【模拟】2014ACM/ICPC亚洲区北京站-A-（Curious Matt）

                          A Curious Matt

                                  Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)

 Problem Description

There is a curious man called Matt.
One day, Matt's best friend Ted is wandering on the non-negative half of the number line.

Matt finds it interesting to know the maximal speed Ted may reach. In order to do so, Matt takes

records of Ted’s position. Now Matt has a great deal of records. Please help him to find out the

maximal speed Ted may reach, assuming Ted moves with a constant speed between two consecutive records.

 

Input

The first line contains only one integer T, which indicates the number of test cases.

For each test case, the first line contains an integer N (2 ≤ N ≤ 10000),indicating the number of

records.

Each of the following N lines contains two integers t_i and x_i (0 ≤ t_i, x_i ≤ 10⁶), indicating the time when this record is taken and Ted’s corresponding position. Note that records may be unsorted by time. It’s guaranteed that all t_i would be distinct.

 

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1), and y is the maximal speed Ted may reach. The result should be rounded to two decimal places.

 

Sample Input

2 3 2 2 1 1 3 4 3 0 3 1 5 2 0

 

Sample Output

Case #1: 2.00 Case #2: 5.00

 

求每个时间段的平均速度，然后输出最大的一个平均速度。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;

struct node{
    int t, pos;
}q[10010];

bool cmp(struct node x, struct node y)
{
    if(x.t<y.t)
        return 1;
    else return 0;
}

int main()
{
    int T, n, i,  k=1;
    double s[10010], maxx;
    scanf("%d", &T);
    while(T--)
    {
        maxx=-1;
        scanf("%d", &n);
        for(i=0; i<n; i++)
        {
            scanf("%d%d", &q[i].t, &q[i].pos);
        }
        sort(q, q+n, cmp);
        for(i=0; i<n-1; i++)
        {
            s[i] = (double) ( (abs)(q[i+1].pos-q[i].pos) )/(q[i+1].t-q[i].t);
            {
                if(s[i]>maxx)
                    maxx=s[i];
            }
        }
        printf("Case #%d: %.2lf\n", k++, maxx);
    }
    return 0;
}