题意：有两个一大一小的同心圆，圆心在原点，大圆外有一小圆，其圆心有一个速度(vx, vy)，如果碰到了小圆会反弹，问该圆在大圆内运动的时间
分析：将圆外的小圆看成一个点，判断该直线与同心圆的交点，根据交点个数计算时间。用到了直线的定义，圆的定义，直线与圆交点的个数。
/************************************************
* Author :Running_Time
* Created Time :2015/10/24 星期六 16:14:33
* File Name :C.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
int dcmp(double x) { //三态函数，减少精度问题
if (fabs (x) < EPS) return 0;
else return x < 0 ? -1 : 1;
}
struct Point { //点的定义
double x, y;
Point (double x=0, double y=0) : x (x), y (y) {}
Point operator + (const Point &r) const { //向量加法
return Point (x + r.x, y + r.y);
}
Point operator - (const Point &r) const { //向量减法
return Point (x - r.x, y - r.y);
}
Point operator * (double p) { //向量乘以标量
return Point (x * p, y * p);
}
Point operator / (double p) { //向量除以标量
return Point (x / p, y / p);
}
bool operator < (const Point &r) const { //点的坐标排序
return x < r.x || (x == r.x && y < r.y);
}
bool operator == (const Point &r) const { //判断同一个点
return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;
}
};
typedef Point Vector; //向量的定义
Point read_point(void) { //点的读入
double x, y;
scanf ("%lf%lf", &x, &y);
return Point (x, y);
}
double polar_angle(Vector A) { //向量极角
return atan2 (A.y, A.x);
}
double dot(Vector A, Vector B) { //向量点积
return A.x * B.x + A.y * B.y;
}
double cross(Vector A, Vector B) { //向量叉积
return A.x * B.y - A.y * B.x;
}
double length(Vector A) { //向量长度，点积
return sqrt (dot (A, A));
}
double angle(Vector A, Vector B) { //向量转角，逆时针，点积
return acos (dot (A, B) / length (A) / length (B));
}
double area_triangle(Point a, Point b, Point c) { //三角形面积，叉积
return fabs (cross (b - a, c - a)) / 2.0;
}
Vector rotate(Vector A, double rad) { //向量旋转，逆时针
return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));
}
Vector nomal(Vector A) { //向量的单位法向量
double len = length (A);
return Vector (-A.y / len, A.x / len);
}
Point point_inter(Point p, Vector V, Point q, Vector W) { //两直线交点，参数方程
Vector U = p - q;
double t = cross (W, U) / cross (V, W);
return p + V * t;
}
double dis_to_line(Point p, Point a, Point b) { //点到直线的距离，两点式
Vector V1 = b - a, V2 = p - a;
return fabs (cross (V1, V2)) / length (V1);
}
double dis_to_seg(Point p, Point a, Point b) { //点到线段的距离，两点式
if (a == b) return length (p - a);
Vector V1 = b - a, V2 = p - a, V3 = p - b;
if (dcmp (dot (V1, V2)) < 0) return length (V2);
else if (dcmp (dot (V1, V3)) > 0) return length (V3);
else return fabs (cross (V1, V2)) / length (V1);
}
Point point_proj(Point p, Point a, Point b) { //点在直线上的投影，两点式
Vector V = b - a;
return a + V * (dot (V, p - a) / dot (V, V));
}
bool inter(Point a1, Point a2, Point b1, Point b2) { //判断线段相交，两点式
double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1),
c3 = cross (b2 - b1, a1 - b1), c4 = cross (b2 - b1, a2 - b1);
return dcmp (c1) * dcmp (c2) < 0 && dcmp (c3) * dcmp (c4) < 0;
}
bool on_seg(Point p, Point a1, Point a2) { //判断点在线段上，两点式
return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0;
