2016SDAU课程练习一1006 Problem G

最新推荐文章于 2019-04-21 19:55:38 发布

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本文介绍了一种计算电梯在特定请求列表中所需总时间的方法。考虑到电梯向上和向下移动的时间成本不同，以及每层停留时间，文章提供了一个简单的算法实现。

Problem G

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 28 Accepted Submission(s) : 16

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

目的算总时间，向上一层是6s，想下一层是4s，到了停留5s。

这道题也是比较水，一次提交就ac了，分两种情况讨论，上楼，下楼，分别是对应的楼层间隔乘4或6，再加上停留的5s。

唯一值得注意的地方是电梯是从0层开始的所以a【0】=0。

ac代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<set>
using namespace std;
int main(){
    int a[101],sum=0,n;
    while(scanf("%d",&n)&&n!=0){
        a[0]=0;
        sum=0;
        for(int i=1;i<=n;i++){
            cin>>a[i];
        }
        for(int i=1;i<=n;i++){
            if(a[i]>a[i-1]){
                sum+=(a[i]-a[i-1])*6+5;
            }
            else{
                sum+=(a[i-1]-a[i])*4+5;
            }
        }
        cout<<sum<<endl;
    }
    return 0;
}