| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3880 |
| 2 | jiangly | 3669 |
| 3 | ecnerwala | 3654 |
| 4 | Benq | 3627 |
| 5 | orzdevinwang | 3612 |
| 6 | Geothermal | 3569 |
| 6 | cnnfls_csy | 3569 |
| 8 | jqdai0815 | 3532 |
| 9 | Radewoosh | 3522 |
| 10 | gyh20 | 3447 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | awoo | 161 |
| 2 | maomao90 | 160 |
| 3 | adamant | 156 |
| 4 | maroonrk | 153 |
| 5 | -is-this-fft- | 148 |
| 5 | atcoder_official | 148 |
| 5 | SecondThread | 148 |
| 8 | Petr | 147 |
| 9 | nor | 144 |
| 9 | TheScrasse | 144 |
int highest(long long n) {
int r = 0;
while(n) {
r++;
n >>= 1;
}
return r-1;
}
int dfs(long long n) {
if(n == 0) return 0;
int h = highest(n);
long long n1 = n-(1LL<<h);
long long n2 = (1LL<<(h+1))-n;
int res = 1+dfs(n1);
if(n2 < n) res = min(res, 1+dfs(n2));
return res;
}
What is complexity of dfs(n) and why?
At most log(n) phase n1 will equal to 0. If (n2&(1LL<<h)) and n2<n then when call DFS(n2),the n2 in that DFS will be equal to n and thus it don't call in further. Else it's similar to n1. Function highest work in log(n). So total complexity is O(log(n)^2). Currently I can't think of any case when DFS(n) produce two small DFS and so on. Correct me if I'm wrong.
I thought so, but it's not true. It's probably something like O(2number of blocks with consecutive 1's), but I can't prove it.
DFS(n2)will not be called iff n binary representation is100...0