2016年杭电1008



Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 60547    Accepted Submission(s): 33223

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
   
   
   
   
   
   
   
   

    
    
    
    题意：有一架电梯，上一层楼需要6秒，下一层楼需要4秒，到达指定楼层后停留5秒，假设每次开始时电梯都停留在第0层。输入一个正整数N，表示电梯要停留N次，输入每次要停留的楼层，输出电梯按照指定的顺序走完全程需要花费的时间。
   
   
   
   
   
   
   
   

    
    
    
    
#include <iostream>

using namespace std;

int main()
{
    int D;
    while(cin>>D)
    {
        if(D==0){
            break;
        }
        else{
            int a[D+1];
            a[0]=0;
            int i=1,sum=0;
            while(D--){
                cin>>a[i];
                i++;
            }
            for(int j=0;j<i-1;j++)
            {
                if(a[j+1]>a[j]){
                    sum=sum+(a[j+1]-a[j])*6;
                }
                else if(a[j+1]<a[j]){
                    sum=sum+(a[j]-a[j+1])*4;
                }
            }
            sum=sum+(i-1)*5;
            cout<<sum<<endl;
        }
    }
    return 0;
}