Hi everyone!
Here's another collection of little tricks and general ideas that might make your life better (or maybe worse).
Most of them are somewhat well-known, but perhaps would still be useful to someone.
1. Evaluating polynomial modulo small prime $$$p$$$. Given a polynomial $$$q(x)$$$, it may be evaluated in all possible $$$a \in \mathbb Z_p$$$ in $$$O(p \log p)$$$. To do this, compute $$$q(0)$$$ separately and use chirp Z-transform to compute $$$q(g^0), q(g^1), \dots, q(g^{p-2})$$$, where $$$g$$$ is a primitive root modulo $$$p$$$.
This method can be used to solve 1054H - Epic Convolution.
2. Generalized Euler theorem. Let $$$a$$$ be a number, not necessarily co-prime with $$$m$$$, and $$$k > \log_2 m$$$. Then
where $$$\phi(m)$$$ is Euler's totient. This follows from the Chinese remainder theorem, as it trivially holds for $$$m=p^d$$$.
This fact can be used in 906D - Power Tower.
3. Range add/range sum in 2D. Fenwick tree, generally, allows for range sum/point add queries.
Let $$$s_{xy}$$$ be a sum on $$$[1,x] \times [1,y]$$$. If we add $$$c$$$ on $$$[a, +\infty) \times [b, +\infty)$$$, the sum $$$s_{xy}$$$ would change as
for $$$x \geq a$$$ and $$$y \geq b$$$. To track these changes, we may represent $$$s_{xy}$$$ as
which allows us to split the addition of $$$c$$$ on $$$[a,+\infty) \times [b,+\infty)$$$ into additions in $$$(a;b)$$$:
The solution generalizes 1-dimensional Fenwick tree range updates idea from Petr blog from 2013.
You can check your implementation on eolymp — Чипполино.
4. DP on convex subsets. You want to compute something related to convex subsets of a given set of points in 2D space.
You sort points over bottom-left point $$$O$$$, then over point $$$B$$$ and go through all pairs $$$(A, C)$$$ with two pointers
This can be done with dynamic programming, which generally goes as follows:
- Iterate over possible bottom left point $$$O$$$ of the convex subset;
- Ignore points below it and sort points above it by angle that they form with $$$O$$$;
- Iterate over possible point $$$B$$$ to be the "last" in the convex subset. It may only be preceded by a point that was sorted before it and succeeded by a points that was sorted after it when the points were sorted around $$$O$$$;
- Sort considered points around $$$B$$$, separately in "yellow" and "green" areas (see picture);
- Iterate over possible point $$$C$$$ which will succeed $$$B$$$ in the convex subset;
- Set of points that may precede $$$B$$$ with a next point $$$C$$$ form a contiguous prefix of points before $$$B$$$;
- The second pointer $$$A$$$ to the end of the prefix is maintained;
- Eventually, for every $$$B$$$, all valid pairs of $$$A$$$ and $$$C$$$ are iterated with two pointers.
This allows to consider in $$$O(n^3)$$$ all the convex subsets of a given set of points, assuming that sorting around every point $$$B$$$ was computed beforehand in $$$O(n^2 \log n)$$$ and is now used to avoid actual second sorting of points around $$$B$$$.
The method may probably be used to solve AtCoder — ConvexScore.
5. Subset sum on segments. Given $$$a_1, \dots, a_n$$$, answer $$$q$$$ queries. Each query is whether $$$a_l, a_{l+1}, \dots, a_r$$$ has a subset of sum $$$w$$$. This can be done with dynamic programming $$$L[r][w]$$$ being the right-most $$$l$$$ such that $$$a_l, \dots, a_r$$$ has a subset with sum $$$w$$$:
This allows to solve the problem in $$$O(nw + q)$$$.
Unfortunately, I forgot the original problem on which I saw this approach.
6. Data structure with co-primality info. There is a structure that supports following queries:
- Add/remove element $$$x$$$ from the set, all prime divisors of $$$x$$$ are known;
- Count the number of elements in the structure that are co-prime with $$$x$$$.
Without loss of generality, we may assume that the numbers are square-free.
Let $$$w(x)$$$ be the number of distinct prime divisors of $$$x$$$ and $$$N_x$$$ be the amount of numbers divisible by $$$x$$$ in the structure. When $$$x$$$ is added or removed from the structure, you need to update $$$2^{w(x)}$$$ values of $$$N_x$$$. Now, having $$$N_x$$$, how to count numbers co-prime with $$$x$$$?
where $$$\mu(d)$$$ is the Möbius function. This formula essentially uses inclusion-exclusion principle, as $$$N_d$$$ counts numbers divisible by $$$d$$$ and we need to count numbers that are not divisible by any divisor of $$$x$$$.
The method was used in 102354B - Yet Another Convolution.
7. Generalized inclusion-exclusion. Let $$$A_1, \dots, A_n$$$ be some subsets of a larger set $$$S$$$. Let $$$\overline{A_i} = S \setminus A_i$$$.
With the inclusion-exclusion principle, we count the number of points from $$$S$$$ that lie in neither of $$$A_i$$$:
assuming the empty intersection to be the full set $$$S$$$. We may split the formula half-way as
This way, we're able to count the number of points from $$$S$$$ that lie in exactly $$$r$$$ set among $$$A_1, \dots, A_n$$$.
