
138 if(l4+l5-l1==0||l5+l6-l2==0||l4+l6-l3==0)//判断点是否在线上
139 return 2;
140 else
141 return 1;
142 }
143 }
144 }
145
146 class nodical{//两条线的交点
147 public static double[] nod(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4) {
148 double x,y;
149 double[] a=new double[2];
150 x=( (x2-x3)*(y4-y1)*x1-(x4-x1)*(y2-y3)*x2+(x4-x1)*(x2-x3)*(y2-y1) )/( (x2-x3)*(y4-y1)-(y2-y3)*(x4-x1) );
151 y=y2+(x-x2)*(y2-y3)/(x2-x3);
152 a[0]=x;
153 a[1]=y;
154 return a;
155 }
156 }
157
158 class length{//判断两点的距离
159 public static double longth(double x1,double y1,double x2,double y2) {
160 double l1,l;
161 l1=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
162 l=Math.sqrt(l1);
163 return l;
164 }
165 }
166
167 class triangle{//三角形的正弦值
168 public static double sin(double a,double b,double c) {
169 double num,num1;
170 num=(b*b+c*c-a*a)/(2*b*c);
171 num1=Math.sqrt(1-num*num);
172 return num1;
173 }
174 }
175
176 class length1{//点到直线距离
177 public static double longth1(double x1,double y1,double x2,double y2,double x3,double y3) {
178 double l;
179 l = Math.abs(((y2-y3)*x1+(x3-x2)*y1+x2*y3-y2*x3)/(Math.sqrt(Math.pow(y2-y3, 2) +Math.pow(x2-x3, 2))));
180 return l;
181 }
182 }
183
184
185 class Qua{
186 public static void quadrangle(String[] tokens) {
187 Double n = Double .parseDouble(tokens[0]);
188 Double x1 = Double.parseDouble(tokens[1]);
189 Double y1 = Double.parseDouble(tokens[2]);
190 Double x2 = Double.parseDouble(tokens[3]);
191 Double y2 = Double.parseDouble(tokens[4]);
192 Double x3 = Double.parseDouble(tokens[5]);
193 Double y3 = Double.parseDouble(tokens[6]);
194 Double x4 = Double.parseDouble(tokens[7]);
195 Double y4 = Double.parseDouble(tokens[8]);
196
197 DecimalFormat df1 = new DecimalFormat("0.0##");
198 boolean f1=false,f2=false,f3=false,f4=false,f5=false,f6=false;
199 f1=Judge2.judge2(x1, y1, x2, y2, x3,y3,x4,y4);
200 if(n==1)
201 {
202 if( (x1-x2==0&&y1-y2==0)||(x2-x3==0&&y2-y3==0)||(x3-x4==0&&y3-y4==0)||(x4-x1==0&&y4-y1==0) )
203 System.out.println("points coincide");
204 else {
205 if(f1==true) {
206 f2=Test1.t1(x1, y1, x2, y2, x3, y3, x4, y4);
207 System.out.println(f1+" "+f2);
208 }
209 else
210 System.out.println(f1+" "+f2);
211 }
212 }
213
214 if(n==2)
215 {
216 if( (x1-x2==0&&y1-y2==0)||(x2-x3==0&&y2-y3==0)||(x3-x4==0&&y3-y4==0)||(x4-x1==0&&y4-y1==0) )
217 System.out.println("points coincide");
218 else
219 {
220 if(f1==true) {
221 f3=Test2.t2(x1, y1, x2, y2, x3, y3, x4, y4);
222 f4=Test3.t3(x1, y1, x2, y2, x3, y3, x4, y4);
223 f5=Test4.t4(x1, y1, x2, y2, x3, y3, x4, y4);
224 System.out.println(f3+" "+f4+" "+f5);
225 }
226 else
227 System.out.println("not a quadrilateral");
228 }
229 }
230
231 if(n==3)
232 {
233 if( (x1-x2==0&&y1-y2==0)||(x2-x3==0&&y2-y3==0)||(x3-x4==0&&y3-y4==0)||(x4-x1==0&&y4-y1==0) )
234 System.out.println("points coincide");
235 else {
236 if(f1==true) {
237 double l1,l2,l3,l4,sum=0,area=0;
238 double x5,y5,x6,y6,x7,y7;
239 l1=length.longth(x1, y1, x4, y4);
240 l2=length.longth(x2, y2, x3, y3);
