Sum of digits of a factorial

Question

Link to the original problem

It\'s not a homework question. I just thought that someone might know a real solution to this problem.

I was on

Answer 1

Assume you have big numbers (this is the least of your problems, assuming that N is really big, and not 10000), and let's continue from there.

The trick below is to factor N! by factoring all n<=N, and then compute the powers of the factors.

Have a vector of counters; one counter for each prime number up to N; set them to 0. For each n<= N, factor n and increase the counters of prime factors accordingly (factor smartly: start with the small prime numbers, construct the prime numbers while factoring, and remember that division by 2 is shift). Subtract the counter of 5 from the counter of 2, and make the counter of 5 zero (nobody cares about factors of 10 here).

compute all the prime number up to N, run the following loop

for (j = 0; j< last_prime; ++j) {
  count[j] = 0;
  for (i = N/ primes[j]; i; i /= primes[j])
    count[j] += i; 
}

Note that in the previous block we only used (very) small numbers.

For each prime factor P you have to compute P to the power of the appropriate counter, that takes log(counter) time using iterative squaring; now you have to multiply all these powers of prime numbers.

All in all you have about N log(N) operations on small numbers (log N prime factors), and Log N Log(Log N) operations on big numbers.

and after the improvement in the edit, only N operations on small numbers.

HTH