1008. Elevator

题目： pat甲级

思路： 模拟题 ，按照数据输入的顺序依次按照题目的条件计算就可以了。注意：同楼层连续，也需要计算该层等待时间，如：2 3 3.结果应该是28

目的：循环

题目描述：

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:

41

参考代码：

#include <iostream>
using namespace std;

int main()
{
  int n;
  while(cin>>n)
  {
    int i;
    int floor,preFloor=0,sum=0;
    for(i=0;i<n;i++)
    {
      cin>>floor;
      if(floor>preFloor)
          sum+=(floor-preFloor)*6+5;
      else if (floor<preFloor)
        sum+=(preFloor-floor)*4+5;
                  else
                       sum+=5;
      preFloor=floor;
    }
    cout<<sum<<endl;
  }
  return 0;
}

tip：

和hdu 1008 描述一样，只是输入数据不同。