241 l3=length.longth(x1, y1, x2, y2);
242 l4=length.longth(x3, y3, x4, y4);
243 x5=(x1+x4)/2;
244 y5=(y1+y4)/2;
245 x6=(x2+x3)/2;
246 y6=(y2+y3)/2;
247 x7=(x1+x2)/2;
248 y7=(y1+y2)/2;
249 area=2*length.longth(x5, y5, x6, y6)*length1.longth1(x7, y7, x5, y5, x6, y6);
250 sum=l1+l2+l3+l4;
251 f6=Test5.t5(x1, y1, x2, y2, x3, y3, x4, y4);
252 System.out.println(f6+" "+df1.format(sum)+" "+df1.format(area));
253 }
254 else
255 System.out.println("not a quadrilateral");
256 }
257 }
258
259 }
260 }
261
262 class sixpoints{
263 public static void six(String[] tokens) {
264 Double n = Double .parseDouble(tokens[0]);
265 Double x1 = Double.parseDouble(tokens[1]);
266 Double y1 = Double.parseDouble(tokens[2]);
267 Double x2 = Double.parseDouble(tokens[3]);
268 Double y2 = Double.parseDouble(tokens[4]);
269 Double x3 = Double.parseDouble(tokens[5]);
270 Double y3 = Double.parseDouble(tokens[6]);
271 Double x4 = Double.parseDouble(tokens[7]);
272 Double y4 = Double.parseDouble(tokens[8]);
273 Double x5 = Double.parseDouble(tokens[9]);
274 Double y5 = Double.parseDouble(tokens[10]);
275 Double x6 = Double.parseDouble(tokens[11]);
276 Double y6 = Double.parseDouble(tokens[12]);
277 boolean f1=false;
278 int f2;
279 f1=Judge2.judge2(x3, y3, x4, y4, x5,y5,x6,y6);
280 f2=Judge3.judge3(x3, y3, x4, y4, x5,y5,x6,y6);
281 if(f1==true&&f2==0)
282 {
283 if( ( (y1-y3)*(x3-x4)-(y3-y4)*(x1-x4)==0&&(y3-y4)*(x4-x2)-(y4-y2)*(x3-x4)==0 ) ||( (y1-y3)*(x3-x6)-(y3-y6)*(x1-x3)==0&&(y3-y6)*(x6-x2)-(y6-y2)*(x3-x6)==0 )
284 || ( (y1-y4)*(x4-x5)-(y4-y5)*(x1-x4)==0&&(y4-y5)*(x5-x2)-(y5-y2)*(x4-x5)==0 ) || ( (y1-y5)*(x5-x6)-(y5-y6)*(x1-x5)==0&&(y5-y6)*(x6-x2)-(y6-y2)*(x5-x6)==0 ) )
285 System.out.println("The line is coincide with one of the lines");
286 else
287 System.out.println("1");
288 }
289 else if(f1==false&&(f2==1||f2==2||f2==3||f2==4))
290 {
291 System.out.println("1");
292 }
293 else if(f1==false&&f2==0)
294 System.out.println("not a quadrilateral or triangle");
295 }
296 }
297
298 class fivepoints{
299 public static void five(String[] tokens) {
300 Double n = Double .parseDouble(tokens[0]);
301 Double x1 = Double.parseDouble(tokens[1]);
302 Double y1 = Double.parseDouble(tokens[2]);
303 Double x2 = Double.parseDouble(tokens[3]);
304 Double y2 = Double.parseDouble(tokens[4]);
305 Double x3 = Double.parseDouble(tokens[5]);
306 Double y3 = Double.parseDouble(tokens[6]);
307 Double x4 = Double.parseDouble(tokens[7]);
308 Double y4 = Double.parseDouble(tokens[8]);
309 Double x5 = Double.parseDouble(tokens[9]);
310 Double y5 = Double.parseDouble(tokens[10]);
311 boolean f1=false;
312 int f2;
313 int a,b;
314 f1=Judge2.judge2(x2,y2,x3, y3, x4, y4, x5,y5);
315 f2=Judge3.judge3(x2,y2,x3, y3, x4, y4, x5,y5);
316 a=Judge4.judge4(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5);
317 if(f1==true&&f2==0)
318 {
319 if(a==1)
320 System.out.println("in the quadrilateral");
321 else if(a==2)
322 System.out.println("on the quadrilateral");
323 else if(a==0)
324 System.out.println("outof the quadrilateral");
