How to solve for(M,N) it the equation of the form M*c1+c2=N*c3+c4, using Extended euclidean algo. Here c1,c2,c3,c4 are known constants.
# | User | Rating |
---|---|---|
1 | tourist | 3690 |
2 | jiangly | 3647 |
3 | Benq | 3581 |
4 | orzdevinwang | 3570 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | Radewoosh | 3509 |
8 | ecnerwala | 3486 |
9 | jqdai0815 | 3474 |
10 | gyh20 | 3447 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 173 |
2 | awoo | 164 |
3 | adamant | 163 |
4 | TheScrasse | 160 |
5 | nor | 157 |
6 | maroonrk | 156 |
7 | -is-this-fft- | 152 |
8 | Petr | 146 |
8 | orz | 146 |
10 | pajenegod | 145 |
How to solve for(M,N) it the equation of the form M*c1+c2=N*c3+c4, using Extended euclidean algo. Here c1,c2,c3,c4 are known constants.
Name |
---|
Convert to c1M + c3( - N) = c4 - c2. This is now in the form ax + by = c, solving for x, y, which can be done using Extended Euclidean algorithm.
Thanks, can you please share a useful link? :)
Got it, it is a Diophantine equation, Thanks again :)
I bet it is for yesterday's contest ;)
See tags for the question
I didn't see it :)