Editorial for COCI '06 Contest 4 #4 Zbrka

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This problem is solved using dynamic programming. Let $f(n, c)$ $f (n, c)$ be the number of sequences of length $n$ $n$ with confusion $c$ $c$ .

The number of such sequences that start with the number $1$ $1$ is $f(n-1, c)$ $f (n - 1, c)$ because the $1$ $1$ does not affect the confusion of the rest of the sequence and it makes no difference if we use numbers $1 \dots n-1$ $1 \dots n - 1$ or $2 \dots n$ $2 \dots n$ .

If $1$ $1$ is the second number, then $f(n, c) = f(n-1, c-1)$ $f (n, c) = f (n - 1, c - 1)$ because whichever element is first, it will form a confused pair with the $1$ $1$ . It's easy to see that the complete relation is:

$\displaystyle f(n,c) = \sum_{i=0}^{n-1} f(n-1,c-i)$ $f (n, c) = \sum_{i = 0}^{n - 1} f (n - 1, c - i)$

The time complexity of a direct implementation of this formula (using dynamic programming) would be $\mathcal O(N^2 C)$ $O (N^{2} C)$ , which is too slow.

We need to note that $f(n, c) = f(n, c-1) + f(n-1, c) - f(n-1, c-n)$ $f (n, c) = f (n, c - 1) + f (n - 1, c) - f (n - 1, c - n)$ , which leads to an $\mathcal O(NC)$ $O (N C)$ solution. It is also possible to cut down on the memory used by keeping only two rows of the matrix used for calculations at any time.

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