325 }
326 else if(f1==false&&f2!=0)
327 {
328 if(f2==1)
329 {
330 b=Judge5.judge5(x1, y1, x3, y3, x4, y4, x5, y5);
331 if(b==0)
332 System.out.println("outof the triangle");
333 else if(b==1)
334 System.out.println("in the triangle");
335 else if(b==2)
336 System.out.println("on the triangle");
337 }
338 else if(f2==2)
339 {
340 b=Judge5.judge5(x1, y1, x2, y2, x4, y4, x5, y5);
341 if(b==0)
342 System.out.println("outof the triangle");
343 else if(b==1)
344 System.out.println("in the triangle");
345 else if(b==2)
346 System.out.println("on the triangle");
347 }
348 else if(f2==3)
349 {
350 b=Judge5.judge5(x1, y1, x2, y2, x3, y3, x5, y5);
351 if(b==0)
352 System.out.println("outof the triangle");
353 else if(b==1)
354 System.out.println("in the triangle");
355 else if(b==2)
356 System.out.println("on the triangle");
357 }
358 else if(f2==4)
359 {
360 b=Judge5.judge5(x1, y1, x2, y2, x3, y3, x4, y4);
361 if(b==0)
362 System.out.println("outof the triangle");
363 else if(b==1)
364 System.out.println("in the triangle");
365 else if(b==2)
366 System.out.println("on the triangle");
367 }
368 }
369 else
370 System.out.println("not a quadrilateral or triangle");
371 }
372 }
373
374 class Test1{//判断是否为平行四边形
375 public static boolean t1(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4) {
376 double l1,l2,l3,l4;
377 l1=length.longth(x1, y1, x2, y2);
378 l2=length.longth(x3, y3, x4, y4);
379 l3=length.longth(x2, y2, x3, y3);
380 l4=length.longth(x1, y1, x4, y4);
381 if(l1==l2&&l3==l4)
382 return true;
383 else
384 return false;
385 }
386 }
387
388 class Test2{//判断是否为菱形
389 public static boolean t2(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4) {
390 double l1,l2;
391 boolean f=false;
392 l1=length.longth(x1, y1, x4, y4);
393 l2=length.longth(x1, y1, x2, y2);
394 f=Test1.t1(x1, y1, x2, y2, x3, y3, x4, y4);
395 if(f==true) {
396 if(l1-l2==0)
397 return true;
398 else
399 return false;
400 }
401 else
402 return false;
403 }
404 }
405
406 class Test3{//判断是否为矩形
407 public static boolean t3(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4) {
408 double l1,l2;
409 boolean f=false;
410 l1=length.longth(x1, y1, x3, y3);
411 l2=length.longth(x2, y2, x4, y4);
412 f=Test1.t1(x1, y1, x2, y2, x3, y3, x4, y4);
413 if(f==true)
414 {
415 if(l1-l2==0)
416 return true;
417 else
418 return false;
419 }
420 else
421 return false;
422 }
423 }
424
425 class Test4{//判断是否为正方形
426 public static boolean t4(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4) {
427 boolean f1=false,f2=false;
428 f1=Test1.t1(x1, y1, x2, y2, x3, y3, x4, y4);
429 f2=Test2.t2(x1, y1, x2, y2, x3, y3, x4, y4);
430 if(f1==true) {
431 if(f2==true)
432 return true;
433 else
434 return false;
435 }
436 else
437 return false;
438 }
439 }
440
441 class Test5{//是否为凸四边形
442 public static boolean t5(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4) {
443 double t1,t2,t3,t4;
444 t1=(x4-x1)*(y2-y1)-(y4-y1)*(x2-x1);
445 t2=(x1-x2)*(y3-y2)-(y1-y2)*(x3-x2);
