几何题，然后解一下方程就行，然后注意浮点误差。
具体思路看注释。
CO
/*
PROGRAM: $PROGRAM
AUTHOR: Su Jiao
DATE: 2010-4-14
DESCRIPTION:
$DESCRIPTION
*/
#include <stdio.h>
#include <math.h>
#define g 10.0f
#define zero(n) (-1e-9<=n&&n<=1e-9)
#define oo 1e16
struct Point
{
double x,y,z;
Point(){}
Point(double _x,double _y,double _z):x(_x),y(_y),z(_z){}
};
Point C,N,S,V;
double R;
bool check(double t)
{
if (t<0) return false;
Point M(S.x+V.x*t,S.y+V.y*t,S.z+V.z*t-(g*t*t)/2);
Point d(M.x-C.x,M.y-C.y,M.z-C.z);
double distance_sqr=d.x*d.x+d.y*d.y+d.z*d.z;
if (distance_sqr<R*R) return true;
else return false;
}
int main()
{
scanf(“%lf %lf %lf %lf %lf %lf %lf\n”
“%lf %lf %lf %lf %lf %lf\n”,
&C.x,&C.y,&C.z,&N.x,&N.y,&N.z,&R,
&S.x,&S.y,&S.z,&V.x,&V.y,&V.z);
/*
M=S+V*t+(0,0,-(g*t^2)/2)
(M-C)*N=0
distance(M,C)<R
有解?HIT:MISSED
(M-C)*N=0
=> (Mx-Cx)*(Nx)+(My-Cy)*(Ny)+(Mz-Cz)*(Nz)=0
=> (Sx+Vx*t-Cx)*(Nx)+(Sy+Vy*t-Cy)*(Ny)+(Sz+Vz*t-(g*t^2)/2-Cz)*(Nz)=0
=> (-g*Nz)/2*t^2
+(Vx*Nx+Vy*Ny+Vz*Nz)*t
+(Sx*Nx+Sy*Ny+Sz*Nz-Cx*Nx-Cy*Ny-Cz*Nz)
=0
*/
bool have_answer=false;
double a=(-g*N.z)/2,
b=(V.x*N.x+V.y*N.y+V.z*N.z),
c=(S.x*N.x+S.y*N.y+S.z*N.z-C.x*N.x-C.y*N.y-C.z*N.z);
double t;
if (!zero(a))
{
double delta=b*b-4.0*a*c;
if (delta>0||zero(delta))
{
if (zero(delta)) delta=0;
double t1=(-b+sqrt(delta))/(2*a),
t2=(-b-sqrt(delta))/(2*a);
if (t1>=0)
if (check(t=t1))
have_answer=true;
if (t2>=0)
if (check(t=t2))
have_answer=true;
}
}
else
{
if (!zero(b))
if (check(t=-c/b))
have_answer=true;
}
if (have_answer) printf(“HIT\n”);
else printf(“MISSED\n”);
return 0;
}