poj1920 Towers of Hanoi

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;
}