Hi Folks,I am learning competitive programming now a days.I have started solving recursion/dp based problems.I am trying this problem link. Please help me to find out recursion in the above problem.
Thanks
→ Pay attention
→ Streams
→ Top rated
| # | User | Rating |
|---|---|---|
| 1 | tourist | 3985 |
| 2 | jiangly | 3814 |
| 3 | jqdai0815 | 3682 |
| 4 | Benq | 3529 |
| 5 | orzdevinwang | 3526 |
| 6 | ksun48 | 3517 |
| 7 | Radewoosh | 3410 |
| 8 | hos.lyric | 3399 |
| 9 | ecnerwala | 3392 |
| 9 | Um_nik | 3392 |
→ Top contributors
| # | User | Contrib. |
|---|---|---|
| 1 | cry | 169 |
| 2 | maomao90 | 162 |
| 2 | Um_nik | 162 |
| 4 | atcoder_official | 160 |
| 5 | djm03178 | 158 |
| 6 | -is-this-fft- | 157 |
| 7 | adamant | 155 |
| 8 | Dominater069 | 154 |
| 8 | awoo | 154 |
| 10 | luogu_official | 151 |
→ Find user
→ Recent actions
Codeforces (c) Copyright 2010-2024 Mike Mirzayanov
The only programming contests Web 2.0 platform
Server time: Dec/23/2024 10:25:26 (f1).
Desktop version, switch to mobile version.
Supported by
It is not a DP problem. Find the leftmost digit that is unlucky. If there is such a digit let its position be i. If the digit is less than 3 then make it 3 and fill all positions on the right with 3s. Otherwise, if it is less than 5 then make it 5 and fill the positions on the right with 3s. If the digit is greater than 5 find the rightmost 3 on its left at position j. Make this 3 a 5 and fill the positions on its right with 3s. If there is no such 3, then print s+1 3s where s is the length of N. If there is no unlucky digit find the last 3, change it to 5 and fill the positions on its right with 3s. If there is no 3 then print s+1 3s.