2179D Approach

1. Decoding the Problem Statement

The Setup: You have a set of numbers from $0$ to $2^n - 1$.

If $n = 2$, you have ${0, 1, 2, 3}$.
If $n = 3$, you have ${0, 1, \ldots, 7}$.

The Score ($S(p)$): You need to arrange these numbers in a line (a permutation). The “score” is calculated by doing a cumulative AND operation. Think of the bits as light switches.

You start with the first number. Some lights are ON.
You AND it with the second number. The & operator can only turn lights OFF; it can never turn them back on.
You AND that result with the third number. More lights might turn off.
The “Score” is just the total number of lights ON at every single step added together.

The Goal:

Maximize the Score: Keep the lights ON for as long as possible.
Lexicographically Minimal: If there are multiple ways to get the high score, pick the sequence that comes first alphabetically (i.e., prefer smaller numbers at the start of the array).

2. Understand the “Cost Function”

You want to maximize:

\[S(p) = \sum_{i=0}^{2^n-1} \text{popcount}(p_0 \land p_1 \land \ldots \land p_i)\]

Let’s look at the term $A_i = (p_0 \land p_1 \land \ldots \land p_i)$.

$A_0 = p_0$
$A_1 = p_0 \land p_1$
$A_2 = p_0 \land p_1 \land p_2$

Key Observation: The value of $A_i$ can only decrease or stay the same as $i$ increases. The bits can turn from 1 to 0, but once a bit becomes 0, it stays 0 forever. To maximize the sum of popcounts, you want $A_i$ to stay as large as possible for as long as possible.

3. The First Element ($p_0$)

Since $A_i$ is formed by AND-ing everything up to $p_i$, the first element $p_0$ sets the “ceiling” for all future values.

If $p_0$ has a 0 at some bit position, every subsequent term $A_1, A_2, \ldots$ will also have a 0 there.
To maximize the sum, you want $p_0$ to have as many 1s as possible.
The largest available number is $2^n - 1$ (which is 11...1 in binary).
Hint 1: You should definitely start with $p_0 = 2^n - 1$.

4. The Logic (How to solve it)

To get the maximum score, you want to lose the “lights” (bits) as slowly as possible. Since all numbers in the permutation must be distinct, you are forced to lose at least 1 bit at certain steps.

The optimal strategy is to lose exactly one bit at a time.

The “Greedy” Strategy for Minimal Order: To make the sequence “alphabetically smallest,” you should always pick the smallest possible number that satisfies the condition of losing just one bit.

To get the Sacred Permutation (Maximum Score), we must “milk” the current AND-value for as long as possible before allowing it to drop.

Algorithm:

Start with curr $= 2^n - 1$.
Phase A (The Milk): Look for all unused numbers $X$ that satisfy (curr & X) == curr.
- These numbers act as “fuel” to keep the current score alive without dropping any bits.
- To be lexicographically minimal, pick these in increasing order.
Phase B (The Drop): Once no such $X$ exists, we are forced to drop a bit.
Repeat until curr becomes $0$.
Print the rest of the numbers sorted.

Example N=3 (0 to 7)

Start: 7 $(111)$.
- “Milk” 7: Any unused $X$ where $7 \land X == 7$? Only 7 itself. (Printed).
Drop: Remove bit 2 $(100)$. New curr $= 3$ $(011)$.
- “Milk” 3: Any unused $X$ where $3 \land X == 3$? Need bits 0 and 1. (e.g., 3, 7, 11…).
- Only 3 is available in range. Print 3.
Drop: Remove bit 1 $(010)$. New curr $= 1$ $(001)$.
- “Milk” 1: Any unused $X$ where $1 \land X == 1$? Need bit 0.
- Available: 1, 5.
- Sort them: 1 then 5.
- Print 1. Then Print 5. (This is the step I missed!)
Drop: Remove bit 0 $(001)$. New curr $= 0$ $(000)$.
End: Print remaining sorted $(0, 2, 4, 6)$.

Final Sequence: 7, 3, 1, 5, 0, 2, 4, 6

Approach for the C++ Implementation Below

This code generates a permutation of numbers from $2^n-1$ down to $0$ that maximizes the cumulative AND score and is lexicographically minimal.

Key Steps:

Start with the Maximum:
- Begin with the largest number, $2^n-1$ (all bits set), as the first element. This ensures the highest possible starting value for the AND sequence.
Bitwise Drops:
- For each bit position from high to low, subtract $2^i$ from the current value to drop one bit at a time, and print the new value.
- Mark each used value to avoid repeats.
Milking the Current AND:
- For each new curr, print all unused numbers $j$ such that (curr & j) == curr. These numbers do not reduce the current AND value and are printed in increasing order for lex minimality.
Finish with Remaining Numbers:
- After all possible drops and milking, print any numbers that haven’t been used yet, in increasing order.

Why does this work?

By always dropping only one bit at a time and milking the current AND value, the sequence keeps the AND result as high as possible for as long as possible.
Printing unused numbers in increasing order ensures the permutation is lexicographically minimal among all optimal solutions.

Complexity:

Time: $O(N \cdot 2^N)$, efficient for $N \leq 16$.
Space: $O(2^N)$ for the used array.

This approach guarantees both optimal score and minimal lexicographical order for the permutation.

The Correct Code

This implementation runs in $O(N \cdot 2^N)$, which is perfectly fast enough for $N \leq 16$.

C++ Implementation

#include <bits/stdc++.h>
using namespace std;
#define ll long long

int main(){
    ios::sync_with_stdio(false); 
    cin.tie(nullptr);
    cout.tie(nullptr);
    ll T;
    cin >> T;
    while(T--){
        ll n;
        cin >> n;

        ll max_val = (1LL << n) - 1;
        vector<bool> used(max_val+1,0);

        ll curr = max_val;
        cout<<curr<<" ";
        used[curr] = true;

        for(ll i = n - 1; i >= 0; i--){
            if(curr > 0){
                curr -= (1LL << i);
                cout<<(curr)<<" ";
                used[curr] = true;

                for(ll j = 0; j <= max_val; j++){
                    if(!used[j] && (curr & j) == curr){
                        cout<<j<<" ";
                        used[j] = true;
                    }
                }
            }
        }
        for(int i = 0; i <= max_val; i++){
            if(!used[i]){
                cout << i << " ";
            }
        }
        cout << "\n";

    }
    return 0;
}

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