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.
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.
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.
2
3
2 2
1 1
3 4
3
0 3
1 5
2 0
Case #1: 2.00
Case #2: 5.00
Hint
In the ?rst 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.
#include<bits/stdc++.h>
using namespace std;
const int M = 10000+50;
struct node{
double t;
double d;
}a[M];
bool cmp(const node &x,const node &y){
return x.t < y.t;
}
int main()
{
int t,i,n,j;
while(scanf("%d",&t)!=EOF){
for(j=1;j<=t;j++){
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%lf%lf",&a[i].t,&a[i].d);
}
sort(a+1,a+n+1,cmp);
double ans;
double maxx = 0;
for(i=1;i<n;i++){
ans = fabs(a[i+1].d - a[i].d)*1.0/fabs(a[i+1].t - a[i].t)*1.0;
maxx = max(maxx,ans);
}
cout<<"Case #"<<j<<": ";
printf("%.2lf\n",maxx);
}
}
}
