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                          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 ≤ 10 6), 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。 
 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 #include<iostream>
 5 #include<cmath>
 6 using namespace std;
 7 
 8 struct node{
 9     int t, pos;
10 }q[10010];
11 
12 bool cmp(struct node x, struct node y)
13 {
14     if(x.t<y.t)
15         return 1;
16     else return 0;
17 }
18 
19 int main()
20 {
21     int T, n, i,  k=1;
22     double s[10010], maxx;
23     scanf("%d", &T);
24     while(T--)
25     {
26         maxx=-1;
27         scanf("%d", &n);
28         for(i=0; i<n; i++)
29         {
30             scanf("%d%d", &q[i].t, &q[i].pos);
31         }
32         sort(q, q+n, cmp);
33         for(i=0; i<n-1; i++)
34         {
35             s[i] = (double) ( (abs)(q[i+1].pos-q[i].pos) )/(q[i+1].t-q[i].t);
36             {
37                 if(s[i]>maxx)
38                     maxx=s[i];
39             }
40         }
41         printf("Case #%d: %.2lf\n", k++, maxx);
42     }
43     return 0;
44 }

 

 

 

转载于:https://wwwblogs/6bing/p/4132320.html

本文标签： 北京站亚洲区ACMICPCcurious

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