I'd be happy if someone explained the solutions of E and J.
http://acm.math.spbu.ru:17249/~ejudge/files/opencup/oc12/gp6/gpce-e.pdf
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3611 |
| 4 | jiangly | 3583 |
| 5 | strapple | 3515 |
| 6 | tourist | 3470 |
| 7 | Radewoosh | 3415 |
| 8 | Um_nik | 3376 |
| 9 | maroonrk | 3361 |
| 10 | XVIII | 3345 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 162 |
| 2 | adamant | 148 |
| 3 | Um_nik | 146 |
| 4 | Dominater069 | 143 |
| 5 | errorgorn | 141 |
| 6 | cry | 138 |
| 7 | Proof_by_QED | 136 |
| 8 | YuukiS | 135 |
| 9 | chromate00 | 134 |
| 10 | soullless | 133 |
I'd be happy if someone explained the solutions of E and J.
http://acm.math.spbu.ru:17249/~ejudge/files/opencup/oc12/gp6/gpce-e.pdf
| Name |
|---|



Problem E.
Let
calc(i, j)be the maximum coverage length of a special symboliin the patternPstarting fromP[j]. Then for each equation of the form A = B + C,calc(A, j) = calc(B, j) + calc(C, j + calc(B, j)), and for equation A = a word over * {a, ..., z} we count it the simple way.To speed up, use a hash table (unordered_map) to memorize
calc(i, j)values.The solution is then
calc(S, 0) == strlen(P).