C. Yet Another Card Deck

You have a card deck of n cards, numbered from top to bottom, i. e. the top card has index 1 and bottom card — index n. Each card has its color: the i-th card has color ai.

You should process q queries. The j-th query is described by integer tj. For each query you should:

find the highest card in the deck with color tj, i. e. the card with minimum index;
print the position of the card you found;
take the card and place it on top of the deck.
Input
The first line contains two integers n and q (2≤n≤3⋅105; 1≤q≤3⋅105) — the number of cards in the deck and the number of queries.

The second line contains n integers a1,a2,…,an (1≤ai≤50) — the colors of cards.

The third line contains q integers t1,t2,…,tq (1≤tj≤50) — the query colors. It’s guaranteed that queries ask only colors that are present in the deck.

Output
Print q integers — the answers for each query.

Example
inputCopy
7 5
2 1 1 4 3 3 1
3 2 1 1 4
outputCopy
5 2 3 1 5
Note
Description of the sample:

the deck is [2,1,1,4,3–,3,1] and the first card with color t1=3 has position 5;
the deck is [3,2–,1,1,4,3,1] and the first card with color t2=2 has position 2;
the deck is [2,3,1–,1,4,3,1] and the first card with color t3=1 has position 3;
the deck is [1–,2,3,1,4,3,1] and the first card with color t4=1 has position 1;
the deck is [1,2,3,1,4–,3,1] and the first card with color t5=4 has position 5.

题意，两个输入分别表示接下来要输入两组数据中数量的个数，
第一个数据表示输入一组数的排列，第二组数据的每一个输入表示对第一个数据的查询查询的内容是第一组数据中排序最靠前的数的位置，并且查询完之后将其放到队首。

思路因为我们每次只查询这个数据最靠前的位置那么相同数字后面的位置我们就可以不管了，因为我们永远不可能用到。那么我们只需要维护一个【50】的数组

#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <algorithm>

using namespace std;

int n,m,ans,flag[60];

int main(){
    memset(flag,0,sizeof(flag));
    int a;
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>a;
        if(!flag[a]){
            flag[a]=i;
        }
        ans = max(ans,a);
    }
    for(int i=1;i<=m;i++){
        cin>>a;
        cout<<flag[a];
        if(i<m)
            cout<<' ';
        else cout<<endl;

        for(int j=1;j<=ans;j++){
            if(flag[j]){
                if(flag[j]<flag[a])
                    flag[j]++;
            }
        }
        flag[a]=1;
    }
}