https://en.wikipedia.org/wiki/Row_echelon_form#Reduced_row_echelon_form

Let n = deg(A), m = deg(B). Treat deg(0) = 0. a[n] != 0, b[m] != 0 R(A, B) := b[m]^n prod_{j=0..m-1} A(b_roots[j]) // counted with multiplicities = (-

Let n = deg(A), m = deg(B). Treat deg(0) = 0. a[n] != 0, b[m] != 0 R(A, B) := b[m]^n prod_{j=0..m-1} A(b_roots[j]) // counted with multiplicities = (-

reverse all the values from qL to qR

intersection between segment ab and line cd 0 if do not intersect, 1 if proper intersect, 2 if segment intersect

intersection point between segment ab and segment cd assuming unique intersection exists

Generate all primes from l to r using segmented sieve in O((r - l) log (r) + sqrt(r))

Sets the kth bit of x to 1 and returns the result

Generate all primes up to limit using sieve of eratosthenes

polynomial representation of f(p)

credit: min_25 takes 0.5s for n = 1e9

use bitset<N> is_prime; to have O(N/64) memory complexity using bitset you can solve upto around N = 10^8 in 1s

number of solutions of the equation x0 xor x1 xor ... x(n - 1) = x s.t. 0 <= xi <= ai O(n * log(MAX))

number of solutions of the equation x0 xor x1 xor ... x(n - 1) = x s.t. 0 <= xi <= ai O(n * log(MAX))

Number of Perfect Matchings of a general graph SQRT of the Permanent of a zero diagonal symmetric matrix Takes ~2s for (38 * 38) matrix

sum of f(i) for 1 <= i <= n pol is the polynomial representation of f(p)

\sum{x=0}^{n}{x^k1 * ((ax + b)/c)^k2} //floor division 0^0 = 1 it solves for all _k1, _k2 s.t. _k1 + _k2 <= k1 + k2 (a * n + b) should fit in long lon

\sum{x=0}^{n}{x^k1 * ((ax + b)/c)^k2} //floor division 0^0 = 1 it solves for all _k1, _k2 s.t. _k1 + _k2 <= k1 + k2 (a * n + b) should fit in long lon

Computes sum_{k=0}^{n-1} x^{(a*k + b) % m} % mod

credit: Min_25

Number of Perfect Matchings of a general graph SQRT of the Permanent of a zero diagonal symmetric matrix Takes ~2s for (38 * 38) matrix

returns {maxflow, mincost}

returns edge chromatic number, ans contains the edge coloring(colors are 1 indexed) supports multiple edges 0 indexed, O(M log M)

returns stirling1st(n, i) for 0 <= i <= n

stirling2nd(i, k) for 0 <= i <= n O(n log^2 n)

returns stirling1st(i, k) for 0 <= i < n O(n log^2 n)

find occurrences of t in s where '?'s are automatically matched with any character res[i + m - 1] = sum_j=0 to m - 1_{s[i + j] * t[j] * (s[i + j] - t[

if sum of the array values = x, then complexity is O(x sqrt(x)) returns the subset ids

if sum of the array values = x, then complexity is O(x sqrt(x))

f*g(s)=sum_{s' $ s} {F(s')*G(s\s')} O(B * B * 2 ^ B)

sum of numbers in [l ,r] less than or equal to k

credit: zimpha compute \sum_{i=1}^{n} sigma0(i) in ~O(n^{1/3}) time. it is also equal to \sum_{i=1}^{n} floor(n / i) takes ~100 ms for n = 1e18

(ccw, cw) tangents from a point that is outside this convex polygon returns indexes of the points ccw means the tangent from Q to that point is in the

Toggles the kth bit of x and returns the result