HDU 1008 Elevator

http://acm.hdu.edu.cn/showproblem.php?pid=1008

Elevator Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 52650 Accepted Submission (s): 29078


Problem DescriptionThe Highest building in We City have only one elevator. A request list is made to with N positive numbers. The numbers denote at which floors the elevator would stop, in specified order. It costs 6 seconds to move the elevator-one floor, and 4 seconds to move-down-one floor. The elevator would stay for 5 seconds on each stop.

For a given request list, you is 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 having to return to the ground floor when the requests was Fu Lfilled.

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

Outputprint the total time, on a, on a, a, and each test case.

Sample Input

1 23 2) 3 10

Sample Output

1741

Test instructions: The elevator on a layer to spend 6s, the next floor to spend four seconds, did not listen to spend five seconds, ask the elevator in the overall process of the total time.

Problem-Solving ideas: 1 each floor to stop 5s, on a floor to 6s, the next floor to 4s.
2) Here is a note for the last point of arrival, but also to stop 5s, so the final should be added n * 5, instead of (n-1) * 5.
3) Start at the beginning of the 0 floor, do not have to go back to the 0 floor.

Complexity Analysis: Time complexity: O (n)
Space complexity: O (n)

#include <iostream>  using namespace std;  int main ()  {      int n,i,sum,tap;      int a[101];      while (CIN >> n && N)      {          sum=0;          Tap=0;          for (i=0;i<n;i++) CIN >> A[i];          for (i=0;i<n;i++)          {              if (tap<a[i]) sum = sum+6* (A[i]-tap) +5;              else if (tap>a[i]) sum = sum+4* (tap-a[i]) +5;              else sum +=5;              Tap=a[i];          }          cout << sum << endl;      }      return 0;  }

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HDU 1008 Elevator