CDZSC_2015寒假新人（1）——基础 B

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

 

思路：简单的计算而已，如果先设它从b=0开始，如果要去的楼层a大于所在的楼层b,那么所用

sum+=(a-b)*6+5（停留需要5），小于sum+=(b-a)*4+5，等于要原地停留5，所以要sum+=5(这是陷阱)

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
struct dangchi
{
    char str[20];
    int sum;
};
int main()
{
#ifdef CDZSC_OFFLINE
    freopen("in.txt","r",stdin);
#endif
    int n,a,b,sum;
    while(scanf("%d",&n)&&n!=0)
    {
        b=0,sum=0;
        while(n--)
        {
            scanf("%d",&a);
            if(a>b)
            {
                sum+=(a-b)*6+5;
                b=a;
            }
            else if(a<b)
            {
                sum+=(b-a)*4+5;
                b=a;
            }
            else
            {
                sum+=5;
            }
        }
        printf("%d
",sum);
    }
    return 0;
}