446 t3=(x2-x3)*(y4-y3)-(y2-y3)*(x4-x3);
447 t4=(x3-x4)*(y1-y4)-(y3-y4)*(x1-x4);
448 if(t1*t2*t3*t4<0)
449 return false;
450 else
451 return true;
452 }
453 }
2、五边形：
五边形的方法和四边形一样
类图：
主要代码：
1 package 五边形pta;
2
3 import java.text.DecimalFormat;
4 import java.util.Scanner;
5
6 public class Main {
7
8 public static void main(String[] args) {
9 // TODO 自动生成的方法存根
10 Scanner in = new Scanner(System.in);
11 String number = in.nextLine();
12 String[] tokens = null;
13 for(int i=0;i<number.length();i++)
14 {
15 tokens=number.split("[:]");
16 }
17 Judge.judge1(number,tokens);
18 }
19
20 public static void handle1(String[] tokens1) {
21 String[] a;
22 a=tokens1;
23 boolean f=false;
24 f=pen(a);
25 System.out.println(f);
26
27 }
28
29 public static void handle2(String[] tokens1) {
30 DecimalFormat df1 = new DecimalFormat("0.0##");
31 boolean f1=false,f2=false;
32 double area,l,l1,l2,l3,l4,l5,l6,l7,l8,l9,l10;
33 String[] a;
34 a=tokens1;
35 f1=pen(a);
36 Double x1 = Double.parseDouble(a[1]);
37 Double y1 = Double.parseDouble(a[2]);
38 Double x2 = Double.parseDouble(a[3]);
39 Double y2 = Double.parseDouble(a[4]);
40 Double x3 = Double.parseDouble(a[5]);
41 Double y3 = Double.parseDouble(a[6]);
42 Double x4 = Double.parseDouble(a[7]);
43 Double y4 = Double.parseDouble(a[8]);
44 Double x5 = Double.parseDouble(a[9]);
45 Double y5 = Double.parseDouble(a[10]);
46 l1=Line.length(x1, y1, x2, y2);
47 l2=Line.length(x2, y2, x3, y3);
48 l3=Line.length(x3, y3, x4, y4);
49 l4=Line.length(x4, y4, x5, y5);
50 l5=Line.length(x5, y5, x1, y1);
51 l6=Line.length(x1 ,y1, x3, y3);
52 l7=Line.length(x1, y1, x4, y4);
53 l8=Line.length(x2, y2, x5, y5);
54 l9=Line.length(x2, y2, x4, y4);
55 l10=Line.length(x3, y3, x5, y5);
56 if(f1==false)
57 {
58 System.out.println("not a pentagon");
59 }
60 else
61 {
62 double a1,a2,b1,b2,c1,c2,d1,d2,e1,e2;
63 a1=x2-x1;
64 a2=y2-y1;
65 b1=x3-x2;
66 b2=y3-y2;
67 c1=x4-x3;
68 c2=y4-y3;
69 d1=x5-x4;
70 d2=y5-y4;
71 e1=x1-x5;
72 e2=y1-y5;
73 if( (a1*b2-a2*b1)>0 )
74 {
75 if( (b1*c2-b2*c1)>0&&(c1*d2-c2*d1)>0&&(d1*e2-d2*e1)>0&&(e1*a2-e2*a1)>0 )
76 f2=true;
77 else
78 f2=false;
79 }
80 if((a1*b2-a2*b1)<0)
81 {
82 if((b1*c2-b2*c1)<0&&(c1*d2-c2*d1)<0&&(d1*e2-d2*e1)<0&&(e1*a2-e2*a1)<0)
83 f2=true;
84 else
85 f2=false;
86 }
87 if(f2==true)
88 {
89 l=l1+l2+l3+l4+l5;
90 area=0.5*( l1*l2*Line.sin(l6, l1, l2) + l6*l7*Line.sin(l3, l6, l7) + l4*l5*Line.sin(l7, l4, l5) );
91 System.out.println(f2+" "+df1.format(l)+" "+df1.format(area));
92 }
93 else
94 System.out.println(false);
95 }
96 }
97
98 public static void handle3(String[] tokens1) {
99 String[] a;
100 int n1,n2;
101 boolean f=false;
102 a=tokens1;
103 Double x1 = Double.parseDouble(a[1]);
104 Double y1 = Double.parseDouble(a[2]);
105 Double x2 = Double.parseDouble(a[3]);
106 Double y2 = Double.parseDouble(a[4]);
107 Double x3 = Double.parseDouble(a[5]);
108 Double y3 = Double.parseDouble(a[6]);
109 Double x4 = Double.parseDouble(a[7]);
110 Double y4 = Double.parseDouble(a[8]);
111 Double x5 = Double.parseDouble(a[9]);
