What are some efficient ways to divide 'a' by 'b' where the range is :
-10^300 <= a,b <= +10^300
# | User | Rating |
---|---|---|
1 | tourist | 3993 |
2 | jiangly | 3743 |
3 | orzdevinwang | 3707 |
4 | Radewoosh | 3627 |
5 | jqdai0815 | 3620 |
6 | Benq | 3564 |
7 | Kevin114514 | 3443 |
8 | ksun48 | 3434 |
9 | Rewinding | 3397 |
10 | Um_nik | 3396 |
# | User | Contrib. |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 156 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
10 | nor | 152 |
What are some efficient ways to divide 'a' by 'b' where the range is :
-10^300 <= a,b <= +10^300
Name |
---|
Since the answer of
a/b
is between 0 and a, we could use binary_search. Complexity:log(a)
, which is around 1000.That's actually since in every iteration of binary search you need to multiply two numbers.
Fortunately there's an easier and faster way in O(n2) (hereafter n is the number of digits), assuming the numeric base d is a small constant (typically 2, 10, or 16).
Start with x = 0, then, for each i from n down to 0, keep incrementing x by di as long as xb ≤ a. This "increment and compare" operation can be done in O(n) by performing it not on x, but on the product xb — you can add bdi to any number by simply appending i zeroes to the d-ary representation of b and performing addition as usual. Running time of this algorithm is O(d·n2). For large values of d you can use binary search inside each iteration of the outer loop, instead of a simple loop which keeps adding di, so the complexity becomes .
There is a way to do it in , where N is the sum of lengths of the numbers. Link. But it's masochistic to implement it (as far as I remember).