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.
→ Pay attention
→ Top rated
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3382 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
→ Top contributors
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 155 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 145 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 134 |
| 9 | maroonrk | 132 |
| 10 | cry | 130 |
→ Find user
→ Recent actions
Codeforces (c) Copyright 2010-2026 Mike Mirzayanov
The only programming contests Web 2.0 platform
Server time: Nov/06/2026 17:26:53 (g2).
Desktop version, switch to mobile version.
Supported by
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 :)