Calculating 1^X + 2^X + … + N^X mod 1000000007

问题:

Is there any algorithm to calculate (1^x + 2^x + 3^x + ... + n^x) mod 1000000007?
Note: a^b is the b-th power of a.

The constraints are 1 <= n <= 10^16, 1 <= x <= 1000. So the value of N is very large.

I can only solve for O(m log m) if m = 1000000007. It is very slow because the time limit is 2 secs.

Do you have any efficient algorithm?

There was a comment that it might be duplicate of this question, but it is definitely different.

回答1:

You can sum up the series

1**X + 2**X + ... + N**X

with the help of Faulhaber's formula and you'll get a polynomial with an X + 1 power to compute for arbitrary N.

If you don't want to compute Bernoulli Numbers, you can find the the polynomial by solving X + 2 linear equations (for N = 1, N = 2, N = 3, ..., N = X + 2) which is a slower method but easier to implement.

Let's have an example for X = 2. In this case we have an X + 1 = 3 order polynomial:

    A*N**3 + B*N**2 + C*N + D

The linear equations are

      A +    B +   C + D = 1              =  1 
    A*8 +  B*4 + C*2 + D = 1 + 4          =  5
   A*27 +  B*9 + C*3 + D = 1 + 4 + 9      = 14
   A*64 + B*16 + C*4 + D = 1 + 4 + 9 + 16 = 30 

Having solved the equations we'll get

  A = 1/3
  B = 1/2
  C = 1/6
  D =   0 

The final formula is

  1**2 + 2**2 + ... + N**2 == N**3 / 3 + N**2 / 2 + N / 6

Now, all you have to do is to put an arbitrary large N into the formula. So far the algorithm has O(X**2) complexity (since it doesn't depend on N).

回答2:

There are a few ways of speeding up modular exponentiation. From here on, I will use ** to denote "exponentiate" and % to denote "modulus".

First a few observations. It is always the case that (a * b) % m is ((a % m) * (b % m)) % m. It is also always the case that a ** n is the same as (a ** floor(n / 2)) * (a ** (n - floor(n/2)). This means that for an exponent <= 1000, we can always complete the exponentiation in at most 20 multiplications (and 21 mods).

We can also skip quite a few calculations, since (a ** b) % m is the same as ((a % m) ** b) % m and if m is significantly lower than n, we simply have multiple repeating sums, with a "tail" of a partial repeat.

回答3:

I think Vatine’s answer is probably the way to go, but I already typed this up and it may be useful, for this or for someone else’s similar problem.

I don’t have time this morning for a detailed answer, but consider this. 1^2 + 2^2 + 3^2 + ... + n^2 would take O(n) steps to compute directly. However, it’s equivalent to (n(n+1)(2n+1)/6), which can be computed in O(1) time. A similar equivalence exists for any higher integral power x.

There may be a general solution to such problems; I don’t know of one, and Wolfram Alpha doesn’t seem to know of one either. However, in general the equivalent expression is of degree x+1, and can be worked out by computing some sample values and solving a set of linear equations for the coefficients.

However, this would be difficult for large x, such as 1000 as in your problem, and probably could not be done within 2 seconds.

Perhaps someone who knows more math than I do can turn this into a workable solution?

Edit: Whoops, I see Fabian Pijcke had already posted a useful link about Faulhaber's formula before I posted.