思路:水题
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=10010;
const int INF=0x3f3f3f3f;
int cas=1,T;
int n;
struct node
{
int t,x;
bool operator<(const node&a)const
{
return t<a.t;
}
};
node a[N];
int main()
{
//freopen("1.in","w",stdout);
//freopen("1.in","r",stdin);
//freopen("1.out","w",stdout);
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=0;i<n;i++) scanf("%d%d",&a[i].t,&a[i].x);
sort(a,a+n);
double maxsp=0;
for(int i=1;i<n;i++) maxsp=max(maxsp,abs(a[i].x-a[i-1].x)/double(a[i].t-a[i-1].t));
printf("Case #%d: %.2lf\n",cas++,maxsp);
}
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
}
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.
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 Output
Case #1: 2.00
Case #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.