关于汉诺塔的递归，记住一个结论是，转移n个盘子至少需要2^n-1步

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<algorithm>

#include<string>

using namespace std;

int two[100005],pos[100005];

int main()

{

    int n,nn[5],i,j,ans,now,end,mid,a;

    two[0]=1;

    for(i=1;i<=100000;i++)

        two[i]=(two[i-1]*2)%1000000;

    while(~scanf("%d",&n))

    {

        scanf("%d%d%d",&nn[1],&nn[2],&nn[3]);

        for(i=1;i<=3;i++)

        {

            for(j=1;j<=nn[i];j++)

            {

                scanf("%d",&a);

                pos[a]=i;

            }

        }

        ans=0;

        end=now=pos[n];

        printf("%d\n",pos[n]);

        while(n>0)

        {

            if(end!=now)

            {

                ans=(ans+two[n-1])%1000000;

                end=mid;

            }

            n--;

            now=pos[n];

            mid=6-now-end;

        }

        printf("%d\n",ans);

    }

    return 0;

}