hdu1008 Elevator

Elevator

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42914    Accepted Submission(s): 23537

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 23 2 3 10

 

Sample Output

1741
 
题目大意：电梯从0层开始，上一层需要6秒，下一层需要4秒，在所给的层数每层停留5秒，总时间难点：电梯时间运算的过程关键点；理解题意；解题时间：2014,07,29解题思路：因为题目先给出一个n后边跟n组数据，所以电梯停留的时间是n*5，在判断上下楼求总时间********************
#include<stdio.h>int main(){int n,i,t;int a[110];while(scanf("%d",&n),n){for(i=0;i<n;i++)scanf("%d",&a[i]);t=a[0]*6;for(i=1;i<n;i++){if(a[i]>a[i-1])t=t+(a[i]-a[i-1])*6;else if(a[i]==a[i-1]) t=t+0;else t=t+(a[i-1]-a[i])*4;}t=t+n*5;printf("%d\n",t);}return 0;}