HDU 5144 NPY and shot(物理运动学+三分查找)

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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

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-;
double ans(double a)
{
double a1=v*v*sin(a)*sin(a);
double a2=*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=;
double r=pi/;
double midx,midy;
while (r-l>eps)
{
midx=(l+l+r)/;
midy=(l+r+r)/;
if(ans(midx)>ans(midy))
r=midy;
else l=midx;
}
printf("%.2f\n",ans(l));
}
}
return ;
}