HDOJ 1008. Elevator 简单模拟水题

Elevator

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 56481    Accepted Submission(s): 30942

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

 

Author

ZHENG, Jianqiang

 

Source

ZJCPC2004

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

 

        看到有人说没看懂题目= =呃..这真的是高中英语的问题了....

        题目大意是我要坐电梯，给你一个列表，上面有N个正整数，代表电梯要按顺序去的楼层，上楼6分钟，下楼4分钟，每个指定的楼层还要停5分钟，问总共需要多少分钟呢？注意初始在第0层。

        很简单，在线模拟就好了。先算出总共停留的时间5*N，然后判断上楼*6，下楼*4。

        题目没说相邻楼层相同的情况，不管它。

 #include <stdio.h>
 int main() {
     int N, arr[]={};
     while(scanf("%d", &N), N) {
         int res=*N, t;
         for(int i=; i<=N; i++) {
             scanf("%d", arr+i);
             t=arr[i-]-arr[i];
             res+=arr[i-]<arr[i]?-t*:t*;
         }
         printf("%d\n", res);
     }
     return ;
 }

By Black Storm（使用为知笔记）