Subtask 1

For the first subtask, we can encode each beaker with a unique binary number from ~0~ to ~N - 1~. For the ~i^{th}~ bit, we can easily determine if the encoding of the hydrogen cyanide beaker is a ~0~ or a ~1~ by pouring all of the beakers that have a ~1~ in the ~i^{th}~ bit into the cup. If the substance changes colours, then we know that the ~i^{th}~ bit of the hydrogen cyanide beaker is also a ~1~. Otherwise, it is a ~0~. If we repeat this for each bit, then we can determine the full binary encoding of the hydrogen cyanide beaker. Determining the formula for the number of bits for ~N~ beakers is an exercise for the reader.

Time Complexity: ~\mathcal{O}(1)~

Subtask 2

For the second subtask, we can use the same process as subtask 1, except we have ~M~ minutes. We want to divide each of the required measurements into ~M~ minutes while minimizing the maximum number of cups used in any minute. It can be shown that if ~Q~ is the number of cups required for ~N~ beakers in ~1~ minute, then the number of cups required for ~M~ minutes is ~\lceil \frac{Q}{M} \rceil~.

Time Complexity: ~\mathcal{O}(1)~

Be careful with precision if you decide to use floating point calculations.