Codeforces 1629 C. Mexico array - simple greed

This way

The question ：

Give you a length of n Array of a And an empty array b, You can choose the front at a time k individual , Find out the MEX, Plug into the b At the back of .
To make b The dictionary order is the largest , Ask what the answer is .

Answer key ：

A lot about the dictionary order is greed , So since this problem needs the largest dictionary order , Then we must be greedy when we pick up , Then enumerate what each number is , And then look at the present a If you can find , If you can't find it, it's it .
At the same time, pay attention to maintenance a The number remaining in .

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
int a[N];
set<int>s[N];
vector<int>ans;
int main()
{
    
    int t;
    scanf("%d",&t);
    while(t--){
    
        ans.clear();
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]),s[a[i]].insert(i);
        int l=1;
        while(l<=n){
    
            int mx=-1,i;
            for(i=0;i<=n;i++){
    
                if(s[i].empty())break;
                mx=max(mx,*s[i].begin());
            }
            if(mx==-1){
    
                s[a[l]].erase(s[a[l]].begin());
                ans.push_back(0);
                l++;
                continue;
            }
            ans.push_back(i);
            while(l<=mx)s[a[l]].erase(s[a[l]].begin()),l++;
        }
        printf("%d\n",ans.size());
        for(int i:ans)
            printf("%d ",i);
        printf("\n");
    }
    return 0;
}