Subtask 1

The number of possible inputs is very small in this subtask. Here are some solutions:

• The only time when the digits infinitely repeat is when ~B=3~ and ~A \neq 3~, so we can just check for this specific case and use something like the to_string(int) method in C++ to count the number of digits past the decimal.
• Hardcode every possible input.

Subtask 2

~B~ is always a multiple of ~10~, so the decimal will never repeat infinitely, so we can convert the fraction to a string and just check the number of digits. Note that we cannot simply just count the number of digits in ~B~, as the fraction may not be completely reduced. Make sure to be careful about precision (i.e. using arbitrary precision integers).

Time Complexity: ~\mathcal{O}(\log(\max(a,b)))~ per test case.

Subtask 3

This subtask exists to reward solutions that would get full marks but fail due to precision or overflow errors.

Subtask 4

One possible solution is to observe that a finite decimal value for ~\frac{A}{B}~ can only get so long with the input constraints, so we can perform long division to find the value of ~\frac{A}{B}~ and assume that the answer is infinite if our decimal gets too long.

Time Complexity: ~\mathcal{O}(\log^2(\max(a,b)))~ per test case.

Alternatively, start by assuming that ~\frac{A}{B}~ is fully reduced (if it is not the case, reduce the fraction first). Next, notice the following:

• The only way for the decimal to be finite is if ~B = 2^a \cdot 5^b~ for some ~a,b \in \mathbb Z~ where ~a,b \ge 0~.
• Again let ~B = 2^a \cdot 5^b~ where ~a,b \in \mathbb Z~ and ~a,b \ge 0~. The number of digits to the right of the decimal is always ~\max(a,b)~. We can show that this is true with the following:
  • Let ~c=a-\min(a,b)~ and ~d=b-\min(a,b)~. We can rewrite ~B~ as ~10^{\min(a,b)} \cdot 2^c \cdot 5^d~ and notice that either ~c=0~ or ~d=0~ (or both).
  • Every power of ~10~ (the ~10^{\min(a,b)}~ portion) adds ~1~ additional digit by 'shifting' the current decimal to the right.
  • Every additional power of ~2~ or ~5~ (either ~c~ or ~d~ respectively) will always increase the number of digits by ~1~ as the fraction is already reduced.
  • Thus, if ~c \neq 0~ the number of digits is ~\min(a,b)+c~, otherwise the number of digits is ~\min(a,b)+d~, which is equivalent to ~\max(a,b)~.

Overall, this solution leads to a simpler implementation.

Time Complexity: ~\mathcal{O}(\log(\max(a,b)))~ per test case.