hdu 1165 Eddy's research II

最新推荐文章于 2019-08-02 18:01:31 发布

risker_2013

最新推荐文章于 2019-08-02 18:01:31 发布

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分类专栏： HDU 数学文章标签： ACM hdu 递推

本文链接：https://blog.youkuaiyun.com/xzf19930910/article/details/10752157

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本文介绍了一种经典的递归函数——Ackermann函数，并通过分析其特性，提出了针对特定参数范围的有效计算方法。通过对递归过程的观察，文章推导出了简化公式，实现了快速计算。

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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): 2576 Accepted Submission(s): 943

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:

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

Sample Output

大致题意：给你m,n，求出A（m，n）

思路：本来是直接递归，果断TLE，后来优化也不行，无奈看了题解，才明白需要推出递归公式。

A(1,n)=A(0,A(1,n-1))=A(1,n-1)+1 那么可以发现，这个其实是一个等差数列

而且很明显有 A(1,0)=A(0,1)=2,A(1,1)=A(0,A(1,0))=A(1,0)+1=3

那么就可以推出 A(1,n)=n+2

同理可以推出 A(2,n)=2*n+3

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

#include<iostream>
#include <cstdio>
#include<cstring>
#include<algorithm>
#include <cmath>
using namespace std;

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

   }

int main()
{
 int m,n;
 while(~scanf("%d%d",&m,&n))
 {
     cout<<A(m,n)<<endl;
 }
    return 0;
}