I am trying to solve this using dynamic programming with complexity of O ( N ^ 2 ) which will give TLE because N <= 10 ^ 5.
№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 3690 |
2 | jiangly | 3647 |
3 | Benq | 3581 |
4 | orzdevinwang | 3570 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | Radewoosh | 3509 |
8 | ecnerwala | 3486 |
9 | jqdai0815 | 3474 |
10 | gyh20 | 3447 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | maomao90 | 174 |
2 | awoo | 164 |
3 | adamant | 163 |
4 | TheScrasse | 159 |
5 | nor | 157 |
6 | maroonrk | 155 |
7 | -is-this-fft- | 152 |
8 | Petr | 146 |
8 | orz | 146 |
10 | BledDest | 145 |
I am trying to solve this using dynamic programming with complexity of O ( N ^ 2 ) which will give TLE because N <= 10 ^ 5.
Название |
---|
First, store an array b, b[i] = log2(a[i]) to get rid of big numbers.
Now, calculate the following DP:
DP[1]=0
DP[i]=min(DP[j]+b[i]) for all i - k < = j < i (note that the problem is now changed to addition instead of multiplication becasue log(a * b) = log(a) + log(b)
Store the best path, and calculate the product of all A[i]'s that occur in the path.
You can solve the above DP in O(nlogn) by maintaining a set of the last k DP values.
Code