This Solution got AC, but it gave wrong output for the input below:
3 1
a
b
c
aaaaa
The program gave output: YES but I think the output should be: NO
Am I right? If I'm, then the data set may b too weak! :(
# | User | Rating |
---|---|---|
1 | tourist | 3556 |
2 | wxhtxdy | 3520 |
3 | Radewoosh | 3409 |
4 | Benq | 3368 |
5 | mnbvmar | 3280 |
6 | ecnerwala | 3278 |
7 | LHiC | 3276 |
8 | sunset | 3264 |
9 | maroonrk | 3159 |
10 | TLE | 3145 |
# | User | Contrib. |
---|---|---|
1 | Errichto | 189 |
2 | Radewoosh | 177 |
3 | tourist | 173 |
4 | antontrygubO_o | 172 |
5 | Vovuh | 168 |
6 | PikMike | 166 |
7 | rng_58 | 157 |
8 | majk | 156 |
9 | farmersrice | 153 |
9 | Um_nik | 153 |
This Solution got AC, but it gave wrong output for the input below:
3 1
a
b
c
aaaaa
The program gave output: YES but I think the output should be: NO
Am I right? If I'm, then the data set may b too weak! :(
Name |
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Well, it seems you're right. According to your code strings like 'aaa..a' have hash equels 0. So, HASH("baaaa") == HASH("b") and so on.
Unfortunatly, testset doesn't cover this case.
HASH("baaaa") == HASH("b") * BASE^4
Right?
But actually HASH("b") == HASH("aaaab"). That's why it's good to use BASE = 5 and a = 1, b = 2 and c = 3.
BTW, this is my first way when I use a as 0, b as 1 and c as 2 and it seems I was really really really lucky with that :D :D :D
Wow, actually I compare the strings not only by the hash value but by the length too, so in my case it doesn't depend on luck. Sorry for the misunderstanding! :)