SDAU课程练习--problemG(1006)

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题目描述

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

题目大意

简单的水题，电梯上升一层楼需要6s，下降需要4s，到达目的楼层后停留5s。给出目的楼层的列表。求需要的时间。
使用模拟法,模拟电梯的运行过程求得总时间。

AC代码

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    //freopen("date.in","r",stdin);
    //freopen("date.out","w",stdout);
    ios::sync_with_stdio(false);
    int N,m,sum,tem;
    while(cin>>N&&N!=0)
    {
        sum=0;
        tem=0;
        for(int i=0;i<N;i++)
        {
            cin>>m;
            if(m>tem)
                sum+=(6*(m-tem));
            else sum+=(4*(tem-m));
            tem=m;
        }
        cout<<sum+N*5<<endl;
    }
}

SDAU课程练习--problemG(1006)

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原文地址：http://www.cnblogs.com/liuzhanshan/p/5352966.html