Here is a Problem: 2069B - Set of Strangers Here is Soln: 306760751
| № | Пользователь | Рейтинг |
|---|---|---|
| 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 |
| Страны | Города | Организации | Всё → |
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|---|---|---|
| 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 |
Here is a Problem: 2069B - Set of Strangers Here is Soln: 306760751
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Bro tune third question 2069C - Beautiful Sequence,dp ka implementation kaise kiya hai yeh bata sakta hai , like approach mujhe bhi aagayi thi but i don't know how to implement it
good subsequences are of form 12222....3. count of 2s >= 1 and start me 1 and end me 3.
// dp[i][1] bolega no of subsequences of form "1" till i // dp[i][2] bolega no of subsequences of form "122...." till i // dp[i][3] bolega no of subsequences of form "12....3" till i
// dp[i][1] ,agar a[i] == 1, dp[i][1] = 1 + dp[i-1][1], nhi toh base dp[i][1] = dp[i-1][1]
// dp[i][2], agar a[i] != 2 toh dp[i][2] = dp[i-1][2] // but if a[i] == 2 // i-1 tak jo already 12... bangayi ume 2 jod de, ya mat jod. total 2 * dp[i-1][2] hojayengi // i-1 tak jo no of form "1" ki subsequences hai unme 2 jod de, toh dp[i — 1][1] // so dp[i][2] = dp[i — 1][1] + 2 * dp[i — 1][2]
//dp[i][3], i-1 tak jo already 12.... bangayi unme 3 jod de bas, dp[i][3] = dp[i-1][2]