Explanation lies in the fact that for a fixed $$$Y$$$, we may use PIE directly:
then if summing up over all possible $$$Y$$$, each set $$$X$$$ will always have $$$(-1)^{m-r}$$$ coefficient and will occur for $$$\binom{m}{r}$$$ sets $$$Y$$$.
8. Finding roots of polynomials over $$$\mathbb Z_p$$$. You're given $$$q(x)$$$. You want to find all $$$a \in \mathbb Z_p$$$, such that $$$q(a)=0$$$.
This is done in two steps. First, you compute
to get rid of non-linear or repeated linear factors of $$$q(x)$$$, as
Second, you pick random $$$a$$$ and compute
This will filter roots of $$$h(x)$$$ by whether they're quadratic residues if $$$a$$$ is added to them or not.
Quadratic residues make up $$$\frac{p-1}{2}$$$ of numbers in $$$\mathbb Z_p$$$ and are distributed uniformly, so you'll have at least $$$\frac{1}{2}$$$ chance to get non-trivial divisor of $$$h(x)$$$. This is particularly useful when you want to solve e.g. $$$x^2 \equiv a \pmod p$$$, which can be done in $$$O(\log p)$$$ with this algorithm:
Generally, the probability of getting a divisor of $$$h(x)$$$ of degree $$$k$$$ for $$$\deg h = n$$$ can be expressed as $$$2^{-n}\binom{n}{k}$$$, thus on average this method nearly halves the degree of $$$h(x)$$$ in a single iteration. From this follows that the expected complexity of the algorithm is $$$O(n^2 \log p)$$$ if naive multiplication is used or $$$O(n \log^2 n \log np)$$$ if one uses FFT-based multiplication and half-GCD.
The method is called Berlekamp–Rabin algorithm and can be generalized to find all factors of $$$q(x)$$$ over $$$\mathbb Z_p$$$ (see this comment).
You can check your implementation on Library Judge — Sqrt Mod.
9. Matching divisible by $$$m$$$. You're given a weighted bipartite graph and you need to check if there exists a perfect matching that sums up to the number that is divisible by $$$m$$$. In other words, whether there exists a permutation $$$\sigma_1, \dots, \sigma_n$$$ such that
For this, we introduce matrices $$$R^{(0)}, \dots, R^{(m-1)}$$$ such that
where $$$A_{ij}$$$ is weight between $$$i$$$ in the first part and $$$j$$$ in the second part, $$$x_{ij}$$$ is a random number when there is an edge between $$$i$$$ and $$$j$$$ or $$$0$$$ otherwise, and $$$\omega$$$ is a root of unity of degree $$$m$$$. The determinants of such matrices is then
where $$$N(\sigma)$$$ is a parity of $$$\sigma$$$. If you sum them up, you get
But at the same time,
Thus, a summand near $$$\sigma_1, \dots, \sigma_n$$$ will be non-zero only if $$$A_{1\sigma_1} + \dots + A_{n \sigma_n}$$$ sums up to the number divisible by $$$m$$$.
Therefore, the property can be checked in $$$O(mn^3)$$$.
The method was used in CSAcademy — Divisible Matching.
10. Eigenvectors of circulant matrix. Let $$$A$$$ be a matrix such that each of its rows is a cyclic shift of the previous one (see circulant matrix). Let the first column be $$$a_0, \dots, a_{n-1}$$$ and $$$A(x) = a_0 + a_1 x + \dots + a_{n-1} x^{n-1}$$$. Then the eigenvalues of $$$A$$$ are
where $$$\omega$$$ is an $$$n$$$-th root of unity. Correspondingly, $$$k$$$-th eigenvector is of form
In particular it means that the determinant of such matrix is
and multiplication by its inverse may be found with pointwise division after DFT of degree $$$n$$$.
These facts may be used to solve 102129G - Permutant and 901E - Cyclic Cipher.
11. Knapsack with repetitions. You have $$$n$$$ item types, there are $$$a_i$$$ items of type $$$i$$$, having weight $$$b_i$$$ and cost $$$c_i$$$. What is the maximum cost you may get with having total weight at most $$$w$$$? This is solvable in $$$O(nw)$$$. The transition formula looks like
To compute it quickly enough, you should divide $$$d[i-1]$$$ into groups having the same remainder modulo $$$b_i$$$, after which the maximum is taken on contiguous segments of the same width rather than with steps of $$$b_i$$$, and can be computed with monotonic queue.
12. Reverses and palindromes. Given strings $$$S$$$ and $$$T$$$, is it possible to reverse some non-intersecting substrings of $$$S$$$ to obtain $$$T$$$?
In other words, we need to check if $$$S$$$ may be represented as
such that
where $$$B^\top$$$ is reversed string $$$B$$$. To check this, one may use operation $$$\operatorname{mix}(S, T)$$$ such that
Key fact here is that $$$\operatorname{mix}(A, A^\top)$$$ gives a palindrome of even length and is invertible operation.
Correspondingly, $$$\operatorname{mix}(A, A)$$$ may be perceived as a concatenation of $$$|A|$$$ palindromes of length $$$2$$$.
That being said, checking that $$$T$$$ is obtained from $$$S$$$ by reversing some of its substrings is equivalent to checking whether $$$\operatorname{mix}(S, T)$$$ can be split in palindromes of even length, which is doable in $$$O(n \log n)$$$ with palindromic tree.
This method was used in 906E - Reverses.