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 ?
№ | Пользователь | Рейтинг |
---|---|---|
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 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 156 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
10 | nor | 152 |
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 ?
Название |
---|
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...