HDU：1165  Eddy's research II_杭电0j1165-优快云博客

本文链接：https://blog.youkuaiyun.com/runningtortoises/article/details/8261011

本文介绍了一种高效计算Ackermann函数的方法，针对不同参数m的值给出了具体的计算公式，并通过预计算的方式存储中间结果来避免重复计算，实现了快速求解。

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Eddy's research II

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1446 Accepted Submission(s): 514

Problem Description

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:
HDU：1165 <wbr> <wbr>Eddy's <wbr>research <wbr>II

HDU：1165 <wbr> <wbr>Eddy's <wbr>research <wbr>II

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).

Input

Each 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.

Output

For each value of m,n, print out the value of A(m,n).

Sample Input

1 3 2
4

Sample Output

5 11

Author

eddy

Recommend

刚开始用递归，这是最容易想到的方法，但是严重超时，后来想到用数组存下数据然后递归，但是发现还是不行，最后还是在网上搜到的方法，写的实在太好了！

m=0时 A(m,n)=n+1;

m=1时 A(m,n)=A(0,A(1,n-1))=A(1,n-1)+1=A(1,n-2)+1+1=……=n+2;

m=2时 A(m,n)=n*2+3

m=3时 A(m,n)=A(2,A(m,n-1))=A(m,n-1)*2+3

其实当m=3的时候还可以递推，

#include<stdio.h>
int a[4][1000010]={0};
int main()
{
int m,n;
for(n=0;n<=1000000;n++)
  a[0][n]=n+1;
for(n=0;n<=1000000;n++)
  a[1][n]=n+2;
for(n=0;n<=1000000;n++)
  a[2][n]=2*n+3;
a[3][0]=5;
for(n=1;n<=25;n++)
  a[3][n]=2*a[3][n-1]+3;
while (scanf("%d %d",&m,&n)!=EOF)
       printf ("%d\n",a[m][n]);
return 0;
}