<h1 id="heading">题目</h1>
<blockquote>
<h2 id="description">Description</h2>
<blockquote>
As is known, Ackermann function plays an important role in the sphere of theoretical computer science. However, in the other hand, the dramatic fast increasing pace of the function caused the value of Ackermann function hard to calcuate.
Ackermann function can be defined recursively as follows: 
<div class='image'><a href='http://7xjob4.com1.z0.glb.clouddn.com/9c2f962ffc49f0b7257be9cd13bc6b79' target='_blank' rel='noopener noreferrer'><img src="http://7xjob4.com1.z0.glb.clouddn.com/9c2f962ffc49f0b7257be9cd13bc6b79" /></a></div>
Now Eddy Gives you two numbers: m and n, your task is to compute the value of A(m,n) .This is so easy problem,If you slove this problem,you will receive a prize(Eddy will invite you to hdu restaurant to have supper).
 
</blockquote>
<h2 id="input">Input</h2>
<blockquote>
Each line of the input will have two integers, namely m, n, where 0 &lt; m &lt; =3. 
Note that when m&lt;3, n can be any integer less than 1000000, while m=3, the value of n is restricted within 24. 
Input is terminated by end of file.
</blockquote>
<h2 id="output">Output</h2>
<blockquote>
For each value of m,n, print out the value of A(m,n).
</blockquote>
<h2 id="sample-input">Sample Input</h2>
<blockquote>
1 3 
2 4
</blockquote>
<h2 id="sample-output">Sample Output</h2>
<blockquote>
5 
11
</blockquote>
</blockquote>
<h1 id="heading-1">题解</h1>
看题目应该是 记忆化搜索 
但是直接搜会导致栈溢出 
找下规律可以发现 <code>m</code> 为 <code>0</code> 、 <code>1</code> 、 <code>2</code> 时是 等差数列 
因此可以在 <code>O(1)</code> 的时间内得到答案
而对于 <code>m==3</code> 如果直接递归展开仍然会栈溢出 
但是可以手动将它展开一步,这样最多也只会递归 <code>24</code> 层
<h1 id="heading-2">代码</h1>
<pre style="background-color:#fff">/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/

#include &lt;iostream&gt;
#include &lt;cstring&gt;
#include &lt;cstdio&gt;
#include &lt;cmath&gt;
#include&lt;algorithm&gt;
using namespace std;

const int maxn = 5000001;
const int maxm = 4;

int dp(int m,int n) {
 if(m == 0)
 return n + 1;
 if(m == 1)
 return n + 2;
 if(m == 2)
 return 3 + 2 * n;
 if(n == 0)
 return dp(m - 1,1);
 return 2 * dp(3,n - 1) + 3;
}

int main() {
 int m,n;
 while(cin &gt;&gt; m &gt;&gt; n) {
 cout &lt;&lt; dp(m,n) &lt;&lt; endl;
 }
 return 0;
}
</pre>

HDU 1165.Eddy's research II