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Help!
By DEAMN, history, 4 years ago, In English

  • I can't think of a DP for this task help plz Find the number of sequences (a1, a2,..., an) such that a[i]^2 + a[i + 1]^2 + ... + a[n]^2 = k a[i] <= a[i + 1] <= ... <= a[n]

  • T is testcase

  • n is quantity of number
  • k is a[i]^2 + a[i + 1]^2 + ... + a[n]^2 = k

  • 1 <= T <= 1e5

  • 1 <= n <= 100
  • 1 <= k <= 5000
  • example
  • input
  • 1
  • 4 243
  • output
  • 9

  • Explanation

  • first 1 1 4 15 -> (1 * 1) + (1 * 1) + (4 * 4) + (15 * 15) = 243 — second 1 3 8 13 — third 1 7 7 12 — fouth 3 3 9 12 — fifth 3 4 7 13 — sixth 3 7 8 11 — seventh 4 5 9 11 — eighth 5 5 7 12 — nineth 7 7 8 9
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Comments Comments (4)
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4 years ago, # |
← Rev. 2   Vote: I like it +2 Vote: I do not like it

Consider DP[LEN][SUM][LAST] — number of non-decreasing sequences with length LEN, last element — LAST, and sum of squares — SUM.

$$$DP[LEN][SUM][LAST] = \sum_{i = 0}^{LAST}DP[LEN - 1][SUM - LAST^2][i] $$$
$$$ANSWER[n][k] = \sum_{i = 0}^{\sqrt{k}}DP[n][k][i]$$$

To calculate sum - for every pair {LEN, SUM} build cumulative sum arrays.

Let:

$$$K = max\{k\}, N = max\{n\}$$$

Note:

$$$ LAST^2 \leq K $$$

Total complexity:

$$$O(N \cdot K \cdot \sqrt{K} + T \cdot \sqrt{K})$$$
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4 years ago, # |
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You can also try think like this-> Suppose dp[i][j] is the number of sorted sequences having last term i and has sum of squares of its elements j. Then it's obvious

dp[i][j] = dp[1][j-i*i] + dp[2][j-i*i] + dp[3][j-i*i] ...dp[i][j-i*i]

Now try to think about the initial conditions, I leave this task to you :)

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    4 years ago, # ^ |
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    dp[1][j * j] = 1?

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      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      Pretty close...

      It should be dp[1][j] = 1 (j>=1) (Why?) also I think there more conditions, Can u figure it out?

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