Problem Description

Jiaoshou selected a course about “openGL” this semester. He was quite interested in modelview, which is a part of “openGL”. Just using three functions, it could make the model to move, rotate and largen or lessen. But he was puzzled with the theory of the modelview. He didn’t know a vertex after several transformations where it will be.

Now, He tells you the position of the vertex and the transformations. Please help Jiaoshou find the position of the vertex after several transformations.

Input

The input will start with a line giving the number of test cases, T.
Each case will always begin with “glBegin(GL_POINTS);”.Then the case will be followed by 5 kinds of function.
1. glTranslatef(x,y,z);
  This function will translate the vertex(x’,y’,z’) to vertex(x+x’,y+y’,z+z’).
2. glRotatef(angle,x,y,z);
  This function will turn angle radians counterclockwise around the axis (0,0,0)->(x,y,z).
3. glScalef(x,y,z);
  This function wiil translate the vertex(x’,y’,z’) to vertex(x*x’,y*y’,z*z’).
4. glVertex3f(x,y,z);
  This function will draw an initial vertex at the position(x,y,z). It will only appear once in one case just before “glEnd();”. In openGL, the transformation matrices are right multiplied by vertex matrix. So you should do the transformations in the reverse order. 
5. glEnd();
  This function tells you the end of the case.
In this problem angle,x,y,z are real numbers and range from -50.0 to 50.0. And the number of functions in each case will not exceed 100.

Output

For each case, please output the position of the vertex after several transformations x,y,z in one line rounded to 1 digits after the decimal point , separated with a single space. We guarantee that x,y,z are not very large.

Sample Input

Sample Output

2.0 1.0 3.0

Author

Water Problem SecKill Expert

Source

HDU “Valentines Day” Open Programming Contest 2010-02-14

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <cstring>
 5 using namespace std;
 6
 7 int cnt;
 8 char operat[105][105];
 9 double ang,x,y,z,xx,yy,zz;
10
11 struct Matrix
12 {
13     double m[4][4];
14 };
15
16 Matrix mult(Matrix a,Matrix b)//矩阵乘法
17 {
18     Matrix c;
19     for(int i=0;i<4;i++)
20     {
21         for(int j=0;j<4;j++)
22         {
23             c.m[i][j]=0.0;
24             for(int k=0;k<4;k++)
25                 c.m[i][j]+=a.m[i][k]*b.m[k][j];
26         }
27     }
28     return c;
29 }
30
31 Matrix Translate(int i)//平移变换
32 {
33     sscanf(operat[i],"glTranslatef(%lf,%lf,%lf);",&x,&y,&z);
34     Matrix tmp={1,0,0,0,0,1,0,0,0,0,1,0,x,y,z,1};
35     return tmp;
36 }
37 Matrix RatioTranslate(int i)//局部比例变换
38 {
39     sscanf(operat[i],"glScalef(%lf,%lf,%lf);",&x,&y,&z);
40     Matrix tmp={x,0,0,0,0,y,0,0,0,0,z,0,0,0,0,1};
41     return tmp;
42 }
43
44 Matrix RotateTranslate(int i)//绕通过原点的任意轴旋转变换
45 {
46     sscanf(operat[i],"glRotatef(%lf,%lf,%lf,%lf);",&ang,&x,&y,&z);
47     double t=sqrt(x*x+y*y+z*z);
48     double n1=x/t;//cos(a),a是与x轴的夹角
49     double n2=y/t;//cos(b),b是与y轴的夹角
50     double n3=z/t;//cos(c),c是与z轴的夹角
51     double S=sin(ang),C=cos(ang);
52     Matrix tmp={n1*n1+(1-n1*n1)*C,n1*n2*(1-C)+n3*S,n1*n3*(1-C)-n2*S,0,
53     n1*n2*(1-C)-n3*S,n2*n2+(1-n2*n2)*C,n2*n3*(1-C)+n1*S,0,
54     n1*n3*(1-C)+n2*S,n2*n3*(1-C)-n1*S,n3*n3+(1-n3*n3)*C,0,
55     0,0,0,1};
56     return tmp;
57 }
58
59 Matrix solve()
60 {
61     Matrix ret={1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1};
62     sscanf(operat[cnt-2], "glVertex3f(%lf,%lf,%lf);",&xx,&yy,&zz);
63     for(int i=cnt-3;i>0;i--)
64     {
65         if(operat[i][2]=='T')
66             ret=mult(ret,Translate(i));
67         else if(operat[i][2]=='S')
68             ret=mult(ret,RatioTranslate(i));
69         else if(operat[i][2]=='R')
70             ret=mult(ret,RotateTranslate(i));
71     }
72     return ret;
73 }
74
75 int main()
76 {
77     int t;
78     scanf("%d",&t);
79     getchar();
80     while(t--)
81     {
82         cnt=0;
83         while(1)
84         {
85             gets(operat[cnt++]);
86             if(operat[cnt-1][2]=='E')
87                 break;
88         }
89         Matrix ans=solve();
90         printf("%.1lf %.1lf %.1lf\n",xx*ans.m[0][0]+yy*ans.m[1][0]+zz*ans.m[2][0]+ans.m[3][0],
91             xx*ans.m[0][1]+yy*ans.m[1][1]+zz*ans.m[2][1]+ans.m[3][1],
92             xx*ans.m[0][2]+yy*ans.m[1][2]+zz*ans.m[2][2]+ans.m[3][2]);
93     }
94     return 0;
95 }