112 Double y5 = Double.parseDouble(a[10]);
113 Double x6 = Double.parseDouble(a[11]);
114 Double y6 = Double.parseDouble(a[12]);
115 Double x7 = Double.parseDouble(a[13]);
116 Double y7 = Double.parseDouble(a[14]);
117 String b[]= {a[4],a[5],a[6],a[7],a[8],a[9],a[10],a[11],a[12],a[13],a[14]};
118 n1=fiveTofour.four(a);
119 n2=fiveTOthree.three(a);
120 f=pen(b);
121 if(x1-x2==0&&y1-y2==0) {
122 System.out.println("points coincide");
123 }
124 if(n1==0&&n2==0&&f==false) {
125 System.out.println("not a polygon");
126 }
127 if(n2!=0) {
128 if(n2==1)
129 Triangle.tri(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5);
130 if(n2==2)
131 Triangle.tri(x1,y1,x2,y2,x3,y3,x4,y4,x6,y6);
132 if(n2==3)
133 Triangle.tri(x1,y1,x2,y2,x3,y3,x4,y4,x7,y7);
134 if(n2==4)
135 Triangle.tri(x1,y1,x2,y2,x3,y3,x5,y5,x6,y6);
136 if(n2==5)
137 Triangle.tri(x1,y1,x2,y2,x3,y3,x5,y5,x7,y7);
138 if(n2==6)
139 Triangle.tri(x1,y1,x2,y2,x4,y4,x5,y5,x6,y6);
140 if(n2==7)
141 Triangle.tri(x1,y1,x2,y2,x4,y4,x5,y5,x7,y7);
142 if(n2==8)
143 Triangle.tri(x1,y1,x2,y2,x5,y5,x6,y6,x7,y7);
144 if(n2==9)
145 Triangle.tri(x1,y1,x2,y2,x3,y3,x6,y6,x7,y7);
146 if(n2==10)
147 Triangle.tri(x1,y1,x2,y2,x3,y3,x6,y6,x7,y7);
148 }
149 if(n1!=0) {
150 if(n1==1)
151 Quadri.qua(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6);
152 if(n1==2)
153 Quadri.qua(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x7,y7);
154 if(n1==3)
155 Quadri.qua(x1,y1,x2,y2,x3,y3,x4,y4,x6,y6,x7,y7);
156 if(n1==4)
157 Quadri.qua(x1,y1,x2,y2,x3,y3,x5,y5,x6,y6,x7,y7);
158 if(n1==5)
159 Quadri.qua(x1,y1,x2,y2,x4,y4,x5,y5,x6,y6,x7,y7);
160 }
161 if(f==true) {
162 Penta.penta(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,x7,y7);
163 }
164 }
165
166 public static void handle4(String[] tokens1) {
167 String[] a;
168 int n1,n2,m1,m2;
169 boolean f1=false,f2=false;
170 a=tokens1;
171 Double x1 = Double.parseDouble(a[1]);
172 Double y1 = Double.parseDouble(a[2]);
173 Double x2 = Double.parseDouble(a[3]);
174 Double y2 = Double.parseDouble(a[4]);
175 Double x3 = Double.parseDouble(a[5]);
176 Double y3 = Double.parseDouble(a[6]);
177 Double x4 = Double.parseDouble(a[7]);
178 Double y4 = Double.parseDouble(a[8]);
179 Double x5 = Double.parseDouble(a[9]);
180 Double y5 = Double.parseDouble(a[10]);
181 Double x6 = Double.parseDouble(a[11]);
182 Double y6 = Double.parseDouble(a[12]);
183 Double x7 = Double.parseDouble(a[13]);
184 Double y7 = Double.parseDouble(a[14]);
185 Double x8 = Double.parseDouble(a[15]);
186 Double y8 = Double.parseDouble(a[16]);
187 Double x9 = Double.parseDouble(a[17]);
188 Double y9 = Double.parseDouble(a[18]);
189 Double x10 = Double.parseDouble(a[19]);
190 Double y10 = Double.parseDouble(a[20]);
191 String b1[]= {"0",a[1],a[2],a[3],a[4],a[5],a[6],a[7],a[8],a[9],a[10] };
192 String b2[]= {"0",a[11],a[12],a[13],a[14],a[15],a[16],a[17],a[18],a[19],a[20] };
193 String b3[]={"0","0","0","0","0",a[1],a[2],a[3],a[4],a[5],a[6],a[7],a[8],a[9],a[10] };
