Heard about Josephus algorithm:)

Do you mean every operation is remove a card which is at top and then move the top card to the bottom, right?

I think can use std::list(C++) or LinkedList(Java) to simulation operation. Because Them are based on a data structure--linked list. It can insert and delete element on $$$O(1)$$$.
So the total complexity is $$$O(n+\frac{n}{2}+\frac{n}{4}+\frac{n}{8}+...)$$$.
Because $$$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$$$ is small then 2.
So the complexity is small then $$$O(2n)$$$. Ignore the coefficient is $$$O(n)$$$.
We can also use mathematical methods. Here is oeis result.

Here is O(log(n)) sol,Idea is that after each cycle half of elements gets removed and also difference between remaining elements will gets double.Now keep track of first element which remains after each cycle,difference bw elements. Also as we are doing cyclic so if last element will be removed then in next iteration we will skip first element of remaining list and if last element remains then we need to remove first element of remaining list,this can done by keeping start variable which will be zero if first element is removed and will be 1 if 2nd element is removed and for next iteration we can count no of elements from start to last if they are odd that means last element will be removed and if they are even last element will survive and accordingly we can find first remaining and start for next iteration.

#include <bits/stdc++.h>
using namespace std;
int main(){
    int n;cin>>n;
    if(n==1){
        cout<<1<<endl;
        return 0;
    }
    int start=0,diff=1,first_remain=2,n1;
    while(1){
        //cout<<n<<" "<<start<<" "<<diff<<" "<<first_remain<<endll;;
        if(n==2||n==1){
            cout<<first_remain<<endl;
            break;
        }   
        n1=n-(n+1-start)/2;
        diff*=2;
        if((n-start)&1){
            start=1;
        }
        else{
           first_remain+=diff;
           start=0;
        }
        n=n1;
    }
    return 0;
}