杭电-PID1008-Elevator

最新推荐文章于 2019-07-14 07:23:30 发布

原创最新推荐文章于 2019-07-14 07:23:30 发布 · 297 阅读

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#杭电 #acm #1008 #elevator

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本文解析了一道经典的算法题——杭电PID1008电梯问题。题目要求计算完成一系列楼层请求所需的总时间，考虑到电梯上行、下行及停留的时间成本。通过给出的代码示例，展示了如何高效解决此类问题。

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杭电-PID1008-Elevator

好吧，其实这道题是水过去的。。

Problem Description

The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.

For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.

Input

There are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed.

Output

Print the total time on a single line for each test case.

Sample Input

1 2
3 2 3 1
0

Sample Output

17
41

有一点需要注意的是，同层的电梯需要等待5秒，这一点记住就很好AC了。

Code：

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n;
    while (scanf("%d", &n) != EOF && n != 0)
    {
        int sum = 0;
        int now = 0;
        int next;
        int i;
        for (i = 0; i < n; i++)
        {
            scanf("%d", &next);
            if (now < next)
            {
                sum = sum + 6 * (next - now);
            }
            else if (now > next)
            {
                sum = sum + 4 * (now - next);
            }
            sum = sum + 5;
            now = next;
        }
        printf("%d\n", sum);
    }
    return 0;
}