Hi,
How can I find the sum of digits in a factorial of a number N, where N can be in range [1, 2000]? Can it be done without resorting to BigNum libraries?
# | User | Rating |
---|---|---|
1 | jiangly | 3976 |
2 | tourist | 3815 |
3 | jqdai0815 | 3682 |
4 | ksun48 | 3614 |
5 | orzdevinwang | 3526 |
6 | ecnerwala | 3514 |
7 | Benq | 3482 |
8 | hos.lyric | 3382 |
9 | gamegame | 3374 |
10 | heuristica | 3357 |
# | User | Contrib. |
---|---|---|
1 | cry | 169 |
2 | -is-this-fft- | 166 |
3 | Um_nik | 161 |
3 | atcoder_official | 161 |
5 | djm03178 | 157 |
6 | Dominater069 | 156 |
7 | adamant | 154 |
8 | luogu_official | 152 |
9 | awoo | 151 |
10 | TheScrasse | 147 |
Name |
---|
Think of logarithms.
Can you please elaborate. I am looking for a Log(N) complexity. The algorithm given below by sbakic uses string multiplication, which won't be Log(N).
-deleted-
Why did you delete your answer? I think it can actually work: if we know the number D of digits of the factorial F = N! and its log L = log(F), we can do some binary search in the factorial digits since log(x) is unique. For example, to find the most significant digit of the factorial we will try every 0 ≤ i ≤ 9; when we know that the most significant digit is i - 1. After that, we store the "current log" and do it again for the second most significant digit and for the third and so on... Overall complexity would be .
This is the algorithm for factorial of a number. Next step is trivial.
You're using a bignum library...
Can you give problem link, please?