[Ybtoj Chapter 1 Example 1] Wrong arrangement [Recursion]

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Problem solving ideas

First set $f\left[n\right]$ means$n$ number of solutions that satisfy the meaning of the question.
First, set a position k, if we put n at position$(k\ne n)$ , then$(n-1)$ kinds of placement methods Next, we consider the placement of k.

• k is placed in the position of n:
  if it is placed this way, the remaining unordered elements will be left $(n-2)$ , which is
  equivalent to putting that$(n-2)$ elements are staggered, the number of schemes is $f [n − 2] $.
• k is not placed at the position of n: in
  this case, the disordered elements will have $(n-1)$ , which is
  equivalent to putting$(n-1)$ elements are staggered, the number of schemes is$f[n-1]$ .

So in $In$ the case of$($$n$$-$$1$$)$ , there is$f[n-1]+f[n-2]$ kinds of placement methods.
The total number of solutions is$(n-1)\times (f[n-1]+f[n-2])$ .

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Code

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;

long long n,f[110];

int main(){
    
    
	scanf("%lld",&n);
	f[1]=0,f[2]=1;
	for(int i=3;i<=n;i++)
		f[i]=(i-1)*(f[i-1]+f[i-2]);
	printf("%lld",f[n]);
}