Each time Snuke screams, every Person i such that $$$i \neq p_i$$$ gives their ball to Person $$$p_i$$$ simultaneously.

Doesn't that mean, that after the first exchange, $$$4th$$$ and $$$5th$$$ men have to stop exchanging? That's misunderstanding in statement. I counted max loop lengthes.

Consider counting the number of permutations of such $$$(1, \ldots, N)$$$ that it consist of $$$k_1, \ldots, k_m$$$ cyclic permutations. These consist of two factors:

What "cyclic permutation group" does each element belong?

This is equivalent to determining such sequence $$$a_1, a_2, \ldots, a_n$$$ such that:

• Each element $$$a_i$$$ is an integer between $$$0$$$ and $$$m$$$;
• For each $$$j = 1, \ldots, m$$$, there are exactly $$$k_j$$$ elements such that $$$a_i = j$$$.

So this means that element $$$i$$$ belongs a cyclic permutation group $$$a_i$$$, or does not belong to any cyclic permutation if $$$a_i = 0$$$.

How many such sequences are there? Consider a sequence B = (($$$N - \sum k_j$$$ copies of $$$0$$$), ($$$k_1$$$ copies of $$$1$$$), ($$$k_2$$$ copies of $$$2$$$), $$$\ldots$$$, ($$$k_m$$$ copies of $$$m$$$)). This is an $$$n$$$-element sequence, and each permutation of $$$B$$$ correspond to the aforementioned $$$(a_i)$$$ one-to-one. Therefore, the number can be found as:

What does each cyclic permutation look like?

Given a set of elements that forms a cyclic permutation, we have to determine the actual permutation. For simplicity, we consider a cyclic permutation of $$$\lbrace 1, 2, \ldots, Q \rbrace$$$, of length $$$Q$$$.

What should $$$1$$$ map to? It should not map to $$$1$$$ itself, since it will never result in a cyclic permutation of length $$$Q$$$. Therefore there are $$$Q-1$$$ choices. Say $$$1$$$ maps to $$$p_1 \neq 1$$$. What should $$$p_1$$$ map to? It should not map to $$$1$$$, because it becomes a cyclic group of length $$$2$$$; nor should it map to $$$p_1$$$ itself. So there are $$$Q-2$$$ choices. The next element has $$$Q-3$$$ choices for its next mapping direction, the next has $$$Q-4$$$, ... and so on, before determine the destination of the last element, which should now go back to $$$1$$$ (no other option).

Therefore, there are $$$(Q-1)!$$$ options.

But wait, there's more!

To sum up the last two sections, we obtain the following number:

However, in the first chapter we distinguished the cyclic group with the same length. So we have to divide by an additional duplicate-remover.

To do this, let's convert the sequence $$$k_1, \ldots, k_m$$$ into the sequence of occurrences: $$$F_1$$$ elements of $$$k_1, \ldots, k_m$$$ are equal to $$$K_1$$$, $$$F_2$$$ elements are equal to $$$K_2$$$, and so on. Using this notation, the last expression can be re-written as follows:

Now we can divide by the freedom of permutation of same-length cycles:

This is equivalent to the expression in the editorial.

Not a simple cycle, but a simple cycle with paths attached to it.

I used a different approach instead of e == v relationship. I started from leaf nodes and removed all the leaf nodes sequentially. After this, all the remaining nodes in the graph must have a degree of 2.

Somewhat inspired from this of a recent Div 3 contest.

ooh got it thanks!

Thanks for the reply ,understood my mistake! char wont take complete no ‘︿’