杭电多校第六场 Pinball（物理）

最新推荐文章于 2021-11-16 16:32:08 发布

原创最新推荐文章于 2021-11-16 16:32:08 发布 · 409 阅读

0 ·

CC 4.0 BY-SA版权

ACM-数学专栏收录该内容

33 篇文章

订阅专栏

本文介绍了一个弹球在斜面上碰撞的模拟问题，通过解析物理过程，利用垂直和平行斜面的运动分解，实现了碰撞次数的有效计算。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Pinball

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 747 Accepted Submission(s): 331

Problem Description

There is a slope on the 2D plane. The lowest point of the slope is at the origin. There is a small ball falling down above the slope. Your task is to find how many times the ball has been bounced on the slope.

It's guarantee that the ball will not reach the slope or ground or Y-axis with a distance of less than 1 from the origin. And the ball is elastic collision without energy loss. Gravity acceleration g=9.8m/s2.

Input

There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 100), indicating the number of test cases.

The first line of each test case contains four integers a, b, x, y (1 ≤ a, b, -x, y ≤ 100), indicate that the slope will pass through the point(-a, b), the initial position of the ball is (x, y).

Output

Output the answer.

It's guarantee that the answer will not exceed 50.

Sample Input

1 5 1 -5 3

Sample Output

Source

2018 Multi-University Training Contest 6

思路：垂直于斜面上做周期运动，平行于斜面做匀加速运动，用总时间除每一次的周期时间+1

#include <bits/stdc++.h>
using namespace std;

int main(void)
{
    int T;
    double a,b,x,y,g1,g2,v1,v2,v3,t1,t2;
    int ans;
    scanf("%d",&T);
    while(T--) {
        scanf("%lf %lf %lf %lf",&a,&b,&x,&y);
        double ta = atan2(b , a);
        double dn = 9.8 * (y - (-x) * (b / a));//第一次接触斜面的动能
        g1 = 9.8 * cos(ta);//垂直斜面的加速度
        g2 = 9.8 * sin(ta);//平行斜面的加速度
        v1 = sqrt(2 * dn) * cos(ta);//垂直斜面的的初速度
        v2 = sqrt(2 * dn) * sin(ta);//平行斜面的的初速度
        t1 = 2 * (v1 / g1);//一次周期的时间
        v3 = sqrt(2 * g2 * (-x) / cos(ta) + v2 * v2);//平行斜面的末速度
        t2 = (v3 - v2) / g2;//总时间
        ans = (int)(t2 / t1) + 1;
        printf("%d\n",ans);
    }
    return 0;
}
//1.277753 1.751581
//1.277753 4.551381