2015
04-13

# Where am I

I am a ball on a circle table. The table’s surface is so smooth that I can move without energy lost.
Now you are to give me a hit vector, I want to know where I will be after T seconds. Collisions between me and the edge of the table satisfy Conservation of Momentum Theorem , also my speed after collision is as fast as before. Look for the figure below to know it more clearly:

The first line contains a number N, the number of test cases you have to proceed.
Each test case has three lines.First line has three numbers, x1,y1,r1, the center and radius of the table. Second line has x2,y2,r2, the parameters of me (the ball). The third line consists of dx, dy, T: (dx,dy) is the hit vector, T is the time I have moved since you hit me. To explain hit vector: if I am at position (x0,y0) now ,next second I will be at position (x0+dx,y0+dy) without conllision. All input numbers are non-negative integer and less than 10^6.Also you can assume the ball must be in the table.

The first line contains a number N, the number of test cases you have to proceed.
Each test case has three lines.First line has three numbers, x1,y1,r1, the center and radius of the table. Second line has x2,y2,r2, the parameters of me (the ball). The third line consists of dx, dy, T: (dx,dy) is the hit vector, T is the time I have moved since you hit me. To explain hit vector: if I am at position (x0,y0) now ,next second I will be at position (x0+dx,y0+dy) without conllision. All input numbers are non-negative integer and less than 10^6.Also you can assume the ball must be in the table.

2
5 5 10
5 5 1
1 0 30
1 1 10
1 0 2
1 1 100

-1.0 5.0
3.4 -0.5

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define eps 1e-8
struct point
{
point(){}
point(double x, double y) : x(x), y(y){}
double x,y;
}A, B;
double xmult(point p1,point p2,point p0){
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
double dist(point p1,point p2){
return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
double dist2(point p1,point p2){
return ((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
point intersection(point u1,point u2,point v1,point v2){
point ret=u1;
double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))
/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
ret.x+=(u2.x-u1.x)*t;
ret.y+=(u2.y-u1.y)*t;
return ret;
}
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//�����߶���Բ�Ľ����������������е��Ƿ����߶���
void intersection_line_circle(point c,double r,point l1,point l2,point& p1,point& p2){
point p=c;
double t;
p.x+=l1.y-l2.y;
p.y+=l2.x-l1.x;
p=intersection(p,c,l1,l2);
t=sqrt(r*r- dist (p,c)* dist (p,c))/ dist (l1,l2);
p1.x=p.x+(l2.x-l1.x)*t;
p1.y=p.y+(l2.y-l1.y)*t;
p2.x=p.x-(l2.x-l1.x)*t;
p2.y=p.y-(l2.y-l1.y)*t;
}
int t;
double r1, r2, T, pos[2];
point rotate(point v,point p,double angle,double scale){
point ret=p;
v.x-=p.x,v.y-=p.y;
p.x=scale*cos(angle);
p.y=scale*sin(angle);
ret.x+=v.x*p.x-v.y*p.y;
ret.y+=v.x*p.y+v.y*p.x;
return ret;
}
point GetTarget(double TT)
{
point a, b, ans;
intersection_line_circle(A, r1 - r2, B, point(B.x + pos[0], B.y + pos[1]), a, b);
if ((a.x - B.x) * pos[0] >= 0 && (a.y - B.y) * pos[1] >= 0)
ans = a;
else ans = b;
//cout<<ans.x<<' '<<ans.y<<endl;
double dis = dist(ans, B);
//cout<<dis<<endl;
if (dis - TT >= -eps)
return point(B.x + TT * (ans.x - b.x) / dis, B.y + TT * (ans.y - b.y) / dis);
TT -= dis;//ʣ���·��
double cosx = (dist2(A, ans) + dist2(B, ans) - dist2(A, B)) * 0.5 / (dist(A, ans) * dist(B, ans));

double len = cosx * (r1 - r2) * 2;
cosx = acos(cosx);
cosx = acos(-1.0) - cosx * 2;

double tmp = xmult(B, A, ans);
//cout<<"TMP:"<<tmp<<endl;
if (tmp > 0) cosx = -cosx;
long long num = (long long)(TT / len);
TT -= num * len;
//cout<<cosx<<endl;
point ans1 = rotate(ans, A, cosx * num, 1.0);
//cout<<ans.x<<' '<<ans.y<<endl<<ans1.x<<' '<<ans1.y<<endl;
point ans2 = rotate(ans, A, cosx * (num + 1), 1.0);
return point(ans1.x + TT * (ans2.x - ans1.x) / len, ans1.y + TT * (ans2.y - ans1.y) / len);
}
int main()
{
scanf("%d", &t);
while(t--)
{
scanf("%lf%lf%lf", &A.x, &A.y, &r1);
scanf("%lf%lf%lf", &B.x, &B.y, &r2);
scanf("%lf%lf%lf", &pos[0], &pos[1], &T);
T = T * sqrt(pos[0] * pos[0] + pos[1] * pos[1]);
point ans = GetTarget(T);
printf("%.1f %.1f\n", ans.x, ans.y);
}
return 0;
}

1. 看看我们的邻居日、韩，再放眼全球，哪个治国靠的是黄老之术？批孔最凶的时候，我们取得了什么政治文明？兴老灭孔，独尊老子，是不是又走向了极端？天天折腾我们的文化，对制度和统治者不敢发声，这就是你们的本事！

2. 看看我们的邻居日、韩，再放眼全球，哪个治国靠的是黄老之术？批孔最凶的时候，我们取得了什么政治文明？兴老灭孔，独尊老子，是不是又走向了极端？天天折腾我们的文化，对制度和统治者不敢发声，这就是你们的本事！

3. 看看我们的邻居日、韩，再放眼全球，哪个治国靠的是黄老之术？批孔最凶的时候，我们取得了什么政治文明？兴老灭孔，独尊老子，是不是又走向了极端？天天折腾我们的文化，对制度和统治者不敢发声，这就是你们的本事！

4. 看看我们的邻居日、韩，再放眼全球，哪个治国靠的是黄老之术？批孔最凶的时候，我们取得了什么政治文明？兴老灭孔，独尊老子，是不是又走向了极端？天天折腾我们的文化，对制度和统治者不敢发声，这就是你们的本事！

5. 看看我们的邻居日、韩，再放眼全球，哪个治国靠的是黄老之术？批孔最凶的时候，我们取得了什么政治文明？兴老灭孔，独尊老子，是不是又走向了极端？天天折腾我们的文化，对制度和统治者不敢发声，这就是你们的本事！

6. 看看我们的邻居日、韩，再放眼全球，哪个治国靠的是黄老之术？批孔最凶的时候，我们取得了什么政治文明？兴老灭孔，独尊老子，是不是又走向了极端？天天折腾我们的文化，对制度和统治者不敢发声，这就是你们的本事！

7. 看看我们的邻居日、韩，再放眼全球，哪个治国靠的是黄老之术？批孔最凶的时候，我们取得了什么政治文明？兴老灭孔，独尊老子，是不是又走向了极端？天天折腾我们的文化，对制度和统治者不敢发声，这就是你们的本事！

8. 看看我们的邻居日、韩，再放眼全球，哪个治国靠的是黄老之术？批孔最凶的时候，我们取得了什么政治文明？兴老灭孔，独尊老子，是不是又走向了极端？天天折腾我们的文化，对制度和统治者不敢发声，这就是你们的本事！