Hints: Interval Game

Write $x_1$ and $x_2$ in 0-based form: after reading, subtract $1$ from each upper bound so that slack variables $a = l - 1$ and $b = x - r$ satisfy $a + b \le X_1$ and $a_2 + b_2 \le X_2$ in the code’s variables X1 and X2.

For the second interval, the losing pairs for Alice are those whose XOR $u = a_2 \oplus b_2$ equals her chosen first-interval XOR $t = a_1 \oplus b_1$. For each possible $a = i$, the valid $b$ run from $0$ to $X_2 - i$ inclusive; you need the multiset of XOR values $i \oplus b$ over that range.

For fixed $i$, decompose the range of $b$ from $0$ to $X_2 - i$ into $O(\log X_2)$ disjoint blocks whose lengths are powers of two (greedy from large bits). On each block, the XOR values hit a structured set; add $+1$ and $-1$ at two keys in a map del so that a later prefix sum over $c$ recovers the total frequency of pairs with XOR equal to $c$.

Sweep $c = 0, 1, \dots, X_1$. Maintain a running count cnt by adding del[c] at each step. The value $c$ with minimum cnt is a best choice of $t$; the implementation keeps the smallest such $c$ when ties occur.

Realize that optimal $t$ with $(a_1, b_1) = (0, t)$: in 1-based coordinates this is $l_1 = 1$ and $r_1 = x_1 - t$ (the code prints 1 and X1 + 1 - best after the decrement, which is $x_1 - t$ in the original input).

Write the interval $[l, r]$ (inside $[1, x]$) as two independent nonnegative values:

• $a = l - 1$ (how much we can still decrease $l$),
• $b = x - r$ (how much we can still increase $r$).

What game does each of these two values represent?

Answer to Hint 1:
Each value is a Nim heap: in one move you can reduce it to any smaller nonnegative value.

So for one interval, Grundy value is:

$$ g = a \oplus b. $$

For two intervals, total Grundy is xor of all four heaps.

Alice fixes only the first interval, so she fixes only:

$$ t = a_1 \oplus b_1. $$

Then Bob’s interval is random and contributes:

$$ u = a_2 \oplus b_2. $$

Alice loses exactly when $t \oplus u = 0$, i.e. $u=t$.

So we need to minimize:

$$ f(t)=\#\{(a_2,b_2)\mid a_2,b_2\ge 0,\ a_2+b_2\le x_2-1,\ a_2\oplus b_2=t\}. $$

The denominator (all possible second intervals) is fixed, so minimizing $f(t)$ maximizes win probability.

For fixed $t$, let:

$$ u = a_2 \& b_2. $$

Then:

$$ a_2+b_2 = t + 2u. $$

Also bits where $t$ has $1$ cannot appear in $u$, so $u \& t = 0$.

Answer to Hint 5:
For fixed $u$ with $u \& t = 0$, each $1$-bit of $t$ can be assigned to either $a_2$ or $b_2$, independently.

So each valid $u$ gives:

$$ 2^{\operatorname{popcount}(t)} $$

pairs $(a_2,b_2)$.

Therefore, if $t \le n$ where $n=x_2-1$:

$$ f(t) = 2^{\operatorname{popcount}(t)} \cdot \#\{u \mid 0\le u\le \lfloor (n-t)/2\rfloor,\ u\&t=0\}. $$

If $t>n$, then $f(t)=0$ immediately.

How to count

$$ \#\{u\le M \mid u\&t=0\} $$

fast?
Use a standard bit-DP with tight flag over bits from high to low.

Reachable values of $t$ for Alice are all $0..x_1-1$: choose $(a_1,b_1)=(t,0)$, which corresponds to interval:

$$ [l_1, r_1] = [t+1, x_1]. $$

So just scan all $t$ in that range and pick one minimizing $f(t)$.

Complexity per test:

• iterate $t=0..x_1-1$,
• for each $t$, run $O(\log x_2)$ bit-DP.

Total complexity:

$$ O\!\left(x_1\log x_2\right) $$

per test, and linear in total constraints.