HDU1008 Elevator

   
   
   
   

    
    
    
    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
     
     
     
     
     
     
     
     


     
     
     
     
     
     
     
     

      
      
      
      这是一道简单的计数题目，电梯上升一层6秒，每层停5秒，下降时每层4秒，每层停5秒，计算总秒数
     
     
     
     
   
   
   
   
代码如下：

#include<stdio.h>
int main() { int a[1000]; int n; int i; while(scanf("%d",&n)!=EOF) { if(n==0) break;
        a[0]=0; int ans=0; for(i=1;i<=n;i++) {
            scanf("%d",&a[i]); } for(i=1;i<=n;i++) { if(a[i]>a[i-1]) {
                ans=ans+(a[i]-a[i-1])*6; } else if(a[i]<a[i-1]) {
                ans=ans+(a[i-1]-a[i])*4; } }
        printf("%d\n",ans+n*5); } return 0; }