hdu 5144 NPY and shot 物理+三分

NPY and shot

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

NPY is going to have a PE test.One of the test subjects is throwing the shot.The height of NPY is H meters.He can throw the shot at the speed of v0 m/s and at the height of exactly H meters.He wonders if he throws the shot at the best angle,how far can he throw ?(The acceleration of gravity, g, is 9.8m/s2)

 

Input

The first line contains a integer T(T≤10000),which indicates the number of test cases.
The next T lines,each contains 2 integers H(0≤h≤10000m),which means the height of NPY,and v0(0≤v0≤10000m/s), which means the initial velocity.

 

Output

For each query,print a real number X that was rounded to 2 digits after decimal point in a separate line.X indicates the farthest distance he can throw.

 

Sample Input

2 0 1 1 2

 

Sample Output

0.10 0.99

Hint

If the height of NPY is 0,and he throws the shot at the 45° angle, he can throw farthest.

 

Source

BestCoder Round #22

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5144

题意：给你一个高度h，一个初始的速度v　；

　　　你选择一个角度扔使其最远；

思路：三分抛出的角度；

　　　显然函数成一段凸的抛物线形状；

　　　代码有注释；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define mk make_pair
#define eps 1e-7
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e6+10,M=4e6+10,inf=2147483647;
const ll INF=1e18+10,mod=1e9+7;

///   数组大小

double h,v;
double equ(double x)
{
    double G=9.8,vx=v*cos(x/180*pi),vy=v*sin(x/180*pi);
    double t=vy/G;//  到达最高点的时间
    t+=sqrt(2*G*h+vy*vy)/G; //到达顶端时间
    // 2*G*h+vy*vy为到达最下端的速度的平方
    return vx*t;
}

double ternarySearch(double l,double r)
{
    while(r-l>eps)
    {
        double L=(2*l+r)/3;
        double R=(l+2*r)/3;
        double ans1=equ(L);
        double ans2=equ(R);
        if(ans1<ans2)
            l=L;
        else
            r=R;
    }
    return equ(l);
}
int main()
{
    //cout<<sin(90.0/180*pi)<<endl;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf%lf",&h,&v);
        double l=0,r=90.0;
        printf("%.2f
",ternarySearch(l,r));
    }
    return 0;
}