def I(n):
terms = []
for (p,k) in factor(n):
if p % 2 == 0 and k>1:
residues = (1, -1, (1<<k-1)-1, (1<<k-1)+1)
else:
residues = (1, -1)
pk = p**k
coef = n // pk * modInverse(n // pk, pk)
terms.append(map(lambda x: x*coef, residues))
return max(filter(lambda x: x != n-1, (sum(res) % n for res in cross(*terms))))
def solve():
piter = allprimes()
next(piter)
next(piter)
p1 = next(piter)
p2 = next(piter)
nextMag = 10
accum = 0
while p1 <= 1000000:
while nextMag < p1:
nextMag *= 10
accum += p1 + nextMag * ((p2 - p1) * modInverse(nextMag, p2) % p2)
(p1, p2) = (p2, next(piter))
return accum
def solve():
reverseOps = {
"D": Polynomial(0, 3),
"U": Polynomial(Fraction(-1, 2), Fraction(3, 4)),
"d": Polynomial(Fraction(1, 2), Fraction(3, 2)),
}
seq = "UDDDUdddDDUDDddDdDddDDUDDdUUDd"
poly = reduce(lambda accum, op: reverseOps[op](accum), seq[::-1], Polynomial.X)
# The resulting polynomial is of the form $(a/b)x + c/b$. (Note that the
# denominators are equal.) We then need to find the smallest $z > 10^{15}$
# that is in the range of this polynomial, i.e., the smallest $z > 10^{15}$
# such that its preimage $(z - c/b)(b/a) = (zb - c)/a$ is an integer, i.e.,
# such that $a$ divides $zb - c$, i.e., such that $z$ is congruent to
# $cb^{-1}$ _modulo_ $a$.
a = poly.coef(1).numerator
b = poly.coef(1).denominator
c = poly.coef(0).numerator
rem = (c * modInverse(b, a)) % a
return (10 ** 15 // a + (10 ** 15 % a > rem)) * a + rem
def S(p):
inv2 = modInverse(-2, p)
inv3 = modInverse(-3, p)
inv4 = modInverse(-4, p)
return (inv2 * (1 + inv3 * (1 + inv4))) % p