Wrong row problem

http://acm.hdu.edu.cn/showproblem.php?pid=1465

At the beginning of the question, I used the full line to complete the overrun.

1#include"stdio.h"2 intN,count;3 inta[ +];4 intequal ()5 {6     intflag=1;7      for(intI=1; i<=n;i++)8     {9         if(a[i]==i)Tenflag=0; One     } A     returnFlag; - } - voidPermintk) the { -     inttemp; -     if(k==N) -     { +         if(Equal ()) -count++; +     } A      for(inti=k;i<=n;i++) at     { -temp=a[i];a[i]=a[k];a[k]=temp; -Perm (k +1); -temp=a[i];a[i]=a[k];a[k]=temp; -     } - } in Main () - { to      while(SCANF ("%d", &n)! =EOF) +     { -      the          for(intI=1; i<=n;i++) *a[i]=i; $Perm1);Panax Notoginsengprintf"%d\n", count); -Count=0; the     } +}

And then I started thinking about recursive formulas.

Assuming that f (1) and F (2) can be directly mental arithmetic we assume that the former n-1 and n-2 have been lined up.

There are two cases when there are n items

1. Front n-1 Item wrong outfit

2. Front n-2 item wrong outfit

First case

It's easier to n-1*f (n-1) by swapping the nth item with any of the preceding n-1 items

The second case

Well, that's the only thing that's right. Exchange with Nth item but yes, one of the items can be any of the n-2 n-1*f (n-2)

So the recursive formula is f (n) =n-1*f (n-1) +n-1*f (n-2) = (n-1) * (f (n-1) +f (n-2))

This was later found to be the famous wrong-row formula

1#include"stdio.h"2 Main ()3 {4     intN;5      while(SCANF ("%d", &n)! =EOF)6     {7         Long Longa[ -];8a[1]=0; a[2]=1; a[3]=2; 9          for(intI=4; i<=n;i++)Tena[i]= (I-1) * (a[i-2]+a[i-1]); Oneprintf"%i64d\n", A[n]); A     } -}

Wrong row problem