麦芽雪冷萃

Problem Description：

You are messaging in one of the popular social networks via your smartphone. Your smartphone can show at most k most recent conversations with your friends. Initially, the screen is empty (i.e. the number of displayed conversations equals 0).
Each conversation is between you and some of your friends. There is at most one conversation with any of your friends. So each conversation is uniquely defined by your friend.
You (suddenly!) have the ability to see the future. You know that during the day you will receive n messages, the i-th message will be received from the friend with ID (1≤≤).
If you receive a message from in the conversation which is currently displayed on the smartphone then nothing happens: the conversations of the screen do not change and do not change their order, you read the message and continue waiting for new messages.
Otherwise (i.e. if there is no conversation with on the screen):
Firstly, if the number of conversations displayed on the screen is k, the last conversation (which has the position k) is removed from the screen.
Now the number of conversations on the screen is guaranteed to be less than k and the conversation with the friend is not displayed on the screen.
The conversation with the friend appears on the first (the topmost) position on the screen and all the other displayed conversations are shifted one position down.
Your task is to find the list of conversations (in the order they are displayed on the screen) after processing all n messages.

Input Specification:

The first line of the input contains two integers n and k (1≤n,k≤2⋅) — the number of messages and the number of conversations your smartphone can show.
The second line of the input contains n integers ,,…,(1≤≤109), where is the ID of the friend which sends you the i-th message.

Output Specification:

In the first line of the output print one integer m (1≤m≤min(n,k)) — the number of conversations shown after receiving all n messages.
In the second line print m integers ,,…,,, where idsiidsi should be equal to the ID of the friend corresponding to the conversation displayed on the position i after receiving all n messages.

Sample Input1:

7 2
1 2 3 2 1 3 2

Sample Output1:

2
2 1

Sample Input2:

10 4
2 3 3 1 1 2 1 2 3 3

Sample Output2:

3
1 3 2

Note：

In the first example the list of conversations will change in the following way (in order from the first to last message):
[][];
[1][1];
[2,1][2,1];
[3,2][3,2];
[3,2][3,2];
[1,3][1,3];
[1,3][1,3];
[2,1][2,1].
In the second example the list of conversations will change in the following way:
[][];
[2][2];
[3,2][3,2];
[3,2][3,2];
[1,3,2][1,3,2];
and then the list will not change till the end.

解题思路：

我是用vector+set来进行求解的。给定一个大小为n的vector，在该容器中添加m个元素，若容器中已存在该元素则不进行操作，若不存在该元素则将vector中最先进入的元素取出来并添加改元素。最后输出vector的大小并逆序输出vector中的元素即可。

AC代码：

#include<bits/stdc++.h>
using namespace std;
#define Up(i,a,b) for(int i = a; i <= b; i++)
typedef long long ll;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    ll m,n;
    cin >> m >> n;
    vector<int> v;
    set<int> s;
    while(m--)
    {
        int _;
        cin >> _;
        int sz = s.size();
        s.insert(_);
        if(sz != s.size())
        {
            v.push_back(_);
            if(v.size() > n)
            {
                s.erase(v.front());
                v.erase(v.begin());
            }
        }
    }
    bool virgin = true;
    while(v.size() > 0)
    {
        if(virgin)
        {
            cout << v.size() << endl << v.back();
            virgin = false;
        }
        else
        {
            cout << " " << v.back();
        }
        v.pop_back();
    }
    cout << endl;
    return 0;
}