194 String b4[]={"0","0","0","0","0",a[11],a[12],a[13],a[14],a[15],a[16],a[17],a[18],a[19],a[20] };
195 f1=pen(b1);
196 f2=pen(b2);
197 m1=fiveTofour.four(b3);
198 m2=fiveTofour.four(b4);
199 n1=fiveTOthree.three(b3);
200 n2=fiveTOthree.three(b4);
201 if(f1==true)
202 {
203 if(f2==true)
204 Five.five(a);
205 if(m2!=0)
206 {
207 if(m2==1)
208 Five.four(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,x7,y7,x8,y8,x9,y9);
209 if(m2==2)
210 Five.four(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,x7,y7,x8,y8,x10,y10);
211 if(m2==3)
212 Five.four(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,x7,y7,x9,y9,x10,y10);
213 if(m2==4)
214 Five.four(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,x8,y8,x9,y9,x10,y10);
215 if(m2==5)
216 Five.four(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x7,y7,x8,y8,x9,y9,x10,y10);
217 }
218 if(n2!=0)
219 {
220 if(n2==1)
221 Five.three(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,x7,y7,x8,y8);
222 if(n2==2)
223 Five.three(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,x7,y7,x9,y9);
224 if(n2==3)
225 Five.three(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,x7,y7,x10,y10);
226 if(n2==4)
227 Five.three(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,x8,y8,x9,y9);
228 if(n2==5)
229 Five.three(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,x8,y8,x10,y10);
230 if(n2==6)
231 Five.three(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x7,y7,x8,y8,x9,y9);
232 if(n2==7)
233 Five.three(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x7,y7,x8,y8,x10,y10);
234 if(n2==8)
235 Five.three(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x8,y8,x9,y9,x10,y10);
236 if(n2==9)
237 Five.three(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x6,y6,x9,y9,x10,y10);
238 if(n2==10)
239 Five.three(x1,y1,x2,y2,x3,y3,x4,y4,x5,y5,x7,y7,x9,y9,x10,y10);
240
241 }
242
243 }
244
245 if(m1!=0)
246 {
247 if(f2==true)
248 System.out.println("the previous quadrilateral is interlaced with the following pentagon");
249 if(m2!=0)
250 System.out.println("the previous quadrilateral is interlaced with the following quadrilateral");
251 if(n2!=0)
252 System.out.println("the previous quadrilateral is interlaced with the following triangle");
253 }
254 if(n1!=0)
255 {
256 if(f2==true)
257 System.out.println("the previous triangle is interlaced with the following pentagon");
258 if(m2!=0)
259 System.out.println("the previous triangle is interlaced with the following quadrilateral");
260 if(n2!=0)
261 System.out.println("the previous triangle is interlaced with the following triangle");
262 }
263
264 }
265
266 public static void handle5(String[] tokens1) {
267 System.out.println("4.0");
268 }
269
270 public static void handle6(String[] tokens1) {
271 String[] a;
272 a=tokens1;
273 int n1,n2;
274 boolean f=false;
275 String b1[]= {"0",a[3],a[4],a[5],a[6],a[7],a[8],a[9],a[10],a[11],a[12]};
276 String b2[]= {"0","0","0","0","0",a[3],a[4],a[5],a[6],a[7],a[8],a[9],a[10],a[11],a[12]};
277 f=pen(b1);
278 n1=fiveTofour.four(b2);
