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 | 3856 |
2 | jiangly | 3747 |
3 | orzdevinwang | 3706 |
4 | jqdai0815 | 3682 |
5 | ksun48 | 3591 |
6 | gamegame | 3477 |
7 | Benq | 3468 |
8 | Radewoosh | 3462 |
9 | ecnerwala | 3451 |
10 | heuristica | 3431 |
# | User | Contrib. |
---|---|---|
1 | cry | 167 |
2 | -is-this-fft- | 162 |
3 | Dominater069 | 160 |
4 | Um_nik | 158 |
5 | atcoder_official | 157 |
6 | Qingyu | 156 |
7 | djm03178 | 151 |
7 | adamant | 151 |
9 | luogu_official | 150 |
10 | awoo | 147 |
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.