2018多校第6场1013hdu6373Pinball

1 #include
2 #define pi acos(-1)
3 #define eps 1e-10
4 //double add(double a,double b)
5 //{
6 // if(abs(a+b) 7 // return a+b;
8 //}
9 using namespace std;
10 const int maxn=1e5+5;
11 int a,b;
12 struct P
13 {
14 double x,y;
15 P(){}
16 P(double x,double y):x(x),y(y){}
17 P operator+(P p)
18 {
19 return P(x+p.x,y+p.y);
20 }
21 P operator-(P p)
22 {
23 return P(x-p.x,y-p.y);
24 }
25 P operator*(double d)
26 {
27 return P(x*d,y*d);
28 // return add(x*p.x,y*p.y)
29 }
30 double dot(P p)
31 {
32 return x*p.x+y*p.y;
33 }
34 double det(P p)
35 {
36 return x*p.y-y*p.x;
37 }
38 };
39 bool on_seg(P p1,P p2,P q)
40 {
41 return (p1-q).det(p2-q)==0&&(p1-q).dot(p2-q)<=0;
42 }
43 P intersection(P p1,P p2,P q1,P q2)//两线交点
44 {
45 return p1+(p2-p1)*((q2-q1).det(q1-p1)/(q2-q1).det(p2-p1));
46 }
47 double dis(P p,P q)
48 {
49 return sqrt((p.x-q.x)*(p.x-q.x)+(p.y-q.y)*(p.y-q.y));
50 }
51 int main()
52 {
53 int T;
54 scanf("%d",&T);
55 double g=9.80,g1,g2,cn,sn;
56 while(T--)
57 {
58 int a,b,x,y;
59 scanf("%d %d %d %d",&a,&b,&x,&y);
60 cn=1.0*a/sqrt(a*a+b*b);
61 sn=1.0*b/sqrt(a*a+b*b);
62 g1=cn*g;g2=sn*g;
63 P p1,p2,p3,p4,pp;
64 p1.x=-a;p1.y=b;
65 p2.x=0;p2.y=0;
66 p3.x=x;p3.y=y;
67 p4.x=x+1;p4.y=y+1.0*a/b;
68 pp=intersection(p1,p2,p3,p4);
69 double len1=dis(pp,p3),len2=dis(pp,p2);
70 double t1=2*sqrt(2*g1*len1)/g1;
71 double t2=sqrt(2*len2/g2);
72 double ans=t2/t1;
73 ans+=0.5;
74 int an=ans;
75 printf("%d\n",an);
76 }
77 }