(Jizhong) 1597.] [GDKOI2004 Tower of Hanoi (hanoi) [problem] Hanta

(File IO): input: hanoi.in output: hanoi.out
time limit: 1000 ms space constraints: 262144 KB specific restrictions
Goto ProblemSet

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Title Description
old Tower of Hanoi problem is this: with the least number of steps will $N$ disk radii equal to each other from $1$ No. column using $2$ No. column to all mobile $3$ No. column, to the top chunk in the market is always in the process of moving. Now coupled with one condition: do not allow directly to disk from $1$ No. moved to column $3$ No. column, does not allow directly from the disk $3$ No. column to move $1$ Hao column. The radius of the disk by small to large $1$ to $N$ number. With each state $N$ integers, said first $i$ integer representation $i$ numbered column number of the disc is located. then $N=2$ movement scheme is: $(1,1)=>(2,1)=>(3,1)=>(3,2)=>(2,2)=>(1,2)=>(1,3)=>(2,3)=>(3,3)$ the initial state of $0$ step, when a programmed number of steps required state.

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Input
The first line of the input file integer $T(1<=T<=50000)$ , it indicates the number of sets of input data. Next T lines of two integers $N,M(1<=n<=19,0<=M<=$ Movement of the N number of steps required disc).

Output
Output file has $T$ line. For each case, the output integer N number of mobile $N$ disks in $M$ state step, between every two numbers separated by a space, at the end of the line and the line do not have extra spaces.

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Sample input
. 4
2 0
2. 5
. 3 0
. 3. 1

Sample output
. 1. 1
. 1 2
. 1. 1. 1
2. 1. 1

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Data range limit

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Problem-solving ideas
: include $n=3$ cases when can be found: the first $1$ rule of the number that appears is a bit $123321$ , the first $2$ law digit number that appears is $111222333333222111……$ and so on.

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Code

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<iomanip>
#include<cmath>
using namespace std;
int a[6]={1,2,3,3,2,1},t,n,m;
int main(){
    freopen("hanoi.in","r",stdin);
    freopen("hanoi.out","w",stdout);
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
	scanf("%d%d",&n,&m);
	for(int j=1;j<=n;j++)
	{
	    printf("%d ",a[m%6]);
	    m/=3;
	}
	printf("\n");
    }
}