HDU 5112 A Curious Matt 水题

A Curious Matt

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5112

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

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

题意

给你n个时间点和当前的位置，问你哪段的平均速度最大，是多少

题解：

暴力就好了

对着时间排序

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 2000000 + 500
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
//**************************************************************************************struct node
{double t,x;
};
node a[100010];
bool cmp(node a,node b)
{return a.t<b.t;
}
int main()
{int t;scanf("%d",&t);int cas = 1;while(t--){int n;scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d%d",&a[i].t,&a[i].x);sort(a+1,a+1+n,cmp);double ans = 0;for(int i=1;i<n;i++)ans = max(ans,abs(a[i+1].x-a[i].x)/(a[i+1].t-a[i].t));printf("Case #%d: %.2f\n",cas++,ans);}
}

转载于:https://www.cnblogs.com/qscqesze/p/4817568.html