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## [$AcWing$ $432$. $Hanoi$双塔问题](https://www.acwing.com/problem/content/434/)
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### 一.汉若单塔（也称经典汉若塔问题）
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首先，我们先画个 $n=3$ 图：
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要把 $A$ 柱上的 **圆盘** 移到 $C$ 柱上，要求无论在哪个柱子上的圆盘都是大的在下面，小的在上面。求从 $A$ 柱上的圆盘移到 $C$ 柱上要多少步？
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#### 1. 实践
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遇到这种问题，首先想到的是先模拟一次，如下图：
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怎么样？看完过程，是不是清晰了许多？全过程是这样的：
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<table>
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<tr bgcolor=#EFF3F5>
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<th>         步数        </th>
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<th>        操作        </th>
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</tr>
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<tr>
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<td align=center>1</td>
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<td align=center>① A to C</td>
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</tr>
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<tr>
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<td align=center>2</td>
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<td align=center>② A to B</td>
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</tr>
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<tr>
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<td align=center>3</td>
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<td align=center>① C to B</td>
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</tr>
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<tr>
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<td align=center>4</td>
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<td align=center>③ A to C</td>
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</tr>
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<tr>
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<td align=center>5</td>
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<td align=center>① B to A</td>
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</tr>
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<tr>
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<td align=center>6</td>
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<td align=center>② B to C</td>
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</tr>
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<tr>
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<td align=center>7</td>
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<td align=center>① A to C</td>
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</tr>
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</table>
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#### 2. 探索
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>**解释**
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① ：最上面的第$1$个圆饼
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② ：中间的第$2$个圆饼
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③ ：最下面的第$3$个圆饼
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总共$7$步。仔细观察我们可以发现，移动的盘子是对称的，如上表：$①②①③①②①$。
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以$③$为点 ，左右对称。再看看右边$①②①$，也是对称的，这是 $n=2$ 时移动的步数。我们是不是可以推测出如果 $n=4$ 时，$①②①③①②①④①②①③①②①$是移动顺序呢？事实证明的确如此。移动顺序总是上一个序列加上新的盘子再加上一个序列。
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> **注**
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> 我终于懂了汉诺塔，真是太$TM$简单了~
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由此可得递推公式：
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$$\large f[i] =f[i−1] ∗2+1$$
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[刘老师视频讲座](https://www.bilibili.com/video/BV1Rq4y1v7sC)
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#### 3. 运用
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如何运用到$c++$里面呢？
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```cpp {.line-numbers}
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int f[10010];
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int n;
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scanf ("%d", &n);
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for (int i = 1 ; i <= n ; i++)
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f[i] = f[i-1] * 2 + 1;
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printf ("%d\n", f[n]);
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```
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这就是主程序。
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#### 4. 思考
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如何输出中间的步骤？可在评论区回答 `or` 讨论！
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### 二.汉若双塔
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#### 1. 认识
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我们来看看汗若双塔长什么样子吧：
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他就是每种型号多了$1$个盘子。移动起来的步骤也就乘$2$了。
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#### 2. 实现
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```cpp {.line-numbers}
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int f[10010];
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int n;
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scanf ("%d", &n);
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for (int i = 1 ; i <= n ; i++)
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f[i] = f[i-1] * 2 + 1;
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printf ("%d\n", f[n] * 2);
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```
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### 三.加强
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当数据 $n<=200$ 时，我们应该怎么做？ $answer$ ：当然是 **无符号** $long$ $long$ 啦。
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$NO！NO！NO！$ 我们应该开高精度，这才是正解！！！
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康康 $code$ ：
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 100010;
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// 加法高精度
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int a[N], b[N];
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int al, bl;
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void add(int a[], int &al, int b[], int &bl) {
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int t = 0;
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al = max(al, bl);
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for (int i = 1; i <= al; i++) {
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t += a[i] + b[i];
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a[i] = t % 10;
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t /= 10;
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}
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if (t) a[++al] = 1;
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}
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int main() {
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int n;
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cin >> n;
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b[++bl] = 1;
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for (int i = 1; i <= n; i++) {
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add(a, al, a, al);
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add(a, al, b, bl);
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}
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// 最后再乘以2
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add(a, al, a, al);
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for (int i = al; i; i--) cout << a[i];
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return 0;
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}
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```
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