HDU 5935 Car(思维，模拟，精度）

最新推荐文章于 2019-09-22 10:07:15 发布

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最新推荐文章于 2019-09-22 10:07:15 发布

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分类专栏：网络赛练习赛

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本题是一道关于赛车行驶速度的算法题目，要求根据一系列位置记录，推算出赛车以不递减的速度通过这些位置所需的最短时间。文章提供了题目的详细描述及样例输入输出，并附带了解题思路及代码实现。

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Car

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 523 Accepted Submission(s): 185

Problem Description

Ruins is driving a car to participating in a programming contest. As on a very tight schedule, he will drive the car without any slow down, so the speed of the car is non-decrease real number.

Of course, his speeding caught the attention of the traffic police. Police record

N positions of Ruins without time mark, the only thing they know is every position is recorded at an integer time point and Ruins started at

0.

Now they want to know the minimum time that Ruins used to pass the last position.

Input

First line contains an integer

T, which indicates the number of test cases.

Every test case begins with an integers

N, which is the number of the recorded positions.

The second line contains

N numbers

a1,

a2,

⋯,

aN, indicating the recorded positions.

Limits

1≤T≤100

1≤N≤105

0<ai≤109

ai<ai+1

Output

For every test case, you should output 'Case #x: y', where x indicates the case number and counts from 1 and y is the minimum time.

Sample Input

1
3
6 11 21

Sample Output

Case #1: 4

题意：，给出n个点坐标，一辆车以不递减的速度行驶，路过每一个点的时间都是整数，问最短的时间。

题解：

刚上手感觉没法做的样子，后来倒着想。因为每过一个点的时间都是整数，求最短时间，所以最后一段所用的时间一定是1s

这样我们可以求得最后一段的速度。则前一段的速度<=这段的速度，时间=前一段的距离/这段的速度，向上取整，注意精度问题，用分数表示速度。

代码：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=1e5+10;
ll a[N];
int T,n;
int main()
{
    scanf("%d",&T);
    for(int i=1;i<=T;i++)
    {
        printf("Case #%d: ",i);
        scanf("%d",&n);
        for(int j=1;j<=n;j++)
        {
            scanf("%I64d",&a[j]);
        }
        ll output=0;
        ll fenmu,fenzi;
        for(int j=n;j>=1;j--)
        {
            if(j==n)
            {
                output++;
                fenzi=a[j]-a[j-1];
                fenmu=1;
            }
            else
            {
                ll dis=a[j]-a[j-1];
                fenmu*=dis;
                swap(fenzi,fenmu);
                ll tmp=fenzi/fenmu+1;
                if(fenzi%fenmu==0)
                    tmp--;
                output+=tmp;
                fenzi=dis;
                fenmu=tmp;
            }
        }
       printf("%I64d\n",output);
    }
}