HDU/HDOJ 2298 Toxophily 二分法+三分法 第4届百度杯总决赛

 

Toxophily

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 12   Accepted Submission(s) : 4

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Problem Description

The recreation center of WHU ACM Team has indoor billiards, Ping Pang, chess and bridge, toxophily, deluxe ballrooms KTV rooms, fishing, climbing, and so on.
We all like toxophily.

Bob is hooked on toxophily recently. Assume that Bob is at point (0,0) and he wants to shoot the fruits on a nearby tree. He can adjust the angle to fix the trajectory. Unfortunately, he always fails at that. Can you help him?

Now given the object's coordinates, please calculate the angle between the arrow and x-axis at Bob's point. Assume that g=9.8N/m.  

Input

The input consists of several test cases. The first line of input consists of an integer T, indicating the number of test cases. Each test case is on a separated line, and it consists three floating point numbers: x, y, v. x and y indicate the coordinate of the fruit. v is the arrow's exit speed.
Technical Specification

1. T ≤ 100.
2. 0 ≤ x, y, v ≤ 10000.  

Output

For each test case, output the smallest answer rounded to six fractional digits on a separated line.
Output "-1", if there's no possible answer.

Please use radian as unit.  

Sample Input

3
0.222018 23.901887 121.909183
39.096669 110.210922 20.270030
138.355025 2028.716904 25.079551

Sample Output

1.561582
-1
-1

Source

The 4th Baidu Cup final

 

这个题主要是一个公式的推导还有

二分法和三分法的综合运用。。个人感觉这个题比较经典

做法是先三分出一个角度，让抛物线在x=x0出的y取得最大值。

然后再到这个y一下去二分一个合适的值。

 

代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#define PI acos(-1.0)
#define G 9.8
#define eps 1e-12

using namespace std;

double X,Y,v,sita;

double cal(double x)
{
    double t;
    t=X/(v*cos(x));
    return v*sin(x)*t-0.5*G*t*t;
}

double triplediv()
{
    int i;
    double mid1,mid2,left,right,h1,h2;
    left=0,right=0.5*PI;
    for(i=1;i<=100;i++)
    {
        mid1=(2*left+right)/3;
        mid2=(left+2*right)/3;
        h1=cal(mid1);
        h2=cal(mid2);
        if(h1>h2)
        {
            right=mid2;
        }
        else
        {
            left=mid1;
        }
    }
    sita=left;
    return cal(left);
}

void doublediv(double maxy)
{
    int i;
    double left,right,mid,h;
    left=0,right=sita;
    for(i=1;i<=100;i++)
    {
        mid=(left+right)/2;
        h=cal(mid);
        if(h>Y)
            right=mid;
        else
            left=mid;
    }
    printf("%.6lf\n",left);
}

int main()
{
    int t;
    double maxy;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf",&X,&Y,&v);
        maxy=triplediv();
        if(maxy<Y)
            printf("-1\n");
        else
            doublediv(maxy);
    }
    return 0;
}