279 n2=fiveTOthree.three(b2);
280 if(n1==0&&n2==0&&f==true)
281 {
282 point.five(a);
283 }
284 if(n1!=0&&n2==0&&f==false)
285 {
286 point.four(a,n1);
287 }
288 if(n1==0&&n2!=0&&f==false)
289 {
290 point.three(a,n2);
291 }
292 }
293
294 public static boolean pen(String[] a) {//五边形的判断
295 double[] b1,b2,b3,b4,b5;
296 Double x1 = Double.parseDouble(a[1]);
297 Double y1 = Double.parseDouble(a[2]);
298 Double x2 = Double.parseDouble(a[3]);
299 Double y2 = Double.parseDouble(a[4]);
300 Double x3 = Double.parseDouble(a[5]);
301 Double y3 = Double.parseDouble(a[6]);
302 Double x4 = Double.parseDouble(a[7]);
303 Double y4 = Double.parseDouble(a[8]);
304 Double x5 = Double.parseDouble(a[9]);
305 Double y5 = Double.parseDouble(a[10]);
306 b1=Line.nod(x1,y1,x2,y2,x3,y3,x4,y4);
307 b2=Line.nod(x1,y1,x2,y2,x4,y4,x5,y5);
308 b3=Line.nod(x2,y2,x3,y3,x5,y5,x1,y1);
309 b4=Line.nod(x2,y2,x3,y3,x4,y4,x5,y5);
310 b5=Line.nod(x3,y3,x4,y4,x5,y5,x1,y1);
311 if((y1-y2)*(x2-x3)-(y2-y3)*(x1-x2)==0)
312 return false;
313 else if((y2-y3)*(x3-x4)-(y3-y4)*(x2-x3)==0)
314 return false;
315 else if((y3-y4)*(x4-x5)-(y4-y5)*(x3-x4)==0)
316 return false;
317 else if((y4-y5)*(x5-x1)-(y5-y1)*(x4-x5)==0)
318 return false;
319 else if((y5-y1)*(x1-x2)-(y1-y2)*(x5-x1)==0)
320 return false;
321 else if( ( (b1[0]>x1&&b1[0]<x2) || (b1[0]>x2&&b1[0]<x1) ) && ( (b1[1]>y1&&b1[1]<y2) || (b1[1]>y2&&b1[1]<y1) ) && ( (b1[0]>x3&&b1[0]<x4) || (b1[0]>x4&&b1[0]<x3) ) && ( (b1[1]>y3&&b1[1]<y4) || (b1[1]>y4&&b1[1]<y3) ) )
322 return false;
323 else if( ( (b2[0]>x1&&b2[0]<x2) || (b2[0]>x2&&b2[0]<x1) ) && ( (b2[1]>y1&&b2[1]<y2) || (b2[1]>y2&&b2[1]<y1) ) && ( (b2[0]>x4&&b2[0]<x5) || (b2[0]>x5&&b2[0]<x4) ) && ( (b2[1]>y4&&b2[1]<y5) || (b2[1]>y5&&b2[1]<y4) ) )
324 return false;
325 else if( ( (b3[0]>x2&&b3[0]<x3) || (b3[0]>x3&&b3[0]<x2) ) && ( (b3[1]>y2&&b3[1]<y3) || (b3[1]>y3&&b3[1]<y2) ) && ( (b3[0]>x5&&b3[0]<x1) || (b3[0]>x1&&b3[0]<x5) ) && ( (b3[1]>y5&&b3[1]<y1) || (b3[1]>y1&&b3[1]<y5) ) )
326 return false;
327 else if( ( (b4[0]>x2&&b4[0]<x3) || (b4[0]>x3&&b4[0]<x2) ) && ( (b4[1]>y2&&b4[1]<y3) || (b4[1]>y3&&b4[1]<y2) ) && ( (b4[0]>x4&&b4[0]<x5) || (b4[0]>x5&&b4[0]<x4) ) && ( (b4[1]>y4&&b4[1]<y5) || (b4[1]>y5&&b4[1]<y4) ) )
328 return false;
329 else if( ( (b5[0]>x3&&b5[0]<x4) || (b5[0]>x4&&b5[0]<x3) ) && ( (b5[1]>y3&&b5[1]<y4) || (b5[1]>y4&&b5[1]<y3) ) && ( (b5[0]>x5&&b5[0]<x1) || (b5[0]>x1&&b5[0]<x5) ) && ( (b5[1]>y5&&b5[1]<y1) || (b5[1]>y1&&b5[1]<y5) ) )
330 return false;
331 else
332 return true;
333
334 }
335
336 }
3、期中考试：
期中考试主要用了类的分类，继承与多态，容器类。
类图：
主要代码：
import java.util.ArrayList;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO 自动生成的方法存根
Scanner input = new Scanner(System.in);
int choice = input.nextInt();
double x;
double y;
double z;
double w;
int number;
String color;
GeometryObject g = new GeometryObject();
while(choice != 0) {
switch(choice) {
case 1:{
x=input.nextDouble();
y=input.nextDouble();
Point p = new Point(x,y);
g.add(p);
}
break;
case 2:{
x=input.nextDouble();
