Hi everyone!

Here's another collection of little tricks and general ideas that might make your life better (or maybe worse).

Evaluating polynomial modulo small prime $$$p$$$. Given a polynomial $$$q(x)$$$, you may evaluate it modulo $$$p$$$ in all possible arguments. 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 $$$p$$$.

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 $$$m=p^d$$$.

Range add/range sum on a plane. 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 range updates idea from Petr blog from 2013.

DP on convex subsets.

Assume you want to compute something related to convex subsets of a given set of points in 2D space.

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 $$$B$$$.

Knapsack on segments. You're given $$$a_1, \dots, a_n$$$ and need to answer $$$q$$$ queries. Each query is whether $$$a_l, a_{l+1}, \dots, a_r$$$ has a subset 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 $$$w$$$: