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Codeforces #713(Div.3) E: Permutation by Sum [EN]

Problem Overview

  • Consider the permutation 1 to \(n\) called \(P\).
  • The parameters \(l, r, s\) that satisfies \(1 \leq l \leq r \leq n\) and \(1 \leq s \leq \frac{n(n + 1)}{2}\) are given.
  • Find the permutation which satisfies \(P_{l} + P_{l + 1} + ... + P_{r} = s\).
  • Print any permutation of length \(n\) that fits the condition above if such a permutation exists; otherwise, -1.

Problem Explanation

First, consider the minimum and the maximum value we can generate with the length \(r - l + 1\).
Hereafter, we define \(k = r - l + 1\).

Minimum Value
As an arithmetic sequence with first term 1, term number \(m\), and tolerance 1, we can derive the minimum value.
\(min(m) = \frac{m(m + 1)}{2}\)

Maximum Value
As an arithmetic sequence with first term \(x\), term number \(m\), and tolerance -1, we can derive the maximum value.

\(max(x, m) = \frac{m * (2 * x + (m - 1) * -1)}{2}\)
\(max(x, m) = \frac{m(2x - m + 1)}{2}\)

Any number \(s\) that satisfies \(min(k) \leq s \leq max(n, k)\) meet the condition.

First, we prepare the vector \(res\) with size \(n\) to push the results.
Consider in descending order.
Start the for loop from \(n\),
if \(i\) meets the condition \(max(i, k) - s \geq 0\) and \(s - i \geq min(k - 1)\), put \(i\) to the \(res[l + k]\) and replace \(k = k - 1, s = s - i\).
Iterate this until \(k\) becomes 0, and then if \(s = 0\) is achieved, we find the permuation; otherwise, prints -1.
The remaining part of the implementation is to insert the unused numbers into the empty parts of the vector.

Source Code