标签：nbsp 实现 which for reac ssi ted contain ota
A Curious Matt
Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 3058 Accepted Submission(s):
1716
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
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
2
3
2 2
1 1
3 4
3
0 3
1 5
2 0
Sample Output
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.
水题，暴力
实现代码：
#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);
}
}
}
HDU 5112 A Curious Matt (2014ACM/ICPC亚洲区北京站-重现赛)
标签：nbsp 实现 which for reac ssi ted contain ota
原文地址：http://www.cnblogs.com/kls123/p/7895011.html
