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2020 May 30
V
Если у нас есть N вариантов, и мы отмели N-1, то последний можно и не проверять
Ты переучился...
MB
Ты переучился...
уже и пошутить нельзя
2020 May 31
A
u ^ c1 = c2
u ^ c3 = c4
c1,c2,c3,c4 are known values, how to find u(which is unknown) ?
PS: ^ is AND
u ^ c3 = c4
c1,c2,c3,c4 are known values, how to find u(which is unknown) ?
PS: ^ is AND
V🇺
u = c1 ^ c2 = c3 ^ c4
DK
Ajay
u ^ c1 = c2
u ^ c3 = c4
c1,c2,c3,c4 are known values, how to find u(which is unknown) ?
PS: ^ is AND
u ^ c3 = c4
c1,c2,c3,c4 are known values, how to find u(which is unknown) ?
PS: ^ is AND
^ is XOR or POW?
e
^ is XOR or POW?
Ксор, очевидно
A
^ is XOR or POW?
those are AND operations
f
Ajay
u ^ c1 = c2
u ^ c3 = c4
c1,c2,c3,c4 are known values, how to find u(which is unknown) ?
PS: ^ is AND
u ^ c3 = c4
c1,c2,c3,c4 are known values, how to find u(which is unknown) ?
PS: ^ is AND
and is &
MB
Ajay
those are AND operations
You should solve problem for each bit separately.
MB
For each bit, you have 2 rules.
Rule 1.
Exists equation
X & 1 = 0.
Then X = 0.
Rule 2.
Exists equation
X & 1 = 1.
Then X = 1.
Rule 1.
Exists equation
X & 1 = 0.
Then X = 0.
Rule 2.
Exists equation
X & 1 = 1.
Then X = 1.
MB
If both rules applied at least once, than no such U exists.
If only one rule applied, bit value is restored.
Otherwise, both 0 and 1 suffice as bit value.
If only one rule applied, bit value is restored.
Otherwise, both 0 and 1 suffice as bit value.
A
Is there a efficient way of iterating over all the factors pairs of a number (more efficient than sqrt(n))?
DK
this is cbrt(n)
A
this is cbrt(n)
which?
DK
Algo for factorization in time O(N^(1/4)) is well-known
A
does that have a name?
DK
Then you have cbrt(n) pairs of divisors and you can iterate over all of them
DK
Ajay
does that have a name?
Yes, you can find it by first link in google using your initial question
mq
а когда на аткодере обычно рейтинги обновляют?