Think about prefix sums

Process all the queries of type 0 and store the type 1 query in a vector of arrays (say End). Now after getting all the elements, You have some ranges(starting from 1st element). You just have to take Xor all the elements of a particular range with the calculated value.

Lets say we have a variable Xor which is the current value with which we have to take the xor with the i-th element of the given array.

Initially Xor = xor(all X of type 1 query).

Now as you procees forward and leave a range You again take xor of the value proviede for the given range(which ended right before) which will compensate the original Xor taken of all the values provided...

Not clear...?

Have a look at the code and read above strategy again. Hope it is clear now.

#include<bits/stdc++.h>
using namespace std;
#define ios ios_base::sync_with_stdio(0);cin.tie(0)
#define scn(n) scanf("%d",&n)
#define lscn(n) scanf("%lld",&n)
typedef long long ll;
#define pri(n) printf("%d\n",n)
#define lpri(n) printf("%lld\n",n);
#define rep(i,st,ed) for(int i=st;i<ed;i++)
#define var(n) int n; scn(n)
#define F first
#define S second 
#define pb(n) push_back(n)
#define all(x) (x).begin(),(x).end()
const int N=1e6+6;
const ll M=1e9+7;
const ll inf=1e18+5;

vector<int> End[N];

int main()
{
	ios;
	int T;
	cin>>T;
	while(T--)
	{
		vector<int> v;
		v.push_back(0);
		int q;
		cin>>q;
		int Xor = 0;
		while(q--)
		{
			int type,x;
			cin>>type>>x;
			if(type==0)
			{
				v.push_back(x);
			}
			else
			{
				int sz = v.size();
				End[sz].push_back(x);
				Xor ^= x;
			}
		}
		int sz = v.size();
		for(int i = 0;i<sz;i++)
		{
			v[i] ^= Xor;
			for(auto it : End[i+1])
			{
				Xor ^= it;
			}
		}
		sort(all(v));
		for(auto it : v)
			cout<<it<<' ';
		cout<<"\n";
	}		
}

Thanks CodeForces For This Great Community

n = int(input())
l = [0]
pre = [0]*n 
ind = 0
for i in range(n):
    x,y = [int(m) for m in input().split()]
    if x==0:
        l.append(y)
        ind+=1 
    else:
        pre[ind] = y 
xor = 0

for i in range(len(l)-1,-1,-1):
    xor^=pre[i]
    l[i]^=xor 
print(*sorted(l))
...