2222E: Seek the Truth

Let $U = 2^n - 1$ be the all-ones mask of length $n$ .

The whole trick is to start with $S = \{U\}.$ Why? Because the values of $f(0)$ and $f(U)$ become absurdly structured:

\begin{aligned} 0\,&\,c &=&\ 0, & U\,&\,c &=&\ c,\\ 0\,|\,c &=&\ c, & U\,|\,c &=&\ U,\\ 0\oplus c &=&\ c, & U\oplus c &=&\ U\oplus c = \bar c. \end{aligned}

Here $\bar c = U\oplus c$ is the bitwise complement of $c$ inside $n$ bits.

So yes, the problem looks interactive and scary, but the structure is doing half the work for us.

Step 1: Two inserts that almost solve everything

We first output the initial value $a=U$ .

Then we ask:

I 0
I U

Let the returned set sizes be $s_1$ and $s_2$ .

What happens?

If $k=1$ (AND)

After I 0, we insert $0$ , so $S = \{U,0\}, \quad s_1=2.$ After I U, we insert $c$ , so

if $c=U$ , size stays $2$ ,
otherwise size becomes $3$ .

So AND gives:

(s_1,s_2)= \begin{cases} (2,2), & c=U,\\ (2,3), & c\ne U. \end{cases}

If $k=2$ (OR)

After I 0, we insert $c$ , so

if $c=U$ , size stays $1$ ,
otherwise size becomes $2$ . After I U, we insert $U$ again, so nothing changes.

Thus OR gives:

(s_1,s_2)= \begin{cases} (1,1), & c=U,\\ (2,2), & c\ne U. \end{cases}

If $k=3$ (XOR)

After I 0, we insert $c$ , so

if $c=U$ , size stays $1$ ,
otherwise size becomes $2$ . After I U, we insert $\bar c$ , so
if $c=U$ , then $\bar c=0$ and size becomes $2$ ,
otherwise $\bar c\notin\{U,c\}$ , so size becomes $3$ .

Thus XOR gives:

(s_1,s_2)= \begin{cases} (1,2), & c=U,\\ (2,3), & c\ne U. \end{cases}

Step 2: One more query finishes the type

The easy cases are already done:

$(1,1) \Rightarrow (k,c)=(2,U)$
$(1,2) \Rightarrow (k,c)=(3,U)$

The remaining pairs are ambiguous:

$(2,2)$ could be AND with $c=U$ or OR with $c\ne U$
$(2,3)$ could be AND with $c\ne U$ or XOR with $c\ne U$

Now ask: Q 1

This is a sneaky little truth serum.

If $(s_1,s_2)=(2,2)$

AND with $c=U$ gives $S=\{U,0\}$ , so only $U\ge1$ . Hence answer is $1$ .
OR with $c\ne U$ gives $S=\{U,c\}$ , and since $c\ge1$ , answer is $2$ .

So:

$q=1 \Rightarrow (k,c)=(1,U)$
$q=2 \Rightarrow k=2$ and $c\ne U$

If $(s_1,s_2)=(2,3)$

AND gives $S=\{U,0,c\}$ , so the elements $\ge1$ are $U$ and $c$ . Answer $2$ .
XOR gives $S=\{U,c,\bar c\}$ . Since $c\ne U$ and $c\ne0$ , both $c$ and $\bar c$ are positive, so all three elements are $\ge1$ . Answer $3$ .

So:

$q=2 \Rightarrow k=1$ and $c\ne U$
$q=3 \Rightarrow k=3$ and $c\ne U$

At this point, $k$ is fully determined.

Step 3: Recovering $c$

Now the remaining job is just extracting the exact value of $c$ .

