2014
02-09

# 汉诺塔 X

1,2,…,n表示n个盘子．数字大盘子就大．n个盘子放在第１根柱子上．大盘不能放在小盘上．在第１根柱子上的盘子是a[1],a[2],…,a[n]. a[1]=n,a[2]=n-1,…,a[n]=1.即a[1]是最下面的盘子．把n个盘子移动到第3根柱子．每次只能移动1个盘子，且大盘不能放在小盘上．问第m次移动的是哪一个盘子，从哪根柱子移到哪根柱子.例如：n=3,m=2. 回答是 ：2 1 2，即移动的是2号盘，从第1根柱子移动到第2根柱子 。

4
3 2
4 5
39 183251937942
63 3074457345618258570

2 1 2
1 3 1
2 2 3
2 2 3

2011-12-31 20:09:16

mark：递归。

# include <stdio.h>

void gao(long long n, long long time, int s, int e)
{
int m = 6-s-e ;
long long mid = (1LL<<(n-1)) ;
if (time == mid)
{
printf ("%I64d %d %d\n", n, s, e) ;
return ;
}
if (time < mid)
gao(n-1, time, s, m) ;
else
gao(n-1, time-mid, m, e) ;
}

int main ()
{
int T ;
long long n, time ;
scanf ("%d", &T) ;
while (T--)
{
scanf ("%I64d%I64d", &n, &time) ;
gao(n, time, 1, 3) ;
}
return 0 ;
}

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2. 思路二可以用一个长度为k的队列来实现，入队后判断下队尾元素的next指针是否为空，若为空，则出队指针即为所求。