hdu 5826 physics(2016 Multi-University Training Contest 8——积分求解)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5826

physics

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 132    Accepted Submission(s): 92

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

13 73 3 13 10 -12 7 132 31 23 3

 

Sample Output

6.0834.7967.141

 

Author

学军中学

 

Source

2016 Multi-University Training Contest 8

 

题目大意：给n个球的初始位置，速度，方向,问t秒之后的第k小的速度是多少;

解题思路：

积分求解：

具体过程如图所示。

详见代码。

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>using namespace std;#define N 100010double v[N];int x[N],d[N];int main(){    int T;    scanf("%d",&T);    while (T--)    {        int n,c,q,t,k;        scanf("%d%d",&n,&c);        for (int i=1; i<=n; i++)        {            scanf("%lf%d%d",&v[i],&x[i],&d[i]);        }        sort(v+1,v+n+1);        scanf("%d",&q);        for (int i=0; i<q; i++)        {            scanf("%d%d",&t,&k);            double ans=sqrt((1.0*2*c    *t+v[k]*v[k]*1.0)*1.0);            printf ("%.3lf\n",ans);        }    }    return 0;}