Case A: OR with $c\ne U$

The set is $S=\{U,c\}.$ For any threshold $y$ ,

2, & y\le c,\\ 1, & y>c. \end{cases}$$ That is a monotone predicate in $$y$$. So binary search the largest $$y$$ such that $$Q(y)=2$$. That value is exactly $$c$$. Search interval size is $$2^n$$, so this costs exactly $$n$$ queries. ### Case B: AND with $$c\ne U$$ The set is $$S=\{U,0,c\}.$$ For any $$y\ge1$$, $$Q(y)=\begin{cases} 2, & y\le c,\\ 1, & y>c. \end{cases}$$ Same monotone predicate, same binary search, same $$n$$ queries. ### Case C: XOR with $$c\ne U$$ This is the only remotely interesting part. The set is $$S=\{U,c,\bar c\}.$$ Let $$m = \min(c,\bar c).$$ Since $$c\ne U$$ and $$c\ne0$$, we have $$1 \le m \le 2^{n-1}-1.$$ Now look at thresholds $$1\le y\le 2^{n-1}$$. Both $$c$$ and $$\bar c$$ are at least $$y$$ iff $$y\le m$$. Therefore $$Q(y)=3 \iff y\le m,$$ and otherwise $$Q(y)=2$$. So we can binary search the largest $$y$$ with $$Q(y)=3$$, obtaining $$m$$ in exactly $$n-1$$ queries. But that only gives the unordered pair $$\{c,\bar c\}$$. We still need to know which one is the real $$c$$. Here comes the last cute trick: insert `I m`. - If $$c=m$$, then $$m\oplus c = 0$$. Since $$0\notin S$$ in this XOR case, the set size increases to $$4$$. - If $$c=\bar m$$, then $$m\oplus c = U$$, and $$U\in S$$ already, so the size stays $$3$$. Thus one final insert orients the pair:

\text{new size }=4 \Rightarrow c=m, \qquad \text{new size }=3 \Rightarrow c=U\oplus m.

## Query count Let’s count carefully, because interactive problems love screwing you on off-by-one nonsense. - First two inserts: $$2$$ queries. - One `Q 1` if needed: $$1$$ more. - OR / AND recovery: $$n$$ more. - XOR recovery: $$n-1$$ threshold queries + $$1$$ final insert. So the worst case is $$2+1+n = n+3$$ for OR/AND, and $$2+1+(n-1)+1 = n+3$$ for XOR. Exactly on budget. No wasted moves. Very satisfying. ## Complexity Per test case, the number of interactions is at most $$n+3$$, as required. The local computation is just binary search, so time complexity is $$O(n)$$ per test case, with $$O(1)$$ memory. ## Final algorithm summary Let $$U=2^n-1$$, print initial $$a=U$$. 1. Ask `I 0` and `I U`. 2. If the answers are: - $$(1,1)$$: answer `A 2 U`. - $$(1,2)$$: answer `A 3 U`. 3. Otherwise ask `Q 1`. 4. If $$(2,2)$$: - `Q 1 = 1` => answer `A 1 U` - `Q 1 = 2` => OR, binary search $$c$$ from `Q mid` 5. If $$(2,3)$$: - `Q 1 = 2` => AND, binary search $$c$$ from `Q mid` - `Q 1 = 3` => XOR, binary search $$m=\min(c,\bar c)$$ from `Q mid`, then insert `I m` to decide whether $$c=m$$ or $$c=U\oplus m$$. That’s the whole thing. The problem looks like some eldritch interactive horror, but once you bully it with $$0$$ and all-ones, it folds pretty fast.

Seek the Truth

Progressive nudges

Full explanation

Step 1: Two inserts that almost solve everything

What happens?

If $k=1$ (AND)

If $k=2$ (OR)

If $k=3$ (XOR)

Step 2: One more query finishes the type

If $(s_1,s_2)=(2,2)$

If $(s_1,s_2)=(2,3)$

Step 3: Recovering $c$

Case A: OR with $c\ne U$

Generated C++

Seek the Truth

Progressive nudges

Full explanation

Step 1: Two inserts that almost solve everything

What happens?

If k=1k=1k=1 (AND)

If k=2k=2k=2 (OR)

If k=3k=3k=3 (XOR)

Step 2: One more query finishes the type

If (s1,s2)=(2,2)(s_1,s_2)=(2,2)(s1​,s2​)=(2,2)

If (s1,s2)=(2,3)(s_1,s_2)=(2,3)(s1​,s2​)=(2,3)

Step 3: Recovering ccc

Case A: OR with c≠Uc\ne Uc=U

Generated C++

If $k=1$ (AND)

If $k=2$ (OR)

If $k=3$ (XOR)

If $(s_1,s_2)=(2,2)$

If $(s_1,s_2)=(2,3)$

Step 3: Recovering $c$

Case A: OR with $c\ne U$