i am stuck with this dp problem , can any someone explain me the state of dp for this problem ? problem link is Question
Codeforces Round 1133 (Div. 1, Div. 2) is coming soon! Register here.
→ Pay attention
→ Top rated
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3382 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
→ Top contributors
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 155 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 145 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 134 |
| 9 | maroonrk | 132 |
| 10 | cry | 130 |
→ Find user
→ Recent actions
Codeforces (c) Copyright 2010-2026 Mike Mirzayanov
The only programming contests Web 2.0 platform
Server time: Nov/06/2026 21:38:36 (g2).
Desktop version, switch to mobile version.
Supported by
This is a difficult DP problem until you know the trick. When would a triangle be invalid? Only when one of the side lengths is >= 1/2 of the total sum. With this in mind we can count how many assignments are invalid triangles, rather than how many assignments are valid triangles. This problem is significantly easier — we have 2 parts to our state: current sum and current item. At each spot we have 2 choices — add to our sum or don't. Succeed when our value reaches 1/2 of the total array sum. Total time is O(2 choices * 300 indexes * 300^2 max sum). We can recover the answer by taking 3^N — 3 * the result of our DP.
What about the difference property. Difference of 2 sides must be smaller than third
That should be covered by the same logic.
What you said isn't much clear. But I understood is by subset sum problem we can find the number of ways to make every sum. Then we can add the counts for the numbers greater than n/2 Is that right
If you have |A-C|>B and A+B+C=X then A or C must be >= X/2