30 ms·
10^12 < 256^5 ≈ 1.1e+12, which isn't too bad. You could also use 10^118 < 256^49, which wastes less but is in bignum land. But don't you want 10^x to be slight
by josh11b 2y ago
10^12 < 256^5 ≈ 1.1e+12, which isn't too bad. You could also use 10^118 < 256^49, which wastes less but is in bignum land.
But don't you want 10^x to be slightly bigger than 256^y, so you could represent all length-y byte sequences in x-digit number? In this direction, there's 10^53 > 256^22, but that is still in bignum land.