1008. Elevator (20)

最新推荐文章于 2021-08-18 12:42:21 发布

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本文介绍了一个电梯调度算法的具体实现过程，该算法通过计算不同楼层请求之间的移动时间和停靠时间来确定完成所有请求所需的总时间。输入包含一系列正数，每个数表示电梯需要停靠的楼层，输出则是完成这些请求所需的总时间。

Elevator (20)

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 Specification:

Each input file contains one test case. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100.

Output Specification:

For each test case, print the total time on a single line.
Sample Input:

3 2 3 1

Sample Output:

#include <bits/stdc++.h>
using namespace std;
int s[1231];
int main()
{
  int n;
  scanf("%d", &n);
  for(int i=0;i<n;i++)
  {
    scanf("%d", &s[i]);
  }
  int k = 0;//记录当前的层数
  long long int ans = 0;//记录时间
  for(int i=0;i<n;i++)
  {
    if(k<s[i])//向上升
    {
      ans = ans + (s[i] - k) * 6 + 5;
      k = s[i];
    }
    else if(k>s[i])//要向下降
    {
       ans = ans + (k - s[i]) * 4 + 5;
       k = s[i];
    }
    else
    ans += 5;//原地不走时，要停留5s
  }
  printf("%lld\n", ans);
  return 0;
}