Count continuous occurences, restart counting when condition met, exclude conditions met

Hi

I'm trying to count the number of occurences in A column downwards. See attached *xlsx file with description. Column A only has 3 Values. "A", "B" and "C" in no particular pattern. There will be values inserted every day in A column, so the total number of rows is unknown. But it's ok to copy formula downwards ahead of time.
The rules are:
The counting occures in column B (and column C in *.xlsx example file). Values to be counted exists in column A.
When counting value "A", the counting should restart counting downwards when encountering a value of "C".
However, if it encounters a value of "B", it should skip it as if uninterrupted, and continue counting "A" until interrupted by a "C".

So a series of values downwards A column should result in the following in the B column. Please see attached *.xlsx file for example

Excelforum_count_contious_and_skipped - Copy.xlsx

A - 1
A - 2
B -
A - 3
C -
A - 1
C -
A - 1
B -
A - 2
A - 3
A - 4

I've tried Countif and IF(S) and searched a lot, but I could only partially solve it. Is it doable in formulas?

Regards

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Thanks.

This solution works in Excel 2007 and newer:

Please try in B2 and copy down:

Formula:

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and try in C2 and copy down:

Formula:

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You are Welcome!

Thanks for the feedback and glad to have helped.

Without seeing a workbook, I have no idea why =B2>2 returns true when it was less than 2.

=B2>2 returns TRUE if B2>2 or (if B2 = not numeric and B2 is not empty) (B2>2 returns also TRUE if a formula in B2 returns "".)
=B2>2 returns FALSE if B2 equal to or less than B2 or if B2 is empty (if there is a formula in B2 the cell is never empty, even if the formula returns "").

If you want 1 only when the value is a number higher than 2 please try

Formula:

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. If there is in B2 a formula that returns "", this formula will then return 0 in this case.