Elevator HDU
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题目
Elevator
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 71366 Accepted Submission(s): 39206
Problem DescriptionThe 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.
InputThere 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.
OutputPrint the total time on a single line for each test case.
Sample Input1 23 2 3 10
Sample Output1741
题意
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.
1 23 2 3 10
1741
电梯的上下操作都有一定的花费,求一个电梯的操作序列所带来的花费。
思路模拟。。。
代码
#include <iostream>using namespace std;int main(){ int n, a_p, a; long long res; while (cin >> n && n) {a_p = 0; res = 0;for (int i = 1; i <= n; i++) {cin >> a;if (a - a_p >= 0) {res += (a - a_p)*6; }else res += (a_p - a)*4;a_p = a;}res += n*5;cout << res << endl; } return 0;}
0 0
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