Kaosar and Loop (Çözgüdiň düşündirişi)

Solution

The function starts with ans = a_1

Every next step does

Bitwise AND can only remove bits, never add.

Therefore, the final value can never exceed $a[1]$.

So if we want f(p) = k then the simplest and safest choice is: a_1 = k

Once $\text{ans}$ becomes $k$, we just need to avoid removing its bits.

For remaining positions, choose any distinct integers that are \le 10^{18} because we are doing OR of a_1

with all number and finally ANDing them so when we OR we get those bits which we have in $k$ and as a_1=k so

ANDing them we left with $k$ so the answer is any sequence starts with $k$ and have distinct values.

Complexity $O(n)$.

for _ in range(int(input())):
    n,k=map(int,input().split())
    ans=[k]
    cur=1
    while(len(ans)<n):
        if(cur!=k):
            ans.append(cur)
        cur+=1
    print(*ans)