题目描述
Background
Charlie Darkbrown sits in another one of those boring Computer Science lessons: At the moment the teacher just explains the standard Tower of Hanoi problem, which bores Charlie to death!

The teacher points to the blackboard (Fig. 4) and says: "So here is the problem:
There are three towers: A, B and C.
There are n disks. The number n is constant while working the puzzle.
All disks are different in size.
The disks are initially stacked on tower A increasing in size from the top to the bottom.
The goal of the puzzle is to transfer all of the disks from tower A to tower C.
One disk at a time can be moved from the top of a tower either to an empty tower or to a tower with a larger disk on the top.

So your task is to write a program that calculates the smallest number of disk moves necessary to move all the disks from tower A to C."
Charlie: "This is incredibly boring—everybody knows that this can be solved using a simple recursion.I deny to code something as simple as this!"
The teacher sighs: "Well, Charlie, let's think about something for you to do: For you there is a fourth tower D. Calculate the smallest number of disk moves to move all the disks from tower A to tower D using all four towers."
Charlie looks irritated: "Urgh. . . Well, I don't know an optimal algorithm for four towers. . . "
Problem
So the real problem is that problem solving does not belong to the things Charlie is good at. Actually, the only thing Charlie is really good at is "sitting next to someone who can do the job". And now guess what — exactly! It is you who is sitting next to Charlie, and he is already glaring at you.
Luckily, you know that the following algorithm works for n <= 12: 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 (and thereby not moving any of the k disks already on tower D). Do this for all k 2 ∈{1, .... , n} and find the k with the minimal number of moves.
So for n = 3 and k = 2 you would first move 1 (3-2) disk from tower A to tower B using the algorithm for four towers (one move). Then you would move the remaining two disks from tower A to tower D using the algorithm for three towers (three moves). And the last step would be to move the disk from tower B to tower D using again the algorithm for four towers (another move). Thus the solution for n = 3 and k = 2 is 5 moves. To be sure that this really is the best solution for n = 3 you need to check the other possible values 1 and 3 for k. (But, by the way, 5 is optimal. . . )
输入描述:
There is no input.
输出描述:
For each n (1 <= n <= 12) print a single line containing the minimum number of moves to solve the problem for four towers and n disks.

思路：
因为是汉诺塔问题，所以这是一道递归/递推的题目。
1.我们先考虑三个塔的汉诺塔问题，最优秀方案：必然是先挪走n-1个圆盘，然后再挪走圆盘N，
因此可以得出递推方程也就是 d[i]=d[i-1]*2+1;
之所以要乘以2，是因为第一次挪到第二个塔，然后还要挪移回到第三个塔，四个塔也是这样的。
2.四塔问题，我们可以这么思考，首先挪走j个塔，也就是有四个塔可以选择，然后再挪走剩下的n-j个塔，此时有三个塔可以选择，因此这就是我们的状态转移方程：f[i]=min(f[i],f[j]*2+d[n-j]);//i表示当前一共有几个塔，也就是上文所说的n

如果你足够聪明可以打个表：

#include<iostream>
using namespace std;
int f[22]={0,1,3,5,9,13,17,25,33,41,49,65,81};
int main(){
    for(int i=1;i<=12;i++) cout<<f[i]<<endl;
    return 0;
}

滑稽.jpg.

完整AC代码：

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int f[20], d[20];

int main() {
    d[1] = 1;
    for (int i = 2; i <= 12; i++) {
        d[i] = 2 * d[i - 1] + 1;
    }

    memset(f, 0x3f, sizeof(f));
    f[1] = 1;

    for (int i = 2; i <= 12; i++){
        for (int j = 1; j < i; j++) {
            f[i] = min(f[i], 2 * f[j] + d[i - j]);
        }
    }

    for (int i = 1; i <= 12; i++) {
        cout << f[i] << endl;
    }
    return 0;
}