Toxophily
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1429 Accepted Submission(s): 739
Problem Description
The recreation center of WHU ACM Team has indoor billiards, Ping Pang, chess and bridge, toxophily, deluxe ballrooms KTV rooms, fishing, climbing, and so on.
We all like toxophily.
Bob is hooked on toxophily recently. Assume that Bob is at point (0,0) and he wants to shoot the fruits on a nearby tree. He can adjust the angle to fix the trajectory. Unfortunately, he always fails at that. Can you help him?
We all like toxophily.
Bob is hooked on toxophily recently. Assume that Bob is at point (0,0) and he wants to shoot the fruits on a nearby tree. He can adjust the angle to fix the trajectory. Unfortunately, he always fails at that. Can you help him?
Now given the object's coordinates, please calculate the angle between the arrow and x-axis at Bob's point. Assume that g=9.8N/m.
Input
The input consists of several test cases. The first line of input consists of an integer T, indicating the number of test cases. Each test case is on a separated line, and it consists three floating point numbers: x, y, v. x and y indicate the coordinate of
the fruit. v is the arrow's exit speed.
Technical Specification
1. T ≤ 100.
2. 0 ≤ x, y, v ≤ 10000.
Technical Specification
1. T ≤ 100.
2. 0 ≤ x, y, v ≤ 10000.
Output
For each test case, output the smallest answer rounded to six fractional digits on a separated line.
Output "-1", if there's no possible answer.
Please use radian as unit.
Output "-1", if there's no possible answer.
Please use radian as unit.
Sample Input
3 0.222018 23.901887 121.909183 39.096669 110.210922 20.270030 138.355025 2028.716904 25.079551
已知发射点坐标为(0，0)和重力加速度g=9.8，给出目标的坐标和初速度。求可以击中目标的最小仰角。有两种思路。第一种是直接如果可以击中目标。写出公式，化成一元二次方程，把公式内的三角函数所有化成tan，推断[0。PI/2]有无解；另外一种方法就是三分+二分。首先三分仰角，求出轨迹在x处的纵坐标最大值。若纵坐标最大值小于y，则直接输出-1，三分过后[0，r]上就是单调递增的，直接二分就可以。
#include<stack>//推导公式
#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma commment(linker,"/STACK: 102400000 102400000")
#define mset0(t) memset(t,0,sizeof(t))
#define lson a,b,l,mid,cur<<1
#define rson a,b,mid+1,r,cur<<1|1
using namespace std;
const double PI=3.141592653;
const double eps=1e-8;
const int MAXN=500020;
const double g=9.8;
double x,y,v,ans;
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
int tcase;
scanf("%d",&tcase);
while(tcase--)
{
scanf("%lf%lf%lf",&x,&y,&v);
/*if(x==0||y==0) //这个if语句主要是用来特判0 50 10000这样的数据的。但不知道为什么去掉这个也能AC
{
if(x==0&&y==0)
printf("0.000000
");
else if(y==0)
printf("-1
");
else
if(v*v*0.5/g>=y)
printf("%.6lf
",PI/2);
else
printf("-1
");
continue;
}*/
ans=3;
double a=g*x*x;
double b=-2*v*v*x;
double c=2*v*v*y+g*x*x;
double der=b*b-4*a*c;
if(der<0)
{
printf("-1
");
continue;
}
double ans1=atan((-b-sqrt(der))/(2*a));
double ans2=atan((-b+sqrt(der))/(2*a));
if(ans1>=0&&ans1<=PI/2)
ans=min(ans,ans1);
if(ans2>=0&&ans2<=PI/2)
ans=min(ans,ans2);
printf("%.6lf
",ans);
}
return 0;
}
#include<stack>//三分+二分代码
#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma commment(linker,"/STACK: 102400000 102400000")
#define mset0(t) memset(t,0,sizeof(t))
#define lson a,b,l,mid,cur<<1
#define rson a,b,mid+1,r,cur<<1|1
using namespace std;
const double PI=3.141592653;
const double eps=1e-8;
const int MAXN=500020;
const double g=9.8;
double x,y,v,ans;
double geth(double r)
{
return (v*sin(r))*(x/(v*cos(r)))-0.5*g*(x/(v*cos(r)))*(x/(v*cos(r)));
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
int tcase;
scanf("%d",&tcase);
while(tcase--)
{
scanf("%lf%lf%lf",&x,&y,&v);
if(x==0&&y==0)
{
printf("0.000000
");
continue;
}
if(y==0)
{
printf("-1
");
continue;
}
if(x==0)
{
if(v*v*0.5/g>=y)
printf("%.6lf
",PI/2);
else
printf("-1
");
continue;
}
double l=0,r=PI/2;
int cnt=10000;
while(cnt--)
{
double mid=(l+r)/2;
double mmid=(mid+r)/2;
if(geth(mid)>geth(mmid))
r=mmid;
else
l=mid;
}
if(geth(r)<y)
{
printf("-1
");
continue;
}
l=0;
r=r;
cnt=10000;
while(cnt--)
{
double mid=(l+r)/2;
if(geth(mid)>y)
r=mid;
else
l=mid;
}
printf("%.6lf
",r);
}
return 0;
}