2018 National multi-school algorithm winter training Camp practice Competition (first field) D N-order Hanoi deformation

Https://www.nowcoder.com/acm/contest/67/D

Ideas:

Simulate the process manually, the following is the simulation process, in order to indicate the number of steps required to move the disk label

1 1 2 1 1 2

1 1 3 1 1 2

1 1 2 1 1 3

1 1 2 1 1 2

1 1 4 1 1 2

。。。。。。

We found that each occurrence two times 1 will appear once 2, every two times 2 will appear once 3, every two times 3 will appear once 4, every two times 4 will appear once 5 ...

Then we find that if all labels greater than 1 are treated as 1, then the number of times that K step 1 appears is K/1

If all labels greater than 2 are considered 2, then the number of times that K step 2 appears is K/3

If all labels greater than 3 are considered 3, then the number of times that K step 3 appears is k/3^2

If all labels greater than 4 are considered 4, then the number of times that K step 4 appears is k/3^3

。。。。。。

Then why is this possible?

Because of this we can take the mobile cycle as a 6-step, just said each appeared two times I, appear a i+1, so each move two steps to stop (③ and ⑥ process), so k/3^ (i-1) can be expressed through this cycle of steps, take more than 6 I know where I go

Code:

#include <bits/stdc++.h>using namespacestd;#definell Long Long#definePB Push_backVector<int>a[3];intMain () {Ios::sync_with_stdio (false); Cin.tie (0);    ll N,k;  while(cin>>n>>k) {         for(intI=0;i<3; i++) a[i].clear (); llBase=1;  for(intI=1; i<=n;i++){            intT= (k/Base)%6; if(t>2) t=5-T;            A[T].PB (i); Base*=3; }         for(intI=0;i<3; i++){            if(A[i].size ()) for(intJ=a[i].size ()-1; j>=0; j--) cout<<a[i][j]<< (j==0?'\ n':' '); Elsecout<<0<<Endl; }    }    return 0;}

2018 National multi-school algorithm winter training Camp practice Competition (first field) D N-order Hanoi deformation