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# WRDSUM - Editorial

Author: Alexey Zayakin
Tester: Istvan Nagy
Editorialist: Xellos

Hard

# PREREQUISITES:

Multiplication of big integers, polynomial interpolation, modular inverses

# PROBLEM:

The function $F(x)$ returns the smallest integer root of $x$. You're given a big integer $N$ and are supposed to find the sum $F(2)+F(3)+\dots+F(N)$.

# SHORT EXPLANATION:

Compute all roots of $N$ using fast exponentiation, bruteforce multiplication and binary search. The answer can be found using a DP over roots of $N$, where you often need to compute sums of the form $\sum_{x=2}^a k^r$. Use polynomial interpolation to get those sums.

# EXPLANATION:

The first thing to notice is that $F(x)$ can be described much more simply: it's just the $g$-th root of $x$ for the maximum possible $g$ that gives a root that's also an integer.

The other thing is that $N$ is YUGE. That means we can't find the roots using standard methods like factorisation or taking the result in doubles and correcting it if necessary. Now's not the time for that, though - roots come later. Let's just estimate how many of them need to be computed: if $2^g \le N\approx 10^n$, then $g \le n\frac{\log{10}}{\log{2}} \approx 6700$.

Most integers aren't powers of other integers. For such numbers, $F(x)=x$; if we used this as an "estimate" of the answer, we'd get $N(N+1)/2-1$. From that, we need to subtract $x-F(x)$ for all relevant $x$. We subtract them for square $x$, i.e. for $x=y^2$ (and all $y$ from the range $[2,\mathrm{floor}(\sqrt{N})]$), which means that we actually subtract $\sum y^2$ and add the answer for $\mathrm{floor}(\sqrt{N})$; that answer can be computed recursively. We do the same for $x$-s that are powers of 3 (perfect cubes?). Powers of 4... oh wait, a power of 4 is also a power of 2, so we can skip 4.

Here's something that breaks the (very short) pattern. If we tried subtracting $x-F(x)$ for all $x \le N$ that are $r$-th powers and all $r$, then we can subtract it multiple times for the same $x$. The most obvious fix that the situation with $r=4$ suggests is to ignore all $r$ which aren't products of distinct primes, but that's not enough.

Let's take $x$-s that are sixth (but not any higher) powers, so $g=6$. We subtracted $x-F(x)$ for them with $r=2$ and with $r=3$, but we should subtract just once, not twice. If we added for $r=6$ instead of subtracting, the contribution of numbers for which $g=6$ to the answer would be correct.

If we repeat the same idea for larger $r$ and choose what to add/subtract so that the numbers with each $g$ would finally contribute to the answer just the sum of their $F()$ values, we arrive at the inclusion-exclusion formula: if $r$ is a product of an odd number of distinct primes, subtract $x-F(x)$ for all $x=y^r$ and $y \le \mathrm{floor}(\sqrt[r]{N})$; if the number of primes is even, add $x-F(x)$ instead. And if $r$ is not a product of distinct primes, ignore it.

There are two questions arising here apart from the computation of roots: how costly is the recursion and how to compute $\left(\sum y^r\right) \% mod$ for an arbitrary range of $y$ and rather large $r$?

In the first question, we can once again use the fact that $\sqrt[ab]{N}=\sqrt[a]{\sqrt[b]{N}}$ and so $\mathrm{floor}(\sqrt[a]{\mathrm{floor}(\sqrt[b]{N})})=\mathrm{floor}(\sqrt[ab]{N})$; that means we only need to compute the answers for all possible floor(root)s of $N$, even if we may need some of them multiple times. We can store them in an array and compute them just once when necessary, making this into a recursive DP that takes the current root of $N$ as a parameter:

array answer[1..6700] of ?

array roots[2..6700] of N
# compute roots

K = roots[R]

answer[R] = K*(K+1)/2-1 # modulo omitted
for x (x <= 6700, product of distinct primes):
dif = (sum of j**x for 2 <= j <= K)
if (x = product of odd number of primes): answer[R] -= dif



Obvious optimisations such as "only try $xR \le 6700$" apply.

The time complexity of this is given by two things: the complexity of computing the sums of $j^x$ and the number of $x$-s to try (the number of edges in a "recursion tree"). The number of products of distinct primes $\le 6700$ is about 4000, which gives an upper bound of $6700\cdot4000 \approx 27\cdot10^6$, but if we only take $x \le 6700/R$ for all $R$, it drops quickly to around $6700+4000+\frac{4000}{2}+\frac{4000}{3}+\dots \approx 6700+4000\cdot\ln{4000} \approx 40000$ (actually, it's asymptotically around $O(n\log{n})$). That's much better.

The next step is figuring out how to compute $\sum_{j=2}^a j^r$. Good news: this sum is the same for $a$ and $a \% mod$, since $j^r \% mod = (j\% mod)^r$ and $\left(\sum_{j=0}^{mod-1} j^r \right) \% mod = 0$; the second identity follows from the fact that for a prime modulo, all values of $j^r$ in the sum will be distinct, so we can rewrite it as $\sum_{j=0}^{mod-1} j = mod\frac{(mod-1)}{2}$; since $mod$ is odd, this is divisible by it. This also means that we only need to compute the floor(root)s of $N$ modulo $mod$.

