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

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：

1 #include 
      
        2 using namespace std; 3 const double pi=acos(-1.0); 4 const double g=9.8; 5 double v,h; 6 const double eps=1e-8; 7 double ans(double a) 8 { 9     double a1=v*v*sin(a)*sin(a);10     double a2=2*g*h;11     a1=a1+a2;12     a1=sqrt(a1);13     a1=a1/g;14     a1=a1+v*sin(a)/g;15     a1=a1*v*cos(a);16     return a1;17 }18 int main()19 {20     int T;21     while(scanf("%d",&T)!=EOF)22     {23         while(T--)24         {25           scanf("%lf%lf",&h,&v);26           double l=0;27           double r=pi/2;28           double midx,midy;29           while (r-l>eps)30           {31             midx=(l+l+r)/3;32             midy=(l+r+r)/3;33             if(ans(midx)>ans(midy))34                 r=midy;35             else l=midx;36           }37           printf("%.2f\n",ans(l));38         }39     }40     return 0;41 }