Modify array (and search set)

Given a length of N Array of A=[A1,A2,⋅⋅⋅AN], There may be repeated integers in the array .

Now Xiaoming will change it to an array without repeating integers as follows .

Xiao Ming will revise in turn A2,A3,⋅⋅⋅,AN.

When modifying Ai when , Xiao Ming will check Ai Whether in A1∼Ai−1 There has been .

If there has been , Then Xiao Ming will give Ai add 1; If the new Ai Still before , Xiaoming will continue to give Ai Add 1, until Ai Not in the A1∼Ai−1 There has been .

When AN After the above modifications , obviously A There are no duplicate integers in the array .

Now let's give the initial A Array , Please work out the final A Array
Input format
The first line contains an integer N.

The second line contains N It's an integer A1,A2,⋅⋅⋅,AN.

Output format
Output N It's an integer , In turn is the final A1,A2,⋅⋅⋅,AN.

Data range
1≤N≤10^5,
1≤Ai≤10^6
sample input ：
5
2 1 1 3 4
sample output ：
2 1 3 4 5

Answer key ：
Union checking set , The ancestors of initializing numbers at the beginning are themselves , If the output is too , Then the number of ancestors shall be added to his original ancestors 1

#include<bits/stdc++.h>
using namespace std;
int n,a[100005];
int father[1000006];
int findfather(int x){
    
	if(x!=father[x]) father[x]=findfather(father[x]);
	return father[x];
} 
int main(){
    
	cin>>n;
	for(int i=1;i<1e6+2;i++){
    
		father[i]=i;
	}
	for(int i=0;i<n;i++){
    
		int x;
		cin>>x;// use scanf Better, faster ,cin It's easy to time out 
		x=findfather(x);
		cout<<x<<" ";
		father[x]=x+1;
	}
	return 0;
}