I was only able to solve PPDIV (Perfect Power Divisors) for 30 pts by finding the contribution of every Perfect Power till sqrt(n).

What was the approach for 100 pts.

My solution might be more complex, hand-wavy than others but here it is:

You can prove that answer for \displaystyle n = n + \sum_{d=2}^{\infty} \sum_{r=2}^{\infty} (-\mu[d])r^d \left \lfloor \frac{n}{r^d} \right \rfloor

Since r^d < n for only finitely many values, this sum only has finitely many non 0 terms.

For d=2, there will be n^{\frac{1}{2}} values of r which will give non 0 terms.

For d=3, there will be n^{\frac{1}{3}} values of r which will give non 0 terms.

and in general,

For d=i, there will be n^{\frac{1}{t}} values of r which will give non 0 terms.

If we can somehow optimise the calculation for d=2 to O(n^{1/3}), we can solve it in the TL.

The calculation for d=2 is \displaystyle \sum_{r=2}^{\infty} r^2 \left \lfloor \frac{n}{r^2} \right \rfloor

For that we can use the harmonic lemma.

Given an r, we calculate the largest r' in O(1) such that \displaystyle \left \lfloor \frac{n}{r^2} \right \rfloor = \left \lfloor \frac{n}{r'^2} \right \rfloor and add

\displaystyle ((r)^2 + (r+1)^2 + \ldots (r')^2) \left \lfloor \frac{n}{r^2} \right \rfloor to our answer.

Also, we know that \left \lfloor \frac{n}{(r' + 1)^2} \right \rfloor \neq \left \lfloor \frac{n}{r'^2} \right \rfloor.

In this way, we can iterate over all distinct values of \left \lfloor \frac{n}{r^2} \right \rfloor

We can prove that there are only O(n^{\frac{1}{3}}) distinct values of \left \lfloor \frac{n}{r^2} \right \rfloor. (Proof by AC)

You have to do things for perfect squares:

- for perfect squares till 10^12, you can simply find the contribution by iterating 10^6 times.
- For perfect squares from 10^12 to 10^18, you cannot simply iterate because it will take 10^9-10^8 iterations. Therefore, you have to find the sum of perfect squares appearing k times.

I have discussed this problem in detail in this video : PPDIV

We can prove that there are only O(n^{\frac{1}{3}}) distinct values of \left \lfloor \frac{n}{r^2} \right \rfloor. (Proof by AC)

You don’t have to Proof by AC! :)) You can go over all possible values of \left \lfloor \frac{10^{18}}{r^2} \right \rfloor *offline* (I left my computer running for several minutes) and actually count the number of distinct values. You’ll get a value that’s approximately 10^6 ish, which means you should be fine.

you can refer the solution here :