Is there any greedy approach for finding minimum operations required for making a_i to b_i ?
№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3773 |
3 | Radewoosh | 3646 |
4 | ecnerwala | 3624 |
5 | jqdai0815 | 3620 |
5 | Benq | 3620 |
7 | orzdevinwang | 3612 |
8 | Geothermal | 3569 |
8 | cnnfls_csy | 3569 |
10 | Um_nik | 3396 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | Um_nik | 163 |
2 | cry | 161 |
3 | maomao90 | 160 |
4 | -is-this-fft- | 159 |
5 | awoo | 158 |
5 | atcoder_official | 158 |
7 | adamant | 155 |
8 | nor | 154 |
9 | maroonrk | 152 |
10 | Dominater069 | 149 |
Is there any greedy approach for finding minimum operations required for making a_i to b_i ?
Название |
---|
This is weird! I think this is because we can know how many operations are needed to get from $$$1$$$ to $$$b_i$$$ greedily.
can you share your greedy approach ?
i've been trying but not able to come up with any!!
We can always choose $$$x = 1$$$ so that when performing $$$a_i = a_i + \lfloor{\frac{a_i}x\rfloor}$$$ is the same as $$$ a_i = 2a_i$$$.
Let $$$b_i$$$ is the minimum number of the above operation to get from $$$1$$$ to $$$i$$$. Sense $$$i <= 10^3$$$ so we need at most $$$\lceil \log_2{10^3} \rceil$$$ operations, $$$b_i \le 10$$$.
Update:
Ops! forget about it, in the last operation we can't increase by all values in the range $$$[1, a_i)$$$. Only some of them are available.
The editorial provided a DP approach for it, check it out.
It tagged greedy because it's a knapsack problem.
By the way, thanks for sharing a great problem to solve.
But the problem can't be solved greedily. The greedy approach only works with the fractional knapsack (as far as I know).
Tags are sometimes wrong.