Card_Rotation.cpp

// Given a sorted deck of cards numbered 1 to N.

// 1) We pick up 1 card and put it on the back of the deck.

// 2) Now, we pick up another card, it turns out to be card number 1, we put it outside the deck.

// 3) Now we pick up 2 cards and put it on the back of the deck.

// 4) Now, we pick up another card and it turns out to be card numbered 2, we put it outside the deck. ...

// We perform this step until the last card.

// If such an arrangement of decks is possible, output the arrangement, if it is not possible for a particular value of N then output -1.

// Example 1:

// Input:
// N = 4
// Output: 2 1 4 3
// Explanation:
// We initially have [2 1 4 3]
// In Step1, we move the first card to the end. 
// Deck now is: [1 4 3 2]
// In Step2, we get 1. Hence we remove it. Deck 
// now is: [4 3 2]
// In Step3, we move the 2 front cards ony by one 
// to the end  ([4 3 2] -> [3 2 4] -> [2 4 3]).
// Deck now is: [2 4 3]
// In Step4, we get 2. Hence we remove it. Deck 
// now is: [4 3]
// In Step5, the following sequence follows: 
// [4 3] -> [3 4] -> [4 3] -> [3 4]. Deck now 
// is: [3 4]
// In Step6, we get 3. Hence we remove it. Deck 
// now is: [4] Finally, we're left with a single 
// card and thus, we stop. 
 

// Example 2:

// Input:
// N = 5
// Output: 3 1 4 5 2




#include<bits/stdc++.h>
using namespace std;
int main(){
    int n;
    cin>>n;
    queue<int> q;
    vector<int> res;
    for(int i=0;i<n;i++){
        q.push(i);
        res.push_back(i+1);
    }
    vector<int> v(n,0);
    for(int i=0;i<n;i++){
        for(int j=0;j<=i;j++){
            int front=q.front();
            q.pop();
            q.push(front);
        }
        v[q.front()]=res[i];
        q.pop();
    }
    for(int i:v){
        cout<<i<<" ";
    }
}