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.
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 <bits/stdc++.h> 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 }
