CodeTON Round 7 (Div. 1 + Div. 2, Rated, Prizes!)
B
AB Flipping
Codeforces
900
原题
← AC →
翻译中…

You are given a string $s$ of length $n$ consisting of characters $\texttt{A}$ and $\texttt{B}$. You are allowed to do the following operation:

翻译中…
  • Choose an index $1 \le i \le n - 1$ such that $s_i = \texttt{A}$ and $s_{i + 1} = \texttt{B}$. Then, swap $s_i$ and $s_{i+1}$.
翻译中…

You are only allowed to do the operation at most once for each index $1 \le i \le n - 1$. However, you can do it in any order you want. Find the maximum number of operations that you can carry out.

翻译中…

Input

翻译中…

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 1000$). Description of the test cases follows.

翻译中…

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

翻译中…

The second line of each test case contains the string $s$ ($s_i=\texttt{A}$ or $s_i=\texttt{B}$).

翻译中…

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

翻译中…

Output

翻译中…

For each test case, print a single integer containing the maximum number of operations that you can carry out.

翻译中…

Note

翻译中…

In the first test case, we can do the operation exactly once for $i=1$ as $s_1=\texttt{A}$ and $s_2=\texttt{B}$.

翻译中…

In the second test case, it can be proven that it is not possible to do an operation.

翻译中…

In the third test case, we can do an operation on $i=2$ to form $\texttt{ABAB}$, then another operation on $i=3$ to form $\texttt{ABBA}$, and finally another operation on $i=1$ to form $\texttt{BABA}$. Note that even though at the end, $s_2 = \texttt{A}$ and $s_3 = \texttt{B}$, we cannot do an operation on $i=2$ again as we can only do the operation at most once for each index.

翻译中…

样例

Input #1
3
2
AB
4
BBBA
4
AABB
Output #1
1
0
3
InputCopy OutputCopy
题目信息
题号
1896B
时间限制
1 s
内存限制
256 MB
标签
贪心
字符串
双指针

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。