Car HDU - 5935(???)

最新推荐文章于 2020-07-31 17:26:03 发布

coldfresh

最新推荐文章于 2020-07-31 17:26:03 发布

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分类专栏：赛后感想

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本文介绍了一种算法问题，该问题涉及通过记录的位置来计算参与编程竞赛时车辆行驶的最短时间。具体而言，需要根据一系列按整数时间点记录的位置来确定参赛者通过最后一个位置所需的最小时间。

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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
$1≤T≤100$
$1≤N≤10^5$
$0<ai≤10^9$
$a_i<a_{i+1}$
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
这题我只能说没看懂，真的读不懂…………………………………………
简单题

#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 2005
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;
int a[100005];
int n;
int main()
{
    int t;
    int cal=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);

        for(int i=n;i>=1;i--)
            a[i]-=a[i-1];
        long long ans=1;
        long long temp=1;
        long long get;
        for(int i=n-1;i>=1;i--)
        {
            get=(a[i]*temp/a[i+1]);
            if((a[i]*temp)%a[i+1])
                get++;
            ans+=get;
            temp=get;
        }
        printf("Case #%d: %lld\n",cal++,ans);
    }

    return 0;
}