HDU 2298（纯物理加解一元二次方程）

Toxophily

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2555    Accepted Submission(s): 1416

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

 解题报告：

二分三分什么的好麻烦啊。。。。。。。。。。。。。。。。

直接按照做物理题的思路：

代码如下：

#include<bits/stdc++.h>
using namespace std;
const double PI=3.1415926535898;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        double x,y,v;
        scanf("%lf %lf %lf",&x,&y,&v);
        if(x==0)//考虑特殊情况
        {
            if(y==0)
                printf("0.000000\n");
            else
                printf("%lf\n",PI/2);
                continue;
        }
        double t=x*x-4*(4.9)*(x*1.0/v*1.0)*(x*1.0/v*1.0)*(4.9*(x*1.0/v*1.0)*(x*1.0/v*1.0)+y);
        if(t>0)
        {
            double r;
            if(x>sqrt(t))
            {
                r=(x-sqrt(t))/(9.8*(x/v)*(x/v));
            }else
            {
                r=(x+sqrt(t))/(9.8*(x/v)*(x/v));
            }
            printf("%lf\n",atan(r));
        }else
        {
            printf("-1\n");
        }
    }
}

 

 

 

 

转载于:https://www.cnblogs.com/yinbiao/p/8919686.html