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 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
9 | nor | 153 |
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?