HDU 2298 Toxophily(公式/三分+二分)

 

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?

 

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. 

 

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. 

 

Sample Input

3 0.222018 23.901887 121.909183 39.096669 110.210922 20.270030 138.355025 2028.716904 25.079551

 

 

 

已知發射點座標爲(0，0)和重力加速度g=9.8，給出目標的座標和初速度。求能夠擊中目標的最小仰角。有兩種思路。第一種是直接若是能夠擊中目標。寫出公式，化成一元二次方程，把公式內的三角函數全部化成tan，推斷[0。PI/2]有無解；另一種方法就是三分+二分。首先三分仰角，求出軌跡在x處的縱座標最大值。若縱座標最大值小於y，則直接輸出-1，三分事後[0，r]上就是單調遞增的，直接二分就能夠。

 

 

 

#include<stack>//推導公式
#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)    //這個if語句主要是用來特判0 50 10000這樣的數據的。但不知道爲何去掉這個也能AC
        {
            if(x==0&&y==0)
                printf("0.000000\n");
            else if(y==0)
                printf("-1\n");
            else
                if(v*v*0.5/g>=y)
                    printf("%.6lf\n",PI/2);
                else
                    printf("-1\n");
            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\n");
            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\n",ans);
    }
    return 0;
}

 

 

 

#include<stack>//三分+二分代碼
#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\n");
            continue;
        }
        if(y==0)
        {
            printf("-1\n");
            continue;
        }
        if(x==0)
        {
            if(v*v*0.5/g>=y)
                printf("%.6lf\n",PI/2);
            else
                printf("-1\n");
            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\n");
            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\n",r);
        
    }
    return 0;
}