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http://acm.hdu.edu.cn/showproblem.php?pid=1008
Elevator
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52650 Accepted Submission(s): 29078
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
Sample Output
题意：电梯上一层要花6s，下一层楼要花四秒，没听一次要花五秒，请问电梯在整个过程所用的总时间。
解题思路：1）每层楼停5s，上一层楼要6s，下一层楼要4s。
2）这里有个注意点是对于最后一个到达点，也是要停5s的，所以最后应该是加上n * 5，而不是(n - 1) * 5。
3）刚开始是从0层开始的，最后不必回到0层。
复杂度分析： 时间复杂度：o(n)
空间复杂度：o(n)
#include<iostream>
using namespace std;
int main()
{
int n,i,sum,tap;
int a[101];
while(cin >> n && n)
{
sum=0;
tap=0;
for(i=0;i<n;i++) cin >> a[i];
for(i=0;i<n;i++)
{
if(tap<a[i]) sum = sum+6*(a[i]-tap)+5;
else if(tap>a[i]) sum = sum+4*(tap-a[i])+5;
else sum +=5;
tap=a[i];
}
cout << sum << endl;
}
return 0;
}
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HDU 1008 Elevator
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原文地址：http://blog.csdn.net/hellohelloc/article/details/47699667
