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 $$$(a;b)$$$:

const int maxn = 1e3 + 42;

int S[4][maxn][maxn];

void add_point(int z, int x, int Y, int c) {
    for(; x < maxn; x += x & -x) {
        for(int y = Y; y < maxn; y += y & -y) {
            S[z][x][y] += c;
        }
    }
}

// add c on [a, inf) x [b, inf)
void add_suffix(int a, int b, int c) {
    add_point(0, a, b, c);
    add_point(1, a, b, -c * (a - 1));
    add_point(2, a, b, -c * (b - 1));
    add_point(3, a, b, c * (a - 1) * (b - 1));
}

// add c on [x1, x2) x [y1, y2)
void add_range(int x1, int y1, int x2, int y2, int c) {
    add_suffix(x1, y1, c);
    add_suffix(x1, y2, -c);
    add_suffix(x2, y1, -c);
    add_suffix(x2, y2, c);
}

int get_point(int z, int x, int Y) {
    int res = 0;
    for(; x > 0; x -= x & -x) {
        for(int y = Y; y > 0; y -= y & -y) {
            res += S[z][x][y];
        }
    }
    return res;
}

// get sum on [0, x] x [0, y]
int get_prefix(int x, int y) {
    return get_point(0, x, y) * x * y 
         + get_point(1, x, y) * y
         + get_point(2, x, y) * x 
         + get_point(3, x, y);
}

// get sum on (x1, x2] x (y1, y2]
int get_range(int x1, int y1, int x2, int y2) {
    return get_prefix(x2, y2)
         - get_prefix(x1, y2) 
         - get_prefix(x2, y1) 
         + get_prefix(x1, y1);
}

The solution generalizes 1-dimensional Fenwick tree range updates idea from Petr blog from 2013.

You may check your solution 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:

1. Iterate over possible bottom left point $$$O$$$ of the convex subset;
2. Ignore points below it and sort points above it by angle that they form with $$$O$$$;
3. 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 $$$O$$$;
4. Sort considered points around $$$B$$$, separately in "yellow" and "green" areas (see picture);
5. Iterate over possible point $$$C$$$ which will succeed $$$B$$$ in the convex subset;
6. Set of points that may precede $$$B$$$ with a next point $$$C$$$ form a contiguous prefix of points before $$$B$$$;
7. The second pointer $$$A$$$ to the end of the prefix is maintained;
8. 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:

1. Add/remove element $$$x$$$ from the set, all prime divisors of $$$x$$$ are known;
2. 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).

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.