[Ybt] [Basic calculation recurrence class example 1] Wrong arrangement

Misalignment

Topic link: Misalignment

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Title description

Problem solving ideas

This question is an entry-level (a ghost) recursion.
Let $f_n$For nnLegal permutation of$n$ numbers.

We will $n$ number is placed at$For\; k$ positions, there are$n-1$ middle release method ($n!=k$）。

There are two cases for the remaining elements:

• Will $k$ in$On\; n$ , the remaining elements are$n-$The misalignment of$2$ , ie$f_{n-2}$。
• Will $k\; is$ placed in non-$In$ the position of$n$ , it includes$The$ remaining elements within$k$ are$n-$The misalignment of$1$ , namely$f_{n-1}$。

Because we are determining nnConsiderkkafter the position of$n$$In$ the case of$k$ , so$n$ 's case andkkThe case of$k$ satisfies the principle of multiplication.

Because $k$ put$n$ and not put$n$ is two different cases, so these two cases satisfy the principle of addition.

The recurrence formula is: $$f_n=(n+1)(f_{n-1}+f_{n-2})$$
Where$f_1=0$，$f_2=1$。

code

#include<iostream>
#include<cstdio>
#define int long long
using namespace std;

int n;
int f[30];

signed main()
{
    
    
	cin>>n;
	f[1]=0,f[2]=1;
	for(int i=3;i<=n;i++)
		f[i]=(i-1)*(f[i-1]+f[i-2]);
	cout<<f[n]<<endl;
}