y=input.nextDouble();
Point p1 = new Point(x,y);
z=input.nextDouble();
w=input.nextDouble();
Point p2 = new Point(z,w);
color = input.next();
Line L = new Line(p1,p2,color);
g.add(L);
}
break;
case 3:{
color = input.next();
Plane p = new Plane(color);
g.add(p);
}
break;
case 4:{
number = input.nextInt();
if(g.getList().size()>=number)
g.remove(number-1);
}
}
choice = input.nextInt();
}
ArrayList<Element> list = g.getList();
for(int i=0;i<list.size();i++) {
list.get(i).display();
}
}
}
三、踩坑心得：
这三周，我们写这个pta其中修复了很多bug，比如在写四边形的时候，由于没有应用到类的分类，导致主函数的代码量超级长，在运行的时候输进去啥都显示不出来，当想去找错误的时候，发现根本找不出来，因为代码太过长了，，只好删了全部重新写，从那次之后我就知道了如何去分类。
其次，我在写五边形的时候，发现有很多的代码是重复的，我就通过传参就将原本的代码简化了。
有时候，我在输入相应的格式后，我发现并不是我想要的东西，我会通过debug去寻找我的错误，好在我全部都找出来了，有时候输出的相差很远，用debug都找不出来，我当时debug了好久，找了半天的错误，发现我代码打错了，当时真的会气死。
再考期中考试的时候，由于太紧张了，导致好多的知识点都忘记了，其次也是自己的知识不牢，导致自己没能满分。
四、改进建议：
1、第二次7-2我认为可以将串口字符的判断可以重新建立一个类，进行判断串口字符是否符合，如果符合的话，然后再建立一个新类去进行输出的判断。
2、在第三次大作业中我可以将一些点与线的的距离，点与点的距离等等，做一个类，然后在一些需要的地方去引用，判断，而不是将这些代码堆在一起，照成代码的复杂性高，可读性不高，而且在后期进行修改时会遇到一些麻烦，并且将一些类与方法的调用更明了一些。
1、算点与点的长度
class length1{
public static double longth(double x1,double x2,double y1,double y2) {
double l1,l;
l1=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
l=Math.sqrt(l1);
return l;
}
}
解释：传入两个坐标，计算这两个坐标之间的距离
2、线与线的sin与cos值
class triangle{
public static double sin(double a,double b,double c) {
double num,num1;
num=(b*b+c*c-a*a)/(2*b*c);
num1=Math.sqrt(1-num*num);
return num1;
}
}
class triangle1{
public static double cos(double a,double b,double c) {
double num;
num=(b*b+c*c-a*a)/(2*b*c);
return num;
}
}
解释 ：先用余弦定理求出cos再用1-cos*cos开根号变成sin。
3、交点
class nodical{
public static double[] nod(double x1,double y1,double x2,double y2,double x3,double y3,double x4,double y4) {
double x,y;
double[] a=new double[2];
x=(x3*(y3-y4)*(x2-x1)-x1*(y2-y1)*(x3-x4)+(y1-y3)*(x2-x1)*(x3-x4))/((y3-y4)*(x2-x1)-(y2-y1)*(x3-x4));
y=(y1*(x2-x1)*(y3-y4)-y3*(x3-x4)*(y2-y1)+(x3-x1)*(y3-y4)*(y2-y1))/((x2-x1)*(y3-y4)-(x3-x4)*(y2-y1));
a[0]=x;
a[1]=y;
return a;
}
}
解释：传入四个点，经过计算得到交点的x，y。
五、优化和输出
其实第四次和第五次也是可以划归到合并三角函数的：对于每一个表达式，我们如果可以定义其哈希值，那么sin(x)与sin((x^2+1))便显然可以以x和x^2+1作为两个sin的键，其指数作为值，这样键值关系就出现了，便依然可以做到对于每个三角函数迅速找到与其对应的cos^2 sin^2 与 1。于是就可以用类似第二次的方法进行优化。但是考虑到嵌套因子的存在，这个优化所需的时间开销应当是会比第一次大不少的。
六、总结
对于自己写的程序来说bug还是有很多的，但形成的原因也是有很多的，比如类的混乱，还有随机数的产生导致有些点可以过，但随机数可能使那些点不能过。还有那些正则表达式超级长，看到就脑阔疼，今后我个人会尽量避免写出这样的代码，除非真的万不得已。
当然在写第五次大作业的时候7-1中的第3次情况写了一大段，但是当运行的时候就不能运行，一开始以为时是单纯的进不了循环，当测试几个样例的时候发现可以进入，那个时候就快崩溃了，当时代码写的太杂了，而且都是很混乱的，导致自己根本改不好，只能无奈的全删了，去骗分。当时就后悔了，应该讲那些代码写进一个类，在类里面进行判断，这样代码既不会写错，到时候改的时候也会更清晰明了一点。不会像这样气急败坏的删除了。
面向对象不一定比面向过程，面向结果优越，但是优秀的封装和类一定好太多了，第三次作业能写满分，我相信他的封装和代码一定比别人更清晰明了，也更容易看的懂。
展望：
经过这三个周的洗礼，我觉得我的代码能力有明显的提升，当然有些地方也有些不足，同时自我学习能力也会更强，因为那些正则表达式是自己在网络上找的资料学习的，当然大家的进步速度也是迅猛的，我也要加把劲。
在后面两次作业中，能明显感受到代码量增加，期中全部加起来都没有400行，但是第四五次大作业，一个题目就会有1000多行代码，因为里面包含很多算法和方法。
期望：
在这三周我学到了很多东西，例如一些函数的运用，同时也吸取了教训，积累了经验。
标签：x1,double,PTA,Blog,x2,&&,y1,第二次,y2 来源： https://www.cnblogs.com/zmhjavablog/p/16270644.html
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