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A15498. Binary Strings are Fun

编程题 普及/提高-

题目描述

A binary string $^\dagger$ $b$ of odd length $m$ is good if $b_i$ is the median $^\ddagger$ of $b[1,i]^\S$ for all odd indices $i$ ( $1 \leq i \leq m$ ).

For a binary string $a$ of length $k$ , a binary string $b$ of length $2k-1$ is an extension of $a$ if $b_{2i-1}=a_i$ for all $i$ such that $1 \leq i \leq k$ . For example, 1001011 and 1101001 are extensions of the string 1001. String $x=$ 1011011 is not an extension of string $y=$ 1001 because $x_3 \neq y_2$ . Note that there are $2^{k-1}$ different extensions of $a$ .

You are given a binary string $s$ of length $n$ . Find the sum of the number of good extensions over all prefixes of $s$ . In other words, find $\sum_{i=1}^{n} f(s[1,i])$ , where $f(x)$ gives number of good extensions of string $x$ . Since the answer can be quite large, you only need to find it modulo $998\,244\,353$ .

$^\dagger$ A binary string is a string whose elements are either $\mathtt{0}$ or $\mathtt{1}$ .

$^\ddagger$ For a binary string $a$ of length $2m-1$ , the median of $a$ is the (unique) element that occurs at least $m$ times in $a$ .

$^\S$ $a[l,r]$ denotes the string of length $r-l+1$ which is formed by the concatenation of $a_l,a_{l+1},\ldots,a_r$ in that order.

输入格式

Each test contains multiple test cases. The first line contains the number of test cases $t$ ( $1 \le t \le 10^4$ ). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ( $1 \le n \le 2 \cdot 10^5$ ), where $n$ is the length of the binary string $s$ .

The second line of each test case contains the binary string $s$ of length $n$ .

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$ .

输出格式

For each test case, print the answer modulo $998\,244\,353$ .

输入输出样例

输入 #1
6
1
1
1
0
2
11
3
010
9
101101111
37
1011011111011010000011011111111011111
输出 #1
1
1
3
3
21
365

说明/提示

In the first and second test cases, $f(s[1,1])=1$ .

In the third test case, the answer is $f(s[1,1])+f(s[1,2])=1+2=3$ .

In the fourth test case, the answer is $f(s[1,1])+f(s[1,2])+f(s[1,3])=1+1+1=3$ .

$f(\mathtt{11})=2$ because two good extensions are possible: 101 and 111.

$f(\mathtt{01})=1$ because only one good extension is possible: 011.
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