if the edge is in a cycle

G is the condensed graph

baby step - giant step returns minimum integer x such that a^x = b (mod m) a and m are co-prime

discrete log but a and m may not be co-prime

returns any x such that x ^ k = a (mod p), p is a prime 0^0 = 1 mod p Complexity: O(p ^ (1 / 4))

distance from point p to plane

divide by x^e - 1

returns k s.t. fib(i * k) (for i >= 0) are the only fib numbers which are divisible by n

erase from qL to qR indexes so all previous indexes from qR+1 to last will be left shifted qR-qL+1 times

poly c = poly a * poly b add a[i] = x and b[i] = y and it will return c[i] online!

ax + by = __gcd(a, b) returns __gcd(a, b)

ax + by = __gcd(a, b) returns __gcd(a, b)

given are a[i], b[i], p and q: sum(x[i] * a[i]) = p --- (i) sum(x[i] * b[i]) = q --- (ii) x[i]s can only be integers. does a solution exist? **

ax + by = __gcd(a, b) returns __gcd(a, b)

p(x) = \sum{a[i] * C(x, i)}, where C(x, i) = x * (x - 1) * ... * (x - i + 1) / i! given p(0), p(1), ..., p(n) of an n degree poly returns {a[0], a[1],

returns the smallest non-negative integer x s.t l <= a * x mod p <= r IMPORTANT : 0 <= a < p, 0 <= l <= r < p, p is not necessarily prime Complexity:

number of solutions to x^2 = 1 mod p^k

sum{i=0}^{n-1} i^k = 1/(k+1) \sum{i=0}^{k} binom(k+1, i)B[i] n^{k + 1 - i} ans[i] = sum_{p = 0}^{t}{p^i} it assumes 0^0 = 1

returns the number of solutions where x is in the range[minx, maxx] and y is in the range[miny, maxy]

a*x+b*y=c. returns valid x and y if possible. all solutions are of the form (x0 + k * b / g, y0 - k * b / g)

a * x + b * y = c. returns valid x and y if possible.

occurrences of s[p, ..., p + len - 1]

root of the rep. tree containing this node

\sum_{i = 0}^{n - 1}{(a + d * i) / m}, O(log m)

g(x) = x

O(a.size() * b.size()) if gcd.size() - 1 = number of common roots between a and b