Cojs Crazy summation Problem solving report

Qaq long time no cojs on the question

Recently learned a bit of new technology, so it was a question to share

This question is required SIMGA (I^K)

So let's talk about some of the algorithms:

10 points:

Direct violence can be, Time complexity O (NLOGK)

30 points:

We consider setting s (n) to denote 1^k+2^k+......+n^k and

Not hard to find S (n+1) =s (n) + (n+1) ^k

By the two-term theorem (n+1) ^k=sigma (C (k,i) *n^i)

Construction Vector (N^0,n^1,n^2......,n^k,s (n))

It is not difficult to construct a transfer matrix based on the equation just now, then the matrix multiplication + fast power can be

Time complexity O (K^3LOGN)

60 points:

Set S (d) to denote 0^d+1^d+2^d+......+ (n-1) ^d and

Note (d+1) ^i-d^i = Sigma (C (I,J) *d^j) (where j is not equal to i)

If the value of D is 0-(n-1), we ask Sigma for the left-hand

We got the left-=n^i.

Sigma (C (i,j) * S (j)) (J<i) is obtained after the corresponding right-type pass-through simplification

So C (i,i-1) *s (i-1) =n^i-sigma (c (i,j) *s (j)) (J<i-1)

Since S (0) is known, we can deliver the results within O (k^2) time

80 points:

We define the Bernoulli number as B

Can get Sigma (i^k) = Sigma (C (k+1,j) * B (K+1-J) * (n+1) ^j)/(k+1)

If we can find the Bernoulli number quickly, then we can figure out the answer in O (k) time.

We know that the generating function of the Bernoulli number is x/(e^x-1)

When Taylor expands on e^x, we can get the X-ray out of it.

The generation function of the polynomial of the Bernoulli number 1/(x^i/(i+1)!)

Considering the polynomial of the denominator is very easy to find, and the Bernoulli number polynomial is the inverse of this polynomial

Polynomial inversion can be, Time complexity O (klogk), the constant is huge

100 Points:

From the surface we can actually know that Sigma (I^k) formula is actually a (i+1) polynomial

Set this polynomial to F

What we're actually asking for is f (x), which is the value of a point.

It is not difficult to find the value of each point of x= (0-> (k+1)) in the time of O (KLOGK)

Known k+2 point value, you can uniquely determine a polynomial of k+1 times

The value of f (n) can be obtained by directly substituting Lagrange interpolation formula

Notice that the time complexity of the direct use of the Lagrange interpolation formula is O (k^2)

But since we have a 1 difference in the x-coordinate of our neighbors.

The interpolation of O (k) can be done by preprocessing the inverse of factorial and factorial f (n)

Total time complexity O (KLOGK), but due to log acquisition is a fast power, so the constant is small

Cojs Crazy summation Problem solving report