hdu5826physics+数学积分

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
学军中学

完全碰撞，加速度不会改变，直接求t秒后的速度即可。。
a=dv/dt=vi/c
cdt=vidv
积分ct=v^2/2

#include<cstdio>#include<cmath>#include<algorithm>using namespace std;int v[100005],x[100005],d[10005];int main(){    int t;    scanf("%d",&t);    while(t--){        int n,c;        scanf("%d %d",&n,&c);        for(int i=0;i<n;i++) scanf("%d %d %d",&v[i],&x[i],&d[i]);        sort(v,v+n);        int q;        scanf("%d",&q);        while(q--){            int t,k;            scanf("%d %d",&t,&k);            double ans=1.0*c*t+0.5*v[k-1]*v[k-1];            ans*=2;            //printf("%f %f\n",1.0*c*t,ans);            printf("%.3f\n",sqrt(ans));        }    }    return 0;}