B-Suffix Array

• Let C_i = min_{j > i and s_j = s_i} {j - i}
• The B-Suffix Array is equivalent to the suffix array of C_1 C_2 ... C_n
• Detailed proof can be found in “Parameterized Suffix Arrays for Binary Strings • ” http://www.stringology.org/event/2008/p08.html

Infinite Tree

• First, compute the “virual tree” of {1!, 2!, ..., n!}
• Second, to compute the actual cost, use Segment Tree or Fenwick Tree.
• O(m log^2 m)

Domino

• See “Distances in Domino Flip Graphs”

Quadratic Form

• The answer is b^T A^{-1} b, which can be proved by Lagrange Duality.
• All we need is to compute the inverse matrix of the matrix A.

Counting Spanning Trees

• The number of spanning trees is prod_{i >= 2} deg(x_i) deg(y_i)
• Detailed proof can be found in “Enumerative properties of Ferrers graphs”
https://arxiv.org/pdf/0706.2918.pdf

Infinite String Comparision

• Compare the string a^{infty} and b^{infty} directly
• By the Periodicity Lemma, if there is no mismatches in the first a + b - gcd(a, b) characters, the two string are identical

BaXianGuoHai, GeXianShenTong

• For simplicity, we denote the multiplication as +, and exponentiation as *
• Precompute B_{i, j} = 2^ {W * j} * v_i
• To compute sum_{i, j} (sum e'{i, j} * 2^{W * j}) v_i
• = sum_x x sum{i, j} [e'{i, j} = x] B{i, j} = sum_x x Q_x
• To compute sum_x x Q_x • = sum_x (sum_{y >= x} Q_y)
• The overall complexity is O(nm / W + 2^W) • Taking W = 16 yields a fast enough solution

Minimum-cost Flow

• We denote the cost in a network with capacity c and flow f as cost(c, f).
• cost(c, 1) = cost(c * 1/c, 1 / c) * c = cost(1, 1 / c) * c
• For a network with unitary capacity, its cost grows linearly with the flow f, with at most O(m) pieces.
• Thus, we can compute O(m) pieces first, and query in O(log m) time.

1 or 2

• For an edge e=(x, y) where d_x = d_y = 2, add the following edges:
• (x, e) (x', e)
• (y, e') (y', e')
• (e, e')
• The problems is turned into to find a perfect matching in a general graph, which can be solved with Edmond's Algorithm.

Easy Integration

• The value is (n!)^2 / (2n+1)!
• Detailed proof can be found in “Wallis' integrals”.
https://en.wikipedia.org/wiki/Wallis%27_integrals

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