HDU 5112 2014ICPC北京站现场赛 A Curious Matt

A Curious Matt

                                                                     Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
                                                                                               Total Submission(s): 1150    Accepted Submission(s): 613

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

232 21 13 430 31 52 0

 

Sample Output

Case #1: 2.00Case #2: 5.00
Hint
In the first sample, Ted moves from 2 to 4 in 1 time unit. The speed 2/1 is maximal.In the second sample, Ted moves from 5 to 0 in 1 time unit. The speed 5/1 is maximal.
 

 

题意：给你N组记录，每组记录输入时刻和位置，让你求这个人可能的最大速度。

分析：原以为是一道物理题，然而并没有那么难，分析一下样例就知道了，我们只需要对每组数据对时间进行排序，然后利用公式，速度=距离/时间，求得最大速度即可，是个平均速度。

#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <string>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <cctype>#include <numeric>#include <iomanip>#include <bitset>#include <sstream>#include <fstream>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-8)#define inf 0x3f3f3f3f#define ll long long int#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1using namespace std;const int mod = 1000000007;const int Max = 100005;struct node{    double x,t;}a[Max];bool cmp(node u,node v){    return u.t<v.t;}int main(){int T,n,cnt=1;cin>>T;while(T--)    {        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%lf%lf",&a[i].t,&a[i].x);        }        sort(a,a+n,cmp);        double ans=-1;        for(int i=1;i<n;i++)        {            double tmp=fabs(a[i].x-a[i-1].x)/(a[i].t-a[i-1].t);            ans=max(tmp,ans);        }        printf("Case #%d: ",cnt++);        printf("%.2lf\n",ans);    }    return 0;}

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Sample Output

Case #1: 2.00Case #2: 5.00
Hint
In the first sample, Ted moves from 2 to 4 in 1 time unit. The speed 2/1 is maximal.In the second sample, Ted moves from 5 to 0 in 1 time unit. The speed 5/1 is maximal.