6.22周赛 Elevator 电梯模拟

input.txt / output.txt
Statements

The Berland State Building is the highest building in the capital of Berland. Curious Polikarp was studying the principle of operation of an elevator in the Berland State Building for a quite a while. Recently he has finally understood the algorithm behind its operation, in case a person enters the elevator on the floor f and presses the floor buttons e₁, e₂, ..., e_n one by one. The buttons are pressed sequentially but very quickly while the elevator is still located on the floor f. All the pressed buttons are distinct and differ from the floor f. No other button pressings are considered in this problem.

After the buttons e₁, e₂, ..., e_n have been pressed, all of them become highlighted and the elevator starts moving according the following rules:

• The elevator starts moving towards the floor, the button of which is highlighted and pressed first among all highlighted buttons. Say, it's floor/button a.
• If on its way to a the elevator passes the floor b, the button of which is highlighted, it stops there, the light goes out for the button b unhighlighting it, and the floor b is considered visited. Then the elevator continues moving towards the floor a. It is possible that there will be more than one floor such as b on the way to floor a — all these floors will be passed one by one according to the described algorithm.
• Having reached the floor a, the elevator stops there, the light goes out for the button a unhighlighting it, and the floor a is considered visited. Then the elevator starts to move towards the floor, the button of which has been pressed the earliest among the currently highlighted buttons. That floor becomes a new value of a. The elevator continues moving according to the rules described in the previous paragraph. If it's impossible to find a new value for a because there are no highlighted floor buttons, it means that all floors have been visited and the elevator stops.

Now, when the principle of the elevator's operation is clear, Polikarp wants to experiment with the elevator's movements without the elevator itself. He wants to write a program that simulates elevator's operation. Unfortunately, he didn't attend any programming lessons and it's a challenge for him. Can you please help Polikarp and write a program which will simulate movements of the elevator?

Input

The first line of input contains a pair of integers n, f (1 ≤ n, f ≤ 100), where n — amount of pressings made, f — index of the current floor where all these pressings were made. The second line contains distinct integers e₁, e₂, ..., e_n (1 ≤ e_i ≤ 100, e_i ≠ f) — buttons indices in the order they were pressed.

Output

Output all the floors where the elevator stops, in a chronological order of the stops.

Example

Input

4 5
10 9 2 1

Output

9 10 2 1 

Input

4 3
2 4 1 5

Output

2 4 1 5 

题目意思和现实中乘电梯一样，第一行输入当前要到的楼层数量和当前楼层。电梯移动按照先后顺序，但是到达目的的楼层如果遇到可到达的楼层就顺便放人。

思路：

判断好当前楼层与当前要到达的楼层是上升还是下降，如果上升，把可以到达的按照升序输出，如果下降，则把可以到达的按照降序输出。桶排序思想，另开数组标记记录f需到达电梯，lag做标记上升还是下降，处理好最后一个要到达的楼层（换行）。

注意输入输出：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100+10;
int el[maxn],vis[maxn];
int main()
{
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    int n,f,cnt,flag;

    while(cin>>n>>f)
    {
        memset(el,0,sizeof(el));
        memset(vis,0,sizeof(cnt));
        cnt = 1;
        for(int i = 0; i<n; i++)
        {
            cin>>el[i];
            vis[el[i]]=1;
        }

        for(int i=0; i<n; i++)
        {
            if(vis[el[i]])
            {
                if(el[i]>f)
                    flag = 1;
                else
                    flag = 0;
                if(flag==1)
                {
                    for(int j=f; j!=el[i]+1; j++)
                    {
                        if(vis[j]==1)
                        {
                            if(cnt==n)
                            {
                                cout<<j<<endl;
                                vis[j]=0;
                                break;
                            }
                            else
                            {
                                cout<<j<<" ";
                                vis[j]=0;
                            }

                            cnt++;

                        }
                    }
                }

                if(flag==0)
                {
                    for(int j=f; j!=el[i]-1; j--)
                    {
                        if(vis[j]==1)
                        {
                            if(cnt==n)
                            {
                                cout<<j<<endl;
                                vis[j]=0;
                                break;
                            }
                            else
                            {
                                cout<<j<<" ";
                                vis[j]=0;
                            }

                            cnt++;

                        }
                    }

                }

            }
        }
    }
    return 0;
}