题目链接：

https://www.shuzhiduo.com/A/nAJvRgVxJr/acm.hdu.edu.cn/showproblem.php?pid=1165

Eddy’s research II

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5656    Accepted Submission(s): 2045

Problem DescriptionAs 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:

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).

 InputEach line of the input will have two integers, namely m, n, where 0 < m < =3.
Note that when m<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.  OutputFor each value of m,n, print out the value of A(m,n). Sample Input1 32 4 Sample Output511 Authoreddy 分析：开始是想用备忘录加递归来做的，定义一个f[][]的二维数组，初始化全部为-1，每次计算之前看看f[m][n]的值，如果不是-1，则表示A(m,n)还没有计算过，递归计算，然后将得到的值存起来（备忘录法），如果值不等于-1，说明A(m,n)计算过了，直接拿来用即可，这样可以解决大量重复的子问题但是，这样做会超时。。。。。。所以第二种做法，公式推导因为m只有4个值，那么我们想想是不是对每个不同的m，都有一个公式与之对应呢？开始推导当m=0的时候， 代入题目给的递归公式可知A(0,n)=n+1当m=1的时候， A(1,n)=A(0,A(1,n-1)) =A(1,n-1)+1………(n个1累加)直到n=1的时候，A(0,1)=2,所以A(1,n)=n+2当m=2的时候， A(2,n)=A(1,A(2,n-1))=A(2,n-1)+2………(n个2的累加)直到n=1的时候，A(1,1)=3,所以A(2,n)=2*n+3当m=3的时候， A(3,n)=A(2,A(3,n-1))=2*A(3,n-1)+3……..(2的n次方个5相乘，2的n次方减一个3相乘，二者相加)直到n=1的时候，A(2,1)=5,所以A(3,n)=2的n+3次方-3 我觉得我可能没有说清楚，但是我自己懂了，我真的很努力去说清楚了，唉，自己表达能力有限，大家自己推导一下肯定是可以推导出来的（前前后后推导了一个多小时）代码如下：

#include<bits/stdc++.h>
using namespace std;
int f(int n)
{
    int s=1;
    for(int i=1;i<=n;i++)
    {
        s=s*2;
    }
    return s;
}
int solved(int m,int n)
{
    if(m==0)
    {
        return n+1;
    }else if(m==1)
    {
        return n+2;
    }else if(m==2)
    {
        return 2*n+3;
    }else if(m==3)
    {
        return f(n+3)-3;
    }
}
int main()
{
    int n,m;
    while(~scanf("%d %d",&m,&n))
    {
        printf("%d\n",solved(m,n));
    }
    return 0;
}