Palindrome XOR
Contest: Forethought Future Cup - Final Round (Onsite Finalists Only)
Difficulty rating: 2400
Tags: dfs and similar, graphs

You are given a string $$$s$$$ consisting of characters "1", "0", and "?". The first character of $$$s$$$ is guaranteed to be "1". Let $$$m$$$ be the number of characters in $$$s$$$.

Count the number of ways we can choose a pair of integers $$$a, b$$$ that satisfies the following:

- $$$1 \leq a < b < 2^m$$$
- When written without leading zeros, the base-2 representations of $$$a$$$ and $$$b$$$ are both palindromes.
- The base-2 representation of bitwise XOR of $$$a$$$ and $$$b$$$ matches the pattern $$$s$$$. We say that $$$t$$$ matches $$$s$$$ if the lengths of $$$t$$$ and $$$s$$$ are the same and for every $$$i$$$, the $$$i$$$-th character of $$$t$$$ is equal to the $$$i$$$-th character of $$$s$$$, or the $$$i$$$-th character of $$$s$$$ is "?".

Compute this count modulo $$$998244353$$$.

Input format:
The first line contains a single string $$$s$$$ ($$$1 \leq |s| \leq 1\,000$$$). $$$s$$$ consists only of characters "1", "0" and "?". It is guaranteed that the first character of $$$s$$$ is a "1".

Output format:
Print a single integer, the count of pairs that satisfy the conditions modulo $$$998244353$$$.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
For the first example, the pairs in base-2 are $$$(111, 10001), (11, 10101), (1001, 11111)$$$.

Example input:
10110

Example output:
3

Example input:
1?0???10

Example output:
44

Editorial:
Since the leading character of $$$s$$$ is a "1", then that means $$$a < 2^{m-1}$$$ and $$$2^{m-1} \leq b < 2^m$$$. Let's fix the length of $$$a$$$ as $$$k$$$. I'll describe a more general solution, so there might be simpler solutions that work for this specific problem.
Let's make a graph with $$$n+k+2$$$ nodes. The first $$$n$$$ nodes represent the $$$n$$$ bits of $$$b$$$, and the next $$$k$$$ nodes represent the $$$k$$$ bits of $$$a$$$. The last two nodes represent a $$$0$$$ node and $$$1$$$ node (which we will explain later). We want to find the number of ways to color the graph with two colors $$$0$$$ and $$$1$$$ such that they satisfy some conditions.
Let's draw two different types of edges $$$0$$$-edges and $$$1$$$-edges. If two nodes are connected by a $$$0$$$-edge, then that means they must be the same color. If two nodes are connected by a $$$1$$$-edge, then that means they must be a different color.
We will draw some edges as follows:
• Draw a $$$1$$$ edge between the $$$0$$$ node and $$$1$$$ node to represent they must be different colors.
• Draw a $$$0$$$ edge between $$$b_i$$$ and $$$b_{n-i-1}$$$ to represent the palindrome conditions (similarly we can do this for $$$a$$$).
• For the $$$i$$$-th bit, if $$$s_i$$$ is "1", draw a $$$1$$$ edge between $$$a_i$$$ and $$$b_i$$$ (if $$$i > k$$$, we instead draw an edge from $$$b_i$$$ to $$$1$$$). If $$$s_i$$$ is "0", then draw a $$$0$$$ edge between $$$a_i$$$ and $$$b_i$$$. If $$$s_i$$$ is "?", then don't draw any edges, since there are no explicit constraints.
Now, we want to count the number of valid colorings. We want to split the graph into a two colors, which is a bipartite graph. We want all edges that cross the bipartition to be $$$1$$$ edges and all edges within the same bipartition to be $$$0$$$ edges.
To count this, we first collapse all connected components of $$$0$$$ edges, then check if the remaining $$$1$$$ edges form a bipartite graph. If there is a non-bipartite graph, return $$$0$$$ immediately, since this means it's impossible to fulfill the conditions. Otherwise, let $$$C$$$ be the number of connected components. We add $$$2^{C-1}$$$ to our answer. The reason we subtract $$$1$$$ is that the component containing the $$$0$$$ and $$$1$$$ node is fixed to be colored $$$0$$$, but for other components, we are free to color the components in either of two ways.
There are $$$n$$$ different lengths to try, each of which take a linear amount of time to get the count, so the overall time complexity is $$$O(n^2)$$$.