hdoj 5112 A Curious Matt 【签到题】

最新推荐文章于 2018-10-27 15:03:22 发布

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本文介绍了一道关于求解最大瞬时速度的问题，通过记录人物在不同时间点的位置，利用排序和比较相邻记录的方法来找出最大速度。文章提供了一个完整的C++代码实现，并通过样例解释了算法的工作原理。

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A Curious Matt

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)

Total Submission(s): 931 Accepted Submission(s): 487

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 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

Sample Output

Case #1: 2.00
Case #2: 5.00

HintIn 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：给你一个人在一些时刻的位置，让你求出这个人所能达到的最大速度。

水题。。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define MAXN 10000+10
using namespace std;
struct rec
{
    int pos, time;
};
bool cmp(rec a, rec b)
{
    return a.time < b.time;
}
rec num[MAXN];
int main()
{
    int t, N;
    int k = 1;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &N);
        for(int i = 0; i < N; i++)
            scanf("%d%d", &num[i].time, &num[i].pos);
        sort(num, num+N, cmp);
        double ans = 0;
        for(int i = 1; i < N; i++)
        {
            double speed = (num[i].pos - num[i-1].pos) * 1.0 / (num[i].time - num[i-1].time);
            ans = max(ans, fabs(speed));
        }
        printf("Case #%d: %.2lf\n", k++, ans);
    }
    return 0;
}