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 | tourist | 4009 |
2 | jiangly | 3831 |
3 | Radewoosh | 3646 |
4 | jqdai0815 | 3620 |
4 | Benq | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | gamegame | 3386 |
10 | ksun48 | 3373 |
# | User | Contrib. |
---|---|---|
1 | cry | 164 |
1 | maomao90 | 164 |
3 | Um_nik | 163 |
4 | atcoder_official | 160 |
5 | -is-this-fft- | 158 |
6 | awoo | 157 |
7 | adamant | 156 |
8 | TheScrasse | 154 |
8 | nor | 154 |
10 | Dominater069 | 153 |
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 symboli
in the patternP
starting 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)
.