Tutorial uses mobius function to solve this problem. How can we solve this using DP, as I have seen many people use it in their solutions.
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3831 |
3 | Radewoosh | 3646 |
4 | jqdai0815 | 3620 |
4 | Benq | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | gamegame | 3386 |
10 | ksun48 | 3373 |
# | User | Contrib. |
---|---|---|
1 | cry | 164 |
1 | maomao90 | 164 |
3 | Um_nik | 163 |
4 | atcoder_official | 160 |
5 | -is-this-fft- | 158 |
6 | awoo | 157 |
7 | adamant | 156 |
8 | TheScrasse | 154 |
8 | nor | 154 |
10 | Dominater069 | 153 |
Tutorial uses mobius function to solve this problem. How can we solve this using DP, as I have seen many people use it in their solutions.
Name |
---|
Let $$$dp[i]$$$ be the number of ways with $$$\gcd = i$$$.
If you calculate it in descending order of $$$i$$$, $$$dp[i] =$$$ (number of ways with values multiples of $$$i$$$) — $$$\sum_{k=2}^{\lfloor m/i \rfloor} dp[ik]$$$.
You can check out similar techniques in this blog.
Got this bit, thanks! I was looking at your solution 125973986 Just wanted to know if you're storing in kn[i][j] : number of ways after i terms and upto jth multiple of g, or something else?
The former.