| # | User | Rating |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3603 |
| 4 | jiangly | 3583 |
| 5 | turmax | 3559 |
| 6 | tourist | 3541 |
| 7 | strapple | 3515 |
| 8 | ksun48 | 3461 |
| 9 | dXqwq | 3436 |
| 10 | Otomachi_Una | 3413 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 157 |
| 2 | adamant | 153 |
| 3 | Um_nik | 147 |
| 4 | Proof_by_QED | 146 |
| 5 | Dominater069 | 145 |
| 6 | errorgorn | 141 |
| 7 | cry | 139 |
| 8 | YuukiS | 135 |
| 9 | TheScrasse | 134 |
| 10 | chromate00 | 133 |
UPD: i found the problem here http://mirror.codeforces.com/problemset/gymProblem/100589/G
in many problems involving counting some permutations, we use state dp[mask][i] which denotes solution for subproblem where i is the last element used from a given array and the bits in mask variable are set if those values have been taken. i read somewhere that state in a DP solution should uniqely represent a subproblem. so in this case, how is this representation unique?
suppose array is 1, 2, 3, 4, 5, 6.
mask = 15(1111) and i = 4. this permutation is 1234. this will count the permutation where elements from index 1 to 4 have been taken and index 4 is the last element. but for permutation 1324 also, the state will be same. so how this state will give correct answer in such cases.
i remember seeing such a problem about counting permutations in CF GYM. but i cant find it now. it was solved using bitmask.
| Name |
|---|