Anyway, we know that for $r=1$, the sum is $a(a+1)/2-1$ and we can also find a formula for $r=2$: $\frac{a(a+1)(2a+1)}{6}-1$. For bigger $r$, we can use WolframAlpha; in general, the sum for $r$ turns out to be a polynomial $S_r(a)$ of degree $r+1$. And for polynomials with sufficiently small degrees, we can use Lagrangian interpolation: if the first $r+1$ values of $S_r(N)$ are $S_r(1)=y_1, \dots, S_r(r+1)=y_{r+1}$, then we get the formula

$$S_r(a) = \sum_{j=1}^{r+1}\frac{\prod_{k \neq j} (a-k)}{\prod_{k \neq j} (j-k)}y_j = \prod_k (a-k)\sum_{j=1}^{r+1}\frac{(-1)^{r+1-j}}{(j-1)!(r+1-j)!(a-j)}y_j$$

where $k$ runs from $1$ to $r+1$ except $j$ when necessary). The second expression is only valid for $a > r+1$, but that's not a problem - we've already computed the values for $a \le r+1$, so we can use them directly. Otherwise, we can precompute factorials and use modular inverses to compute $S_r(a)$ from the second expression in $O(r\log{mod})$ time.

Now, let's finally get to computing floor(root)s. We'll do that recursively (starting from $N$ as the "1-st root"), multiplying by prime numbers in non-decreasing order - if $p$ is the greatest prime divisor of $r$ with $r=pq$, then $\mathrm{floor}(\sqrt[r]{N})$ is computed as the $p$-th root from $\mathrm{floor}(\sqrt[q]{N})$. This way, most roots will be computed from numbers with fewer digits, which speeds up the computations.

Multiplying two $d$-digit numbers together is possible in $O(d^2)$; it's also possible faster using fast Fourier / Number Theoretic Transform or Karatsuba's algorithm, but that's not necessary here. Using fast exponentiation, which computes $x^e$ by multiplying $x^{e/2}\cdot x^{e/2}$ for even $e$ or $x^{e-1}\cdot x$ for odd $e$, $x^e$ can be found in time $O((ed)^2\log{e})$ (since $x^{e/2}$ has $O(ed)$ digits).

We can now compute roots using binary search. In big integers, a good implementation is by adding $1$-s in binary representation to the current value of the computed root $x$ of $y$; if $x^e > y$, we keep that $1$ in $x$ and otherwise, we replace it by $0$ again. If $y$ has $d$ digits, we only need to try the last $O(d/e)$ digits in the binary representation of $x$, so the time complexity is $O((d/e)d^2\log{e})=O(d^3\log{e}/e)$; the sum over all prime $e \le n$ is $O(d^3\log^2{n})$ and the sum over all relevant $d$ (which are $n,\frac{n}{2},\frac{n}{3},\dots$) is $O(n^3\log^2{n})$, since $\sum \frac{1}{k^3}$ is convergent.

For $d=2016$, that's way too much. We need to reduce the value of $d$ somehow. The key is that $O(d^2)$ multiplication works as long as all numbers in intermediate computation don't exceed the range of integers, which lets us use any base $b$ such that $b^2d \le 2^{64}$ (approximately). For convenience in binary search, let's look for a large base that's a power of 2. We also know that if a number in base $b_1$ has $d_1$ digits, then it has approximately $d_2=\frac{\log{b_1}}{\log{b_2}}d_1$ digits in base $b_2$; with $b_1=10$ and $d_1=2016$, we can estimate that $b=2^{28}$ gives up to $d=250$ digits and satisfies $b^2d \le 2^{64}$. Therefore, we can try $b=2^{28}$ or a slighly smaller power of 2.

If we convert $N$ into base $b$ and take $N\approx b^n$, then all square roots can be computed in time $O(n^3\log^2{n})$; we only need their remainders modulo $mod$, with which we can do the recursion with computing the sums $S()$ in time $O(n\log{n}\cdot n\log{mod})=O(n^2\log{n}\log{mod})$. Precomputation of small values of those sums runs in $O(n^2\log^2{n})$ as well, so the total time complexity is $O(n^2\log{n}(n\log{n}+\log{mod}))$. With $n\approx 250$ and since the $O(n^3\log^2{n})$ factor is only caused by simple multiplication/addition, this passes quite comfortably (actually, it's hard to estimate the actual running time in any way other than by submitting).

A final note: probably everyone who solved this noticed that the modulo is a prime that's commonly used in Number Theoretic Transform. That, however, is not necessary for multiplying 250-digit numbers (and in fact, we need smaller $b$ if we want to use it), so it's just there for misdirecting people. Gr8 b8 m8.

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question asked: 29 Feb '16, 22:03

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last updated: 01 Mar '16, 09:20