首页 > 代码库 > 2014ACM/ICPC亚洲区北京站-A-（Curious Matt）
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.
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 ti and xi (0 ≤ ti, xi ≤ 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 ti 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
232 21 13 430 31 52 0
Sample Output
Case #1: 2.00Case #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 }
2014ACM/ICPC亚洲区北京站-A-（Curious Matt）
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。