hdu5112水题

A Curious Matt

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 2382    Accepted Submission(s): 1346

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
水题 据说是签到题/。。。。就一组一组输入时间与距离，然后排序下时间，求哪两个时间之间的距离最大 也就是速度最大。
#include <iostream>#include <cstdio>#include <algorithm>using namespace std;struct su{    int t,x;}a[10010];bool cmp(su a,su b){    return a.t<b.t;}int main(){    int t,k,n,i;    double maxsum;    scanf("%d",&t);    for (k=1;k<=t;k++)    {        scanf("%d",&n);        for (i=1;i<=n;i++)            scanf("%d%d",&a[i].t,&a[i].x);        sort(a+1,a+n+1,cmp);        maxsum=0;        for (i=2;i<=n;i++)            if (abs(a[i].x-a[i-1].x)/(double)(a[i].t-a[i-1].t)>maxsum)                 maxsum=abs(a[i].x-a[i-1].x)/(double)(a[i].t-a[i-1].t);        printf("Case #%d: %.2lf\n",k,maxsum);    }}