学步园

思路： 首先你得明白三个柱子的汉诺塔 n个disk从A 到 C的最少步数  F(n) = 2^n
-1;
那四个呢？？ 其实题目已经告诉我们解法了 At first k >= 1 disks on tower
A are fixed and the remaining n-k disks are moved from tower A to
tower B using the algorithm for four towers.Then the remaining k
disks from tower A are moved to tower D using the algorithm for
three towers. At last the n - k disks from tower B are moved to
tower D again using the algorithm for four towers
就是先移 n-k个到B（利用四个柱子） 然后移k到D （这时只能用三个柱子 因为B不能用） 然后再把B上的n-k个移到D

count[n] = min (count[n],2*count[n-k] + 2^k - 1) (0
< k <ｎ)

//196K   
0MS
#include <stdio.h>
#include <math.h>
const int inf = 1000;
int Min (int a,int b)
{
    return a
> b ? b : a;
}
int main ()
{
    int
n,k,count[15];
    count[1] =
1;
    for (n = 2;n
<= 12;n ++)
    {
       
count[n] = inf;
       
for (k = 1;k < n;k ++)
       
{
           
int min = 2*count[n-k] + pow(2,(double)k) - 1;
           
count[n] = Min (count[n],min);
       
}
    }
    for (int i =
1;i <= 12;i ++)
       
printf ("%d\n",count[i]);
    return
0;
}