| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3845 |
| 2 | jiangly | 3707 |
| 3 | Benq | 3630 |
| 4 | orzdevinwang | 3573 |
| 5 | Geothermal | 3569 |
| 5 | cnnfls_csy | 3569 |
| 7 | jqdai0815 | 3532 |
| 8 | ecnerwala | 3501 |
| 9 | gyh20 | 3447 |
| 10 | Rebelz | 3409 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | maomao90 | 171 |
| 2 | adamant | 163 |
| 3 | awoo | 161 |
| 4 | nor | 152 |
| 5 | maroonrk | 151 |
| 5 | -is-this-fft- | 151 |
| 7 | TheScrasse | 148 |
| 8 | atcoder_official | 146 |
| 9 | Petr | 145 |
| 10 | pajenegod | 144 |
Hey I can't seem to formulate the DP for [this] ? (http://community.topcoder.com/stat?c=problem_statement&pm=13964&rd=16514&rm=326779&cr=10574855)
I understand that the solution space is reduced to 3^n as compared to 4^n, and that we have to work on individual digits of the sum, but I'm unable to grasp how to take care of the carry and how that would fit in the DP formulation ?
Perhaps you could have f(d, c, z) = the largest achievable value for the last d digits of the lucky sum that has carry c (0-1), and z (0-2) summands which must have 0 for the rest of their digits.