三分角度....

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 $9.8m/s^2$)

Input

The first line contains a integer $T(T \leq 10000)$,which indicates the number of test cases.

The next T lines,each contains 2 integers $H(0 \leq h \leq 10000m)$,which means the height of NPY,and $v_0(0 \leq v_0 \leq 10000m/s)$, 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

/* ***********************************************
Author        :CKboss
Created Time  :2014年12月13日 星期六 23时27分32秒
File Name     :C.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>

using namespace std;

const double eps=1e-6;
const double g=9.8;

double H,V;

double Distan(double degree)
{
	double vx=V*sin(degree);
	double vy=V*cos(degree);

	double time = ( vx+sqrt(vx*vx+2*g*H) ) / g;

	return time*vy;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%lf%lf",&H,&V);

		double low=0.,high=90.;
		double mid1,mid2,ans=0;

		while(fabs(high-low)>=eps)
		{
			mid1=(low*2+high)/3.;
			mid2=(low+high*2)/3.;

			double len1=Distan(mid1);
			double len2=Distan(mid2);

			ans=max(ans,max(len1,len2));

			if(len2+eps>len1)
				low=mid1;
			else high=mid2;
		}

		printf("%.2lf\n",ans);
	}
    return 0;
}