http://www.codechef.com/COOK56/problems/STRAB/
Can someone explain the dp approach except the one given in the editorial ? Also if someone know similar problems , please provide the link :) .
# | User | Rating |
---|---|---|
1 | tourist | 3880 |
2 | jiangly | 3669 |
3 | ecnerwala | 3654 |
4 | Benq | 3627 |
5 | orzdevinwang | 3612 |
6 | Geothermal | 3569 |
6 | cnnfls_csy | 3569 |
8 | jqdai0815 | 3532 |
9 | Radewoosh | 3522 |
10 | gyh20 | 3447 |
# | User | Contrib. |
---|---|---|
1 | awoo | 161 |
2 | maomao90 | 160 |
3 | adamant | 156 |
4 | maroonrk | 153 |
5 | -is-this-fft- | 148 |
5 | atcoder_official | 148 |
5 | SecondThread | 148 |
8 | Petr | 147 |
9 | nor | 144 |
9 | TheScrasse | 144 |
http://www.codechef.com/COOK56/problems/STRAB/
Can someone explain the dp approach except the one given in the editorial ? Also if someone know similar problems , please provide the link :) .
Name |
---|
In fact, you don't need dp in this problem.
Let's count the following thing (just using a loop over bitmasks): how many there are strings of length L than consist only of A and B and are good (not hated, i. e., don't contain any hated subword).
Now let's iterate over all possible positions of non-AB characters (again as a loop over bitmasks). Obviously, if there is any hated word, it will be between two two consecutive non-AB characters. So we can use the thing we calculated earlier to fill all characters between non-AB ones.
thanks ,it will be very helpful if you can you refer someone's implemented code ? sorry to bother , i got it :).
Nice solution, and it seems to be O(n2n) too (don't know why you got downvoted). Why is it not in the editorial?