Here's a summary of Dimitris Andreou's link.

Remember sum of i-th powers, where i=1,2,..,k. This reduces the problem to solving the system of equations

a_{1} + a_{2} + ... + a_{k} = b_{1}

a_{1}^{2} + a_{2}^{2} + ... + a_{k}^{2} = b_{2}

...

a_{1}^{k} + a_{2}^{k} + ... + a_{k}^{k} = b_{k}

Using Newton's identities, knowing b_{i} allows to compute

c_{1} = a_{1} + a_{2} + ... a_{k}

c_{2} = a_{1}a_{2} + a_{1}a_{3} + ... + a_{k-1}a_{k}

...

c_{k} = a_{1}a_{2} ... a_{k}

If you expand the polynomial (x-a_{1})...(x-a_{k}) the coefficients will be exactly c_{1}, ..., c_{k} - see Viète's formulas. Since every polynomial factors uniquely (ring of polynomials is an Euclidean domain), this means a_{i} are uniquely determined, up to permutation.

This ends a proof that remembering powers is enough to recover the numbers. For constant k, this is a good approach.

However, when k is varying, the direct approach of computing c_{1},...,c_{k} is prohibitely expensive, since e.g. c_{k} is the product of all missing numbers, magnitude n!/(n-k)!. To overcome this, perform computations in Z_{q} field, where q is a prime such that n <= q < 2n - it exists by Bertrand's postulate. The proof doesn't need to be changed, since the formulas still hold, and factorization of polynomials is still unique. You also need an algorithm for factorization over finite fields, for example the one by Berlekamp or Cantor-Zassenhaus.

High level pseudocode for constant k:

- Compute i-th powers of given numbers
- Subtract to get sums of i-th powers of unknown numbers. Call the sums b
_{i}.
- Use Newton's identities to compute coefficients from b
_{i}; call them c_{i}. Basically, c_{1} = b_{1}; c_{2} = (c_{1}b_{1} - b_{2})/2; see Wikipedia for exact formulas
- Factor the polynomial x
^{k}-c_{1}x^{k-1} + ... + c_{k}.
- The roots of the polynomial are the needed numbers a
_{1}, ..., a_{k}.

For varying k, find a prime n <= q < 2n using e.g. Miller-Rabin, and perform the steps with all numbers reduced modulo q.

EDIT: The previous version of this answer stated that instead of Z_{q}, where q is prime, it is possible to use a finite field of characteristic 2 (q=2^(log n)). This is not the case, since Newton's formulas require division by numbers up to k.

## Best Solution

So here is O(n log n) solution in java.

This is almost normal merge sort, the whole magic is hidden in merge function. Note that while sorting algorithm remove inversions. While merging algorithm counts number of removed inversions (sorted out one might say).

The only moment when inversions are removed is when algorithm takes element from the right side of an array and merge it to the main array. The number of inversions removed by this operation is the number of elements left from the the left array to be merged. :)

Hope it's explanatory enough.