【递推】【YBTOJ】错排问题

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思路：

设f[i]为A1~Ai的最小划分段数
则 $f[i]=(i-1)\ast (f[i-2]+f[i-1])$

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$Code$:

#include <cstdio>
#include <iostream>

using namespace std;
long long n ,f[10001 ];
int main ()
{
    
    
	scanf("%lld",&n);f[1] = 0,f[2] = 1;//初始化
	for (int i = 3;i <= n; ++i)
	f[i]=(i - 1) * (f[i - 2] + f[i - 1]);
	printf("%lld",f[n]);
    return 0;	
}