小球 -s 常數 積分 ask sts testcases fine arch

physics

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 817 Accepted Submission(s): 454

Problem Description 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 Author 學軍中學 Source 2016 Multi-University Training Contest 8 題目大意：

光滑的水平直線上有n個質量相等的小球，已知每個小球的初始位置，初始速度和方向，每個小球的每個時刻的加速度a都滿足a*v=c，v是該時刻的速度，c是已知的

常數，小球之間的碰撞是完全碰撞(不明白就百度)，然後q個詢問，每次詢問第t秒時速度第k小的小球速度是多少？

題解：

a = dv/dt = C/v

-----> vdv = Cdt

兩邊同時積分v是從v0-vt，t是從0到t

-----> [1/2*v^2] (v0---vt) = Ct (0----t)

-----> v = sqrt(2*C*t+v0^2);

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<cmath>
#include<vector>
using namespace std;

struct node
{
    double v,x,d;
}a[100005];

int T,n,q,t,k;
double c;

bool cmp(node a, node b)
{
    return a.v<b.v;
}
int main()
{
    scanf("%d",&T);
    for(;T>0;T--)
    {
        scanf("%d%lf",&n,&c);
        for(int i=1;i<=n;i++)
            scanf("%lf%lf%lf",&a[i].v,&a[i].x,&a[i].d);

        scanf("%d",&q);
        sort(a+1,a+1+n,cmp);
        for(;q>0;q--)
        {
            scanf("%d%d",&t,&k);
            double v=a[k].v;
            printf("%.3lf\n",sqrt(v*v+2*c*t));

        }
    }

    return 0;
}

hdu 5826 physics (物理數學，積分)