题意：给两个点的x,y坐标以及当前点与他们的角度，求当前点的位置。

题解：每一个点的坐标与角度可以构成一条直线，然后求两直线交点。

View Code 
 1 #include<cstdlib>
 2 #include<cmath>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #define max(a,b) (((a)>(b))?(a):(b))
 6 #define min(a,b) (((a)>(b))?(b):(a))
 7 #define sign(x) ((x)>eps?1:((x)<-eps?(-1):(0))) //符号函数
 8 using namespace std;
 9 const int MAXN=1000;
10 const double eps=1e-8,inf=1e50,PI=acos(-1.0);
11 struct point
12 {
13     double x,y;
14     point(){}
15     point(double _x,double _y){x=_x;y=_y;}
16 };
17 struct line
18 {
19     point a,b;
20     line(){}
21     line(point _a,point _b){a=_a;b=_b;}
22 };
23 inline double xmult(point o,point a,point b)
24 {
25     return (a.x-o.x)*(b.y-o.y)-(b.x-o.x)*(a.y-o.y);
26 }
27 inline double xmult(double x1,double y1,double x2,double y2)
28 {
29     return x1*y2-x2*y1;
30 }
31 inline double dmult(point o,point a,point b)
32 {
33     return (a.x-o.x)*(b.x-o.x)+(a.y-o.y)*(b.y-o.y);
34 }
35 //求直线交点，必须存在交点，或者预判断
36 point line_intersection(line u,line v)
37 {
38     double a1=u.b.y-u.a.y,b1=u.a.x-u.b.x;
39     double c1=u.b.y*(-b1)-u.b.x*a1;
40     double a2=v.b.y-v.a.y,b2=v.a.x-v.b.x;
41     double c2=v.b.y*(-b2)-v.b.x*a2;
42     double D=xmult(a1,b1,a2,b2);
43     return point(xmult(b1,c1,b2,c2)/D,xmult(c1,a1,c2,a2)/D);
44 }
45 void counter (double &x,double &y,double th)
46 {
47 //绕原点逆时针旋转th弧度,顺时针则传入-th
48     double tx=x,ty=y;
49     x=tx*cos(th)-ty*sin(th);
50     y=tx*sin(th)+ty*cos(th);
51     return;
52 }
53 int main()
54 {
55     int T;
56     for(scanf("%d",&T);T;T--)
57     {
58         point a,b,c;
59         line u,v;
60         double d1,d2,x,y;
61         scanf("%lf%lf%lf",&a.x,&a.y,&d1);
62         scanf("%lf%lf%lf",&b.x,&b.y,&d2);
63         x=0,y=-1;
64         counter(x,y,-d1/180.0*PI);
65         u=line(a,point(a.x+x,a.y+y));
66         x=0,y=-1;
67         counter(x,y,-d2/180.0*PI);
68         v=line(b,point(b.x+x,b.y+y));
69         c=line_intersection(u,v);
70         printf("%.4lf %.4lf\n",c.x,c.y);
71     }
72     return 0;
73 }