}
double area_poly(Point *p, int n) { //多边形面积
double ret = 0;
for (int i=1; i<n-1; ++i) {
ret += fabs (cross (p[i] - p[0], p[i+1] - p[0]));
}
return ret / 2;
}
/*
点集凸包，输入点集会被修改
*/
vector<Point> convex_hull(vector<Point> &P) {
sort (P.begin (), P.end ());
P.erase (unique (P.begin (), P.end ()), P.end ()); //预处理，删除重复点
int n = P.size (), m = 0;
vector<Point> ret (n + 1);
for (int i=0; i<n; ++i) {
while (m > 1 && cross (ret[m-1]-ret[m-2], P[i]-ret[m-2]) < 0) m--;
ret[m++] = P[i];
}
int k = m;
for (int i=n-2; i>=0; --i) {
while (m > k && cross (ret[m-1]-ret[m-2], P[i]-ret[m-2]) < 0) m--;
ret[m++] = P[i];
}
if (n > 1) m--;
ret.resize (m);
return ret;
}
struct Circle {
Point c;
double r;
Circle () {}
Circle (Point c, double r) : c (c), r (r) {}
Point point(double a) {
return Point (c.x + cos (a) * r, c.y + sin (a) * r);
}
};
struct Line {
Point p;
Vector v;
double r;
Line () {}
Line (const Point &p, const Vector &v) : p (p), v (v) {
r = polar_angle (v);
}
Point point(double a) {
return p + v * a;
}
};
/*
直线相交求交点，返回交点个数，交点保存在P中
*/
int line_cir_inter(Line L, Circle C, double &t1, double &t2, vector<Point> &P) {
double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y;
double e = a * a + c * c, f = 2 * (a * b + c * d), g = b * b + d * d - C.r * C.r;
double delta = f * f - 4 * e * g;
if (dcmp (delta) < 0) return 0;
if (dcmp (delta) == 0) {
t1 = t2 = -f / (2 * e); P.push_back (L.point (t1));
return -1;
}
t1 = (-f - sqrt (delta)) / (2 * e); P.push_back (L.point (t1));
t2 = (-f + sqrt (delta)) / (2 * e); P.push_back (L.point (t2));
if (dcmp (t1) < 0 || dcmp (t2) < 0) return 0;
return 2;
}
/*
两圆相交求交点，返回交点个数。交点保存在P中
*/
int cir_cir_inter(Circle C1, Circle C2, vector<Point> &P) {
double d = length (C1.c - C2.c);
if (dcmp (d) == 0) {
if (dcmp (C1.r - C2.r) == 0) return -1; //两圆重叠
else return 0;
}
if (dcmp (C1.r + C2.r - d) < 0) return 0;
if (dcmp (fabs (C1.r - C2.r) - d) < 0) return 0;
double a = polar_angle (C2.c - C1.c);
double da = acos ((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d)); //C1C2到C1P1的角？
Point p1 = C1.point (a - da), p2 = C2.point (a + da);
P.push_back (p1);
if (p1 == p2) return 1;
else P.push_back (p2);
return 2;
}
int main(void) {
Circle C1, C2; Point a, b;
double Rm, R, r, x, y, vx, vy;
while (scanf ("%lf%lf%lf%lf%lf%lf%lf", &Rm, &R, &r, &x, &y, &vx, &vy) == 7) {
Rm += r; R += r;
C1 = Circle (Point (0, 0), Rm);
C2 = Circle (Point (0, 0), R);
a = Point (x, y); b = Point (vx, vy);
vector<Point> P1, P2; P1.clear (); P2.clear ();
double t1, t2, t3, t4, ans, speed = sqrt (vx * vx + vy * vy);
int num1 = line_cir_inter (Line (a, b), C1, t1, t2, P1);
int num2 = line_cir_inter (Line (a, b), C2, t3, t4, P2);
if (num2 == 2) {
if (num1 == 2) {
double len = length (P2[0] - P2[1]) - length (P1[0] - P1[1]);
ans = len / speed;
}
else {
ans = length (P2[0] - P2[1]) / speed;
}
}
else {
ans = 0.0;
}
printf ("%.4f\n", ans);
}
return 0;
}