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 }