HDU 4793 Collision

【科普】什么是BestCoder？如何参加？

Collision Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 459    Accepted Submission(s): 180 Special Judge Problem Description There's a round medal fixed on an ideal smooth table, Fancy is trying to throw some coins and make them slip towards the medal to collide. There's also a round range which shares exact the same center as the round medal, and radius of the medal is strictly less than radius of the round range. Since that the round medal is fixed and the coin is a piece of solid metal, we can assume that energy of the coin will not lose, the coin will collide and then moving as reflect. Now assume that the center of the round medal and the round range is origin ( Namely (0, 0) ) and the coin's initial position is strictly outside the round range. Given radius of the medal R m, radius of coin r, radius of the round range R, initial position (x, y) and initial speed vector (vx, vy) of the coin, please calculate the total time that any part of the coin is inside the round range. Please note that the coin might not even touch the medal or slip through the round range.   Input There will be several test cases. Each test case contains 7 integers R m, R, r, x, y, vx and vy in one line. Here 1 ≤ R m < R ≤ 2000, 1 ≤ r ≤ 1000, R + r < |(x, y)| ≤ 20000, 1 ≤ |(vx, vy)| ≤ 100.   Output For each test case, please calculate the total time that any part of the coin is inside the round range. Please output the time in one line, an absolute error not more than 1e -3 is acceptable.   Sample Input 5 20 1 0 100 0 -1 5 20 1 30 15 -1 0   Sample Output 30.000 29.394   Source 2013 Asia Changsha Regional Contest

题意：有一个圆环......圆环外面有一个小圆C，一开始知道C的速度的向量，C如果碰到圆环的内圆，会反弹，假设不会损失能量，问C有多长时间在圆环的外圆和内圆之间。

思路：直接乱搞一下就好了吧。 先判断会不会与圆环的外圆相交，如果相交，再判断是否会与内圆相交。如果不与内圆相交，直接求出第一次与外圆相切的时间和第二次相切的时间就行了。 如果与内圆相交求出第一次与内圆相切的时间就行了，最后是这个时间的两倍，因为整个过程是对称的。

代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define rep(i,a,b) for(int i=(a);i<(b);++i)
#define rrep(i,b,a) for(int i = (b); i >= (a); --i)
#define clr(a,x) memset(a,(x),sizeof(a))
#define LL long long
#define eps 1e-8
using namespace std;
const double PI = 4.0 * atan(1.0);

struct Point
{
    double x,y;
    Point (double x=0,double y=0)
    :x(x),y(y) { }
};

int dcmp(double a,double b)
{
    if (a < b - eps) return -1;
    else if (a > b + eps) return 1;
    else return 0;
}

void input()
{
}

inline double sqr(double x) { return x * x; }

Point operator - (const Point & a,const Point & b)
{
    return Point(a.x-b.x,a.y-b.y);
}

inline double Dot(Point a,Point b) { return a.x * b.x + a.y * b.y ; }
inline double Length(Point a) { return sqrt(Dot(a,a)); }
inline double Len2(Point a) { return Dot(a,a); }
inline double Length(Point a,Point b) { return Length(a-b); }


inline double Angel(Point a,Point b)
{
    double X = (Len2(a) + Len2(b) - Len2(a-b)) / 2 / Length(a) / Length(b);
    return acos(X);
}

bool solve()
{
    double Rm,R,r,x0,y0,vx,vy;
    if (scanf("%lf%lf%lf%lf%lf%lf%lf",&Rm,&R,&r,&x0,&y0,&vx,&vy) != 7) return false;
    double A = vy, B = -vx , C = vx * y0 - vy * x0;
    double distToCenter = abs(C) / sqrt(A*A + B*B);
    Point center = Point(0,0); Point Ap = Point(x0,y0);
    double angel = Angel(Point(vx,vy),center-Ap);
    double cosA = cos(angel);
    double AC = Length(Ap-center);
    A = 1; B = -2.0 * AC * cosA; C = sqr(AC) - sqr(R+r);
    double delta = sqr(B) - 4.0 * A * C;
    if (dcmp(delta,0) <= 0 ||
         dcmp(distToCenter-r,R) >= 0 ||
          dcmp(angel,PI * 0.5) >= 0) {
        puts("0.000");
        return true;
    }

    delta = sqrt(delta);
    double d1 = (-B - delta) / 2, d2 = (-B + delta) / 2;
    double v = Length(Point(vx,vy));
    if (dcmp(distToCenter-r,Rm) >= 0) {
        double t = (d2 - d1) / v;
        printf("%.5lf\n",t);
        return true;
    }
    A = 1; B = -2.0 * AC * cosA; C = sqr(AC) - sqr(Rm+r);
    delta = sqrt(sqr(B) - 4.0 * A * C);
    double dm1 = (-B - delta) / 2;
    double t = (dm1 - d1) / v * 2;
    printf("%.5lf\n",t);
    return true;
}

int main()
{
    #ifdef ACM
        freopen("in.txt", "r", stdin);
    #endif // ACM
    while (solve()) {
       // input();
    }
}