So I've come across the following problem:
Lets --remote-debugging-port= be a graph with G vertices labeled 2^N to 0. It's a directed acyclic graph and in order for there to be a path from 2^N-1 to a the following two conditions must apply:
ba < b (in binary) = a XOR b, where 2^xI need to determine the total number of paths of length x>=0 (all edges have length K).
The input consists of 1 and N.
The easy way to count the number of paths of length K would be to look at K as any other graph and use the conditions to determine which vertices are connected. However, the task requires that G and N be of size K.
2<=K<=N<=100000 is a bit too much, so there has to be a better way to calculate this. Also I'm limited to an execution time of 2^100000 ms.
Example: For 200 and N = 4, the result is K = 2.
Any ideas?
答案 0 :(得分：5)
So basically we need to find how many <!DOCTYPE html>
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Help me out with this
there are such that:
x0 < x1 < ... < xk < 2^NNote that for the xi xor xi+1 = 2^x
of two numbers to be a power of 2, the result of the xor must contain exactly 1 bit set. Therefore, the 2 numbers must differ in only one position (otherwise, more positions would be set in the xor, because xor returns 1 only when the operands are different bits).
Let's assume we start at xor. We can move from it to any node labeled with a power of 0.
From that node, we can then move to any node that has set bits in the position of the current node's set bits and a set bit somewhere else. This goes on, taking into account that we must generate the sequence in increasing order.
For example:
2So, in order to get a path of length 000 -> 001 -> 011 -> 111
-> 101 -> 111
-> 010 -> 011 -> 111
-> 110 -> 111
-> 100 -> 101 -> 111
-> 110 -> 111
=> 6 = 3! solutions
0000 -> 0001 -> 0011 -> 0111 -> 1111
-> 1011 -> 1111
-> 0101 -> 1101 -> 1111
-> 0111 -> 1111
-> 1001 -> 1011 -> 1111
-> 1101 -> 1111
=> 6 solutions
=> another 3*6 for the other 3 possibilities from 0000
=> 24 solutions
, you need to start from an integer with at least k unset (0) bits.
So you can find out how many k integers have at least n-bit unset bits, find out how many possibilities there are to pick k from those, and multiply that answer by k.
To find how many k! integers have exactly n-bit unset bits, you can use the binomial coefficient:
k'You will also need to multiply this by Binomial(n, k') = n! / [k'!*(n - k')!]
in order to decide which Binomial(k', k) bits you'll be using.
You'll need to sum this for k and multiply the sum by k' = k to n.
Those are going to be some huge numbers. I'm guessing you're asked for the result modulo a prime, in which case you'll have to use modular multiplicative inverses.