Elevator (1008. Elevator (20) | ProgrammingAbility Test https://www.patest.cn/contests/pat-a-practise/1008)

The highest building in our city has onlyone elevator（電梯）. A request list is made up with N positive numbers. The numbers denote atwhich floors the elevator will stop, in specified order. It costs 6 seconds tomove the elevator up one floor, and 4 seconds to move down one floor. Theelevator will stay for 5 seconds at each stop.

For a given request list, you are tocompute the total time spent to fulfill the requests on the list. The elevatoris on the 0th floor at the beginning and does not have to return to the groundfloor 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. Allthe numbers in the input are less than 100.

Output Specification:

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

Sample Input:

3 2 3 1

Sample Output:

41

題目翻譯：1008.電梯

在我們的城市裡，最高的建築物裡只有一部電梯。有一份由N個正陣列成的請求列表。這些數表示電梯將會以規定的順序在哪些樓層停下。電梯升高一層需要6秒，下降一層需要4秒。每次停下電梯將花費5秒。

給你一個請求列表，你需要計算出完成列表裡的請求總共花費的時間。一開始電梯在0層。當請求全部完成時，電梯不需要回到底層。

輸入說明：

每個輸入檔案包含一個測試例項。每個例項包含一個正整數N，後面跟著N個數字。所有輸入的數字小於100。

輸出說明：

對於每個測試例項，在一行中輸出總時間。

樣例輸入：

3 2 3 1

樣例輸出：

#include<stdio.h>
#define N 100
int main()
{
	int n,sum=0,i,t,c[N]={0};
	scanf("%d",&n);
	for(i=1;i<n+1;i++){
		scanf("%d",&c[i]);
	}
	for(i=0;i<n;i++){		
		if(c[i+1]>c[i]){
			sum+=(c[i+1]-c[i])*6;
		}
		else sum+=(c[i]-c[i+1])*4; 
		sum+=5;
	}
	printf("%d\n",sum);
	return 0;
}