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 Specification:

Each input file contains one test case. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100.

Output Specification:

For each test case, print the total time on a single line.

Sample Input:

3 2 3 1

Sample Output:

41

翻译:
某城市最高的建筑只有一个电梯,需求列表由N个正数组成,这些数字表示了电梯会按照顺序在在哪些楼层停止,电梯每升一层花费6秒钟,
下降一层花费4秒钟,在每层楼的停留花费5秒钟。对于所给的列表,你需要计算出完成列表中请求的总共的时间,电梯从第0层开始,而
且在完成所有请求后不需要返回到底层。
输入:
包含一个正整数N,后面跟随着N个正数,每一个正数都小于100。
输出:
在单行内输出总时间。
思路:
每次输入一个数就按照和preidnex之间的比较来计算总时间,preindex的值每次都更新为当前的curindex。

#include <iostream>
using namespace std;
int main()
{
	int num = 0;
	int preIdenx = 0;
	int curIndex = 0;
	int cost = 0;
	cin >> num;
	for (int i = 0; i < num;++i) 
	{ 
		cin >> curIndex;
		if (curIndex > preIdenx)
			cost += (curIndex - preIdenx) * 6;
		else
			cost += (preIdenx - curIndex) * 4;
			cost += 5;
			preIdenx = curIndex;
	}
	printf("%d", cost);
	return 0;
}