CF1738E Balance Addicts

Given an integer sequence $a_1, a_2, \dots, a_n$ of length $n$, your task is to compute the number, modulo $998244353$, of ways to partition it into several **non-empty** **continuous** subsequences such that the sums of elements in the subsequences form a **balanced** sequence. A sequence $s_1, s_2, \dots, s_k$ of length $k$ is said to be balanced, if $s_{i} = s_{k-i+1}$ for every $1 \leq i \leq k$. For example, $[1, 2, 3, 2, 1]$ and $[1,3,3,1]$ are balanced, but $[1,5,15]$ is not. Formally, every partition can be described by a sequence of indexes $i_1, i_2, \dots, i_k$ of length $k$ with $1 = i_1 < i_2 < \dots < i_k \leq n$ such that 1. $k$ is the number of non-empty continuous subsequences in the partition; 2. For every $1 \leq j \leq k$, the $j$\-th continuous subsequence starts with $a_{i_j}$, and ends exactly before $a_{i_{j+1}}$, where $i_{k+1} = n + 1$. That is, the $j$\-th subsequence is $a_{i_j}, a_{i_j+1}, \dots, a_{i_{j+1}-1}$. There are $2^{n-1}$ different partitions in total. Let $s_1, s_2, \dots, s_k$ denote the sums of elements in the subsequences with respect to the partition $i_1, i_2, \dots, i_k$. Formally, for every $1 \leq j \leq k$, $$ s_j = \sum_{i=i_{j}}^{i_{j+1}-1} a_i = a_{i_j} + a_{i_j+1} + \dots + a_{i_{j+1}-1}. $$ For example, the partition $[1\,|\,2,3\,|\,4,5,6]$ of sequence $[1,2,3,4,5,6]$ is described by the sequence $[1,2,4]$ of indexes, and the sums of elements in the subsequences with respect to the partition is $[1,5,15]$. Two partitions $i_1, i_2, \dots, i_k$ and $i'_1, i'_2, \dots, i'_{k'}$ (described by sequences of indexes) are considered to be different, if at least one of the following holds. - $k \neq k'$, - $i_j \neq i'_j$ for some $1 \leq j \leq \min\left\{ k, k' \right\}$.

N/A

N/A

For the first test case, there is only one way to partition a sequence of length $ 1 $ , which is itself and is, of course, balanced. For the second test case, there are $ 2 $ ways to partition it: - The sequence $ [1, 1] $ itself, then $ s = [2] $ is balanced; - Partition into two subsequences $ [1\,|\,1] $ , then $ s = [1, 1] $ is balanced. For the third test case, there are $ 3 $ ways to partition it: - The sequence $ [0, 0, 1, 0] $ itself, then $ s = [1] $ is balanced; - $ [0 \,|\, 0, 1 \,|\, 0] $ , then $ s = [0, 1, 0] $ is balanced; - $ [0, 0 \,|\, 1 \,|\, 0] $ , then $ s = [0, 1, 0] $ is balanced. For the fourth test case, there are $ 4 $ ways to partition it: - The sequence $ [1, 2, 3, 2, 1] $ itself, then $ s = [9] $ is balanced; - $ [1, 2 \,|\, 3 \,|\, 2, 1] $ , then $ s = [3, 3, 3] $ is balanced; - $ [1 \,|\, 2, 3, 2 \,|\, 1] $ , then $ s = [1, 7, 1] $ is balanced; - $ [1 \,|\, 2 \,|\, 3 \,|\, 2 \,|\, 1] $ , then $ s = [1, 2, 3, 2, 1] $ is balanced. For the fifth test case, there are $ 2 $ ways to partition it: - The sequence $ [1, 3, 5, 7, 9] $ itself, then $ s = [25] $ is balanced; - $ [1, 3, 5 \,|\, 7 \,|\, 9] $ , then $ s = [9, 7, 9] $ is balanced. For the sixth test case, every possible partition should be counted. So the answer is $ 2^{32-1} \equiv 150994942 \pmod {998244353} $ .