Does anyone knows how to solve this problem ? http://cepc08.ii.uni.wroc.pl/cards.pdf
I've been trying to solve this problem, got some ideas but none of those ideas has really worked. Can anyone give me a hint ?
# | User | Rating |
---|---|---|
1 | tourist | 3993 |
2 | jiangly | 3743 |
3 | orzdevinwang | 3707 |
4 | Radewoosh | 3627 |
5 | jqdai0815 | 3620 |
6 | Benq | 3564 |
7 | Kevin114514 | 3443 |
8 | ksun48 | 3434 |
9 | Rewinding | 3397 |
10 | Um_nik | 3396 |
# | User | Contrib. |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 155 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
10 | djm03178 | 152 |
Does anyone knows how to solve this problem ? http://cepc08.ii.uni.wroc.pl/cards.pdf
I've been trying to solve this problem, got some ideas but none of those ideas has really worked. Can anyone give me a hint ?
Name |
---|
YES is when exist such k, l, n, m ≥ 0 so
c = k × a + l × b
d = m × a + n × b.
NO otherwise.
What's the proof ? I could see this approach, but I wasn't sure if this is right.
counter-example:
c=3 d=7 a=2 b=3
c=0a+1b
d=2a+1b
Yep, one more condition: a × b must divide c × d
UPD This seems to be legit, but in paper I found another: (c = q × a or d = q × a) and (c = w × b or d = w × b).
Can you prove it ?
That paper in russian. Sorry, I'm in lack of time to translate it.