ZOJ Problem Set–2108 Elevator
2012-05-17 09:26
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Time Limit: 2 Seconds Memory Limit: 65536 KB
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: Zhejiang Provincial Programming Contest 2004
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: Zhejiang Provincial Programming Contest 2004
1: #include<iostream>
2: using namespace std;
3: const int UP = 6;
4: const int DOWN = 4;
5: const int STOP = 5;
6: int main(void)
7: {
8: int floors;
9: while(cin>>floors && floors)
10: {
11: int preFloor=0, requestFloor, processTime = 0;
12: while(floors-- && cin>>requestFloor)
13: {
14: processTime += (requestFloor > preFloor ? (requestFloor-preFloor)*UP:(preFloor-requestFloor)*DOWN);
15: processTime += STOP;
16: preFloor = requestFloor;
17: }
18: cout<<processTime<<endl;
19: }
20: return 0;
21: }
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