2049: Ackermann's Function

最新推荐文章于 2021-06-18 09:32:50 发布

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最新推荐文章于 2021-06-18 09:32:50 发布

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分类专栏： joj acm_数学问题文章标签： function primitive input each output pair

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本文介绍了一个使用循环而非递归的方法来实现Ackermann函数的计算。该函数的特点是非原始递归，增长速度非常快。通过预计算的方式填充二维数组，使得在输入限定范围内能够快速输出Ackermann函数的结果。

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2049: Ackermann's Function

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	928	273	Standard

In this easy problem, you are to write a program to calculate the Ackermann's Function A:

A(0, n) = n+1
A(m, 0) = A(m-1, 1)
A(m, n) = A(m-1, A(m, n-1))

Note that the Ackermann's Function is recursive but not primitive recursive which means that it grows very fast.

Input

The input contains several lines, each of which consists of two integers m and n, 0<=m, n<=4. The last line consists of two zeros, which you should not proceed.

Output

For each pair of m and n, print the value of A(m, n) in a single line.

Sample Input

2 3
3 3
0 0

Sample Output

9
61

This problem is used for contest: 10

#include<iostream>
#define N 100000
using namespace std;
int a[5][N]={0};
int main()
{

int m,n;
for(int i=0;i<=4;i++)
   for(int j=0;j<=N;j++)
     {
   if(i==0) a[0][j]=j+1;
   else
   if(j==0) a[i][0]=a[i-1][1];
   else
   a[i][j]=a[i-1][a[i][j-1]];
}
     while(scanf("%d%d",&m,&n),m||n)
{
   cout<<a[m][n]<<endl;
}
return 0;
}

/*A(0, n) = n+1
A(m, 0) = A(m-1, 1)
A(m, n) = A(m-1, A(m, n-1))*/

这道题用递归是写不了的只能用循环了