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题目描述：
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.
输入描述:
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.
输出描述:
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.
输入：
2
3
2 2
1 1
3 4
3
0 3
1 5
2 0
输出：
Case #1: 2.00
Case #2: 5.00
题意：
给出某个时刻对应的速度 求出相邻时刻的平均速度 输出最大值
题解：
排序下时间，求哪两个时间之间的距离最大 也就是速度最大。
代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;

const int maxn = 10000 + 5;

struct point{
    int t,x;
};

point a[maxn];

int cmp(point x,point y){
    return x.t < y.t;
}

int main(){
    int t,n;
    scanf("%d",&t);
    int cas = 1;
    while(t--){
        double maxx = 0;
        scanf("%d",&n);
        for(int i = 1; i <= n; i ++){
            scanf("%d%d",&a[i].t,&a[i].x);
        }
        sort(a + 1,a + n + 1,cmp);
        for(int i = 2; i <= n; i ++){
            maxx = max(maxx,abs(a[i].x - a[i - 1].x) / (double)(a[i].t - a[i - 1].t));
        }
        printf("Case #%d: %.2lf\n",cas ++ ,maxx);
    }
    return 0;
}

本文标签： HDOJMattcurious

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