PAT-PAT (Advanced Level) Practise 1008. Elevator (20) （简单模拟）【一星级】

题目链接：http://www.patest.cn/contests/pat-a-practise/1008

题面：

1008. Elevator (20)

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 

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题目大意：上一层花费6s，下一层花费4s，另每层花费5s，问给定的序列，共花费多久。

解题：直接暴力模拟即可。

代码：

#include 
int main()
{
	int n,cost=0,a[105];
	scanf("%d",&n);
	a[0]=0;
    for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)
	{
		if(a[i]-a[i-1]>0)
			cost+=6*(a[i]-a[i-1]);
		else
			cost+=4*(a[i-1]-a[i]);
	}
	cost+=n*5;
	printf("%d\n",cost);
	return 0;
}