Editorial for Inaho VI

Remember to use this editorial only when stuck, and not to copy-paste code from it. Please be respectful to the problem author and editorialist.

Submitting an official solution before solving the problem yourself is a bannable offence.

Author: Ninjaclasher

Without prior knowledge on the basics of complex numbers, this problem is impossible to solve. For example, it must be known that $$i = \sqrt{-1}$$ and that the rectangular coordinate form of a complex number is $$z = a + bi$$.

For the first subtask, since $$x \le 1$$, we can utilize a simple if statement. If $$x = 0$$, the solution is $$1 + 0i$$. If $$x = 1$$, the solution is simply $$0 + 1i$$.

For the second subtask, some quick Googling will give the solutions to when $$x = 2$$ and when $$x = 3$$. When $$x = 2$$, the solution is $$e^{-\frac{\pi}{2}} \approx 0.20788 + 0i$$. When $$x = 3$$, the solution is $$e^{\frac{\pi i e^{-\frac{\pi}{2}}}{2}} \approx 0.94716 + 0.32076i$$. Alternatively, one can solve for when $$x = 2$$ and $$x = 3$$ manually using Euler's formula.

For the third subtask, one can utilize a for loop that iterates from $$1$$ to $$x$$. Start with $$2$$ floating-point variables, $$a_0$$ set to $$1$$, and $$b_0$$ set to $$0$$. At each iteration of the for loop, it can be found that:

$$a_i = e^{-\frac{\pi b_{i-1}}{2}}\ cos \frac{\pi a_{i-1}}{2}$$ and $$b_i = e^{-\frac{\pi b_{i-1}}{2}}\ sin \frac{\pi a_{i-1}}{2}$$.

Alternatively, one can utilize a complex numbers library in their preferred programming language.

For the last subtask, one can figure out that solution converges to approximately $$0.4383 + 0.3606i$$, which means that after a certain $$x$$, the solution becomes consistent up to $$4$$ decimal places. This means that we can hard code a value instead of looping up to $$x$$. This certain $$x$$ and the proof is left as an exercise for the reader.

Time Complexity: $$O(x)$$