HDU 5144 NPY and shot(物理运动学+三分查找)

NPY and shot

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1035    Accepted Submission(s): 428

 

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

分析：做三分做上瘾了，再来一道，物理数学得学好啊，不然有些题就算算法知道你也写不来啊，都是思维题，典型的质点运动学+三分查找！

下面给出AC代码：

#include <bits/stdc++.h>
using namespace std;
const double pi=acos(-1.0);
const double g=9.8;
double v,h;
const double eps=1e-8;
double ans(double a)
{
    double a1=v*v*sin(a)*sin(a);
    double a2=2*g*h;
    a1=a1+a2;
    a1=sqrt(a1);
    a1=a1/g;
    a1=a1+v*sin(a)/g;
    a1=a1*v*cos(a);
    return a1;
}
int main()
{
    int T;
    while(scanf("%d",&T)!=EOF)
    {
        while(T--)
        {
          scanf("%lf%lf",&h,&v);
          double l=0;
          double r=pi/2;
          double midx,midy;
          while (r-l>eps)
          {
            midx=(l+l+r)/3;
            midy=(l+r+r)/3;
            if(ans(midx)>ans(midy))
                r=midy;
            else l=midx;
          }
          printf("%.2f\n",ans(l));
        }
    }
    return 0;
}