HUD 1465 is not one of the easy series

Original title Link: click here

Problem Solving Ideas:

Should belong to the topic of recursion.

Is that n letters are loaded with the wrong envelope ...

Let's say there are 7 envelopes: a~g

A

_ _ _ _ _ _ _

A

There are 7-1 things wrong with a, first choose a place B

A

B _ _ _ _ _ _

Start to put B, B can put a also can put other, if put a, is left n-2 of the arrangement,

If we put the other assumptions in C, that's the n-1 of the rest.

This can be summed up in the law:

f[n]= (n-1) * (f[n-1] + f[n-2]).

Source:

#include <iostream>#include<stdio.h>using namespacestd;Long Longa[ +]={0,0,1,2};intN;intMain () { for(intI=3; i<= +; i++) A[i]= (I-1) * (a[i-1]+a[i-2]);  while(cin>>N) cout<<a[n]<<Endl; return 0;}

View Code

HUD 1465 is not one of the easy series