杭电1008 Elevator

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Elevator

Time Limit: 2000/1000 MS(Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31446    Accepted Submission(s): 17014

Problem Description

The highestbuilding in our city has only one elevator. A request list is made up with Npositive numbers. The numbers denote at which floors the elevator will stop, inspecified order. It costs 6 seconds to move the elevator up one floor, and 4seconds to move down one floor. The elevator will stay for 5 seconds at eachstop.

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

 

 

Input

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

 

 

Output

Print the totaltime on a single line for each test case. 

 

 

Sample Input

1 2

3 2 3 1

0

 

 

Sample Output

17

41

#include <iostream>using namespace std;int main(){    int n,f[101]={0},i,time=0,total=0;   //n request的个数  f每个请求的层数  i计数 time每次移动的时间    //total 总时间    while(cin>>n&&n!=0)                     //输入request 为0时退出    {        for(i=0;i<n;i++)            cin>>f[i+1];                              //输入每个请求        for(i=0;i<n;i++)        {            if(f[i+1]>f[i])                time=(f[i+1]-f[i])*6;                       //每次移动的时间            else                time=(f[i]-f[i+1])*4;            total+=time;        }        total+=5*n;        cout<<total<<endl;                            //输出总时间        total=0;time=0;                 //计数器归0    }    return 0;}