ACdream区域赛指导赛之手速赛系列(4) C Captain Hammer

题目：

Captain Hammer

Special JudgeTime Limit: 2000/1000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others)

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Problem Description

The Hamjet is a true marvel of aircraft engineering. It is a jet airplane with a single engine so powerful that it burns all of its fuel instantly during takeoff. The Hamjet doesn't have any wings because who needs them when the fuselage is made of a special Wonderflonium isotope that makes it impervious to harm.

Piloting the Hamjet is a not a job for your typical, meek-bodied superhero. That's why the Hamjet belongs to Captain Hammer, who is himself impervious to harm. The G-forces that the pilot endures when taking a trip in the Hamjet are legen-dary.

The Hamjet takes off at an angle of θ degrees up and a speed of V meters per second. V is a fixed value that is determined by the awesome power of the Hamjet engine and the capacity of its fuel tank. The destination is D meters away. Your job is to program the Hamjet's computer to calculate θ given V and D.

Fortunately, the Hamjet's Wondeflonium hull is impervious to air friction. Even more fortunately, the Hamjet doesn't fly too far or too high, so you can assume that the Earth is flat, and that the acceleration due to gravity is a constant 9.8 m/s^2 down.

Input

The first line of the input gives the number of test cases, T.

T lines follow. Each line will contain two positive integers -- V and D.

1 ≤ T ≤ 4500
1 ≤ V ≤ 300
1 ≤ D ≤ 10000

It is guaranteed that each test case will be solvable.

Output

For each test case, output one line containing "Case #x: θ", where x is the case number (starting from 1) and θ is in degrees up from the the horizontal.

If there are several possible answers, output the smallest positive one.

An answer will be considered correct if it is within 10^-6 of the exact answer, in absolute or relative error.

Sample Input

398 98098 490299 1234

Sample Output

Case #1: 45.0000000Case #2: 15.0000000Case #3: 3.8870928

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解题思路：

这是一个抛物线运动，根据中学的物理知识，我们可以推出公式，sin2θ = G*D/(V*V)，我们需要求出θ（0<=θ<=π/2）,需要用二分求，直接用θ的表达式会出现精度问题。

代码：

#include <cstdio>#include <cmath>#include <iostream>using namespace std; typedef double DB;const DB PI = 4 * atan(1.0);const DB G = 9.8;const DB EPS = 1e-10; int main(){    int t, w = 1;    DB v, d, ans;   // printf("%.7f\n",90/PI*asin(0.5));    scanf("%d", &t);    while(t--)    {        scanf("%lf%lf", &v, &d);        DB l = 0, r = PI/2.0, m = G*d/v/v;        while(r-l >= EPS)        {            DB mid = (l + r) / 2.0;            if(2*mid <= PI/2.0)            {                if(sin(2*mid) >= m)                    r = mid;                else                    l = mid;            }            else            {                if(sin(2*mid) >= m)                    l = mid;                else                    r = mid;            }            ans = mid;        }        printf("Case #%d: %.7f\n", w++, ans*180/PI);    }    return 0;}