I was trying to solve COUNT(spoj problem).I am able to solve it using 3D DP but it will give segmentation fault because of the constraints.Can anyone help?
# | User | Rating |
---|---|---|
1 | jiangly | 3976 |
2 | tourist | 3815 |
3 | jqdai0815 | 3682 |
4 | ksun48 | 3614 |
5 | orzdevinwang | 3526 |
6 | ecnerwala | 3514 |
7 | Benq | 3482 |
8 | hos.lyric | 3382 |
9 | gamegame | 3374 |
10 | heuristica | 3357 |
# | User | Contrib. |
---|---|---|
1 | cry | 169 |
2 | -is-this-fft- | 162 |
3 | Um_nik | 161 |
3 | atcoder_official | 161 |
5 | djm03178 | 157 |
5 | Dominater069 | 157 |
7 | adamant | 154 |
8 | luogu_official | 152 |
9 | awoo | 151 |
10 | TheScrasse | 148 |
I was trying to solve COUNT(spoj problem).I am able to solve it using 3D DP but it will give segmentation fault because of the constraints.Can anyone help?
Name |
---|
Always post the link to the problem when you ask for help.
Solution:
dp[n][k]
-- the answer.One can fix the number of ones in final partition:
0
ones --dp[n - k][k]
ways,1
one --dp[n - k][k - 1]
ways,...
n
ones --dp[n - k][0]
ways.So
dp[n][k]
= .Link to the problem : here
How is it different from P(n,k), partitioning ‘n’ elements in ‘k’ sets?