HDU 5112 A Curious Matt 水题

A - A Curious Matt

Time Limit:2000MS Memory Limit:512000KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 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 ≤ 10 6), 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

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.

题意

每次给你两个数，分别表示时间点和现在所在的位置，然后问你在哪个时间段里面，这个人跑的最快，并输出速度

题解

先sort一下，然后取最大就好

代码：

struct node
{
	int x;
	int y;
};
bool cmp(node a,node b)
{
	return a.x<b.x;
}
node kiss[10001];
int main()
{
	int t;
	cin>>t;
	REP_1(ti,t)
	{
		int n;
		cin>>n;
		REP(i,n)
			cin>>kiss[i].x>>kiss[i].y;
		sort(kiss,kiss+n,cmp);
		double ans=0;
		REP_1(i,n-1)
		{
			ans=max(abs((kiss[i].y-kiss[i-1].y)*1.0)/(kiss[i].x-kiss[i-1].x),ans);
		}
		printf("Case #%d: %.2f
",ti,ans);
	}
}