physics HDU - 5826（数学）

最新推荐文章于 2018-11-19 22:44:33 发布

coldfresh

最新推荐文章于 2018-11-19 22:44:33 发布

阅读量288

点赞数

CC 4.0 BY-SA版权

分类专栏：数学水题

本文链接：https://blog.youkuaiyun.com/Coldfresh/article/details/78014834

数学同时被 2 个专栏收录

6 篇文章

订阅专栏

水题

4 篇文章

订阅专栏

本文介绍了一种在平直轨道上多个质量相同的粒子发生弹性碰撞的物理模型，粒子的初始位置、速度及方向已知，粒子间的碰撞为完全弹性碰撞。文章详细探讨了如何根据这些条件来计算指定时间内所有粒子速度的第k小值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

There are n balls on a smooth horizontal straight track. The track can be considered to be a number line. The balls can be considered to be particles with the same mass.

At the beginning, ball i is at position Xi. It has an initial velocity of Vi and is moving in direction $Di.(Di∈−1,1)$
Given a constant $C$ . At any moment, ball its acceleration $Ai$ and velocity $Vi$ have the same direction, and magically satisfy the equation that $Ai * Vi = C.$
As there are multiple balls, they may collide with each other during the moving. We suppose all collisions are perfectly elastic collisions.

There are multiple queries. Each query consists of two integers $t$ and k. our task is to find out the k-small velocity of all the balls t seconds after the beginning.

Perfectly elastic collision : A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision.
Input
The first line contains an integer T, denoting the number of testcases.

For each testcase, the first line contains two integers $n <= 10^5$ and $C <= 10^9$ .
n lines follow. The i-th of them contains three integers $Vi, Xi, Di. Vi$ denotes the initial velocity of ball i. $Xi$ denotes the initial position of ball i. Di denotes the direction ball i moves in.

The next line contains an integer $q <= 10^5$ , denoting the number of queries.
q lines follow. Each line contains two integers $t <= 10^9$ and $1<=k<=n$ .
$1<=Vi<=10^5,1<=Xi<=10^9$
Output
For each query, print a single line containing the answer with accuracy of 3 decimal digits.
Sample Input
1
3 7
3 3 1
3 10 -1
2 7 1
3
2 3
1 2
3 3
Sample Output
6.083
4.796
7.141
也是比较水的题，题目大意是在一条直线上，有n个球，各自有一定的初速度，和方向（也就是说球与球之间会发生碰撞，为弹性碰撞），并且每个球的速度和加速度乘积为一个常数，那天我在的话就把这题给a了，唯一的难点可能就是积分吧，关于碰撞，只要把他当称两球相互穿过就可以了，如果只根据动量守恒定律那么速度是可以抵消的，但是在这里还得遵循能量守恒定律，没有能量损失，并且在质量相等的情况下，两者速度交换即可，因为球都是一样的，就当作相互穿过就好了
现在看积分：

v i * A = C

$v_i*A=C$

d v d t = A = C v

$\frac{dv}{dt}=A=\frac{C}{v}$
移向：

v d v = C d t

$vdv=Cdt$ 两边同时积分有：

\int v x v i v d v = C \int t x 0 d t

$\int_{v_{i}}^{vx}vdv=C\int_{0}^{tx}dt$
得：

v 2 x - v 2 i 2 = C t x

$\frac{v_x^2-v_i^2}{2}=Ct_x$
所以：

v x = v 2 i + 2 C t x - - - - - - - - \sqrt

$v_x=\sqrt {v_i^2+2Ct_x}$
至于询问，只要从小到大排序即可，因为这个速度函数是递增的，经过相同时间，他们的大小关系不会改变
代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<cmath>
#define N 100005
#define mod 530600414
using namespace std;
int a[100005];
int n,c;
int main()
{
    int t;
    int m;
    int x,y;
    int time,k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&c);
        for(int i=1;i<=n;i++)
            scanf("%d%d%d",a+i,&x,&y);
        sort(a+1,a+1+n);
        scanf("%d",&m);

        while(m--)
        {
            scanf("%d%d",&time,&k);
            printf("%.3lf\n",sqrt((double)a[k]*a[k]+2*c*(double)time));
        }
    }
    return 0;
}