Is there an efficient way to calculate the sum of lcm(x,y) for every value x and y such that 1<=x<=n && 1<=y<=n with a given n ?
→ Pay attention
→ Streams
→ Top rated
| # | User | Rating |
|---|---|---|
| 1 | tourist | 3985 |
| 2 | jiangly | 3814 |
| 3 | jqdai0815 | 3682 |
| 4 | Benq | 3529 |
| 5 | orzdevinwang | 3526 |
| 6 | ksun48 | 3517 |
| 7 | Radewoosh | 3410 |
| 8 | hos.lyric | 3399 |
| 9 | ecnerwala | 3392 |
| 9 | Um_nik | 3392 |
→ Top contributors
| # | User | Contrib. |
|---|---|---|
| 1 | cry | 169 |
| 2 | maomao90 | 162 |
| 2 | Um_nik | 162 |
| 4 | atcoder_official | 161 |
| 5 | djm03178 | 158 |
| 6 | -is-this-fft- | 157 |
| 7 | adamant | 155 |
| 8 | awoo | 154 |
| 8 | Dominater069 | 154 |
| 10 | luogu_official | 150 |
→ Find user
→ Recent actions
Codeforces (c) Copyright 2010-2024 Mike Mirzayanov
The only programming contests Web 2.0 platform
Server time: Dec/23/2024 04:23:23 (k3).
Desktop version, switch to mobile version.
Supported by
Mobius Inversion
You can also do dp where you dp[i] is sum of all x * y numbers with gcd(x, y) equal to i, and to transition u find with basic math sum of all x * y with gcd(x, y) >= i by taking sum of all multiples of i and squaring it like sieve then in sieve like fashion subtract out dp values that are multiples of i as that gets rid of overcounting. Obviously you then just take sum of dp[i] / i for all dp values. I'm p sure this dp method and mobius inversion are almost always interchangeable but not sure...