Hdoj 1008.Elevator 题解

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

思路

就是很简单的模拟就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 100000+10;
const int UP = 6;
const int DOWN = 4;
const int STAY = 5;
int a[MAXN];

int cal(int a[], int p)
{
	int delta = a[p] - a[p-1];
	int tmp = 0;
	if(delta > 0)
		tmp += delta * UP;
	else
		tmp += (-delta) * DOWN;
	return tmp;
}	//计算电梯相应的移动所需的时间
int main()
{
	int n;
	int ans;
	while(cin>>n&&n!=0)
	{
		memset(a,0,sizeof(a));
		ans = 0;
		for(int i=1;i<=n;i++)
		{
			cin >> a[i];
			ans += cal(a,i) + STAY;
		}
		cout << ans << endl;
	}
}