HDU - 1165 - Eddy's research II

最新推荐文章于 2025-09-06 09:33:29 发布

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最新推荐文章于 2025-09-06 09:33:29 发布

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文章标签： java

原文链接：http://www.cnblogs.com/sineatos/p/3572584.html

本文介绍了一个基于Ackermann函数的计算问题，通过分析不同输入值下的递归特性，找到了高效的计算方法，并给出了完整的C++实现代码。

先上题目：

Eddy's research II

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2788 Accepted Submission(s): 1019

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

1 3

2 4

Sample Output

　　题意：根据给定的递推式求A(m,n)，看起来没什么难度，但是一开始用打表的方法一直WA，期间也发现了每一行(每一个m为一行)有规律，比较容易看出来的是第零行和第一行，然后第二行的规律是n*2+3，第三行需要是a[i]=a[i-1]*2+3;

上代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #define MAX 1000002
 4 #define LL long long
 5 using namespace std;
 6 
 7 LL a[MAX];
 8 
 9 void deal(){
10     int n;
11     memset(a,0,sizeof(a));
12     n=MAX-2;
13     a[0]=5;
14     for(int i=1;i<=n;i++){
15         a[i]=2*a[i-1]+3;
16     }
17 
18 }
19 
20 int main()
21 {
22     int n,m;
23     //freopen("data.txt","r",stdin);
24     deal();
25     while(scanf("%d %d",&m,&n)!=EOF){
26         if(m==0) printf("%d\n",n+1);
27         else if(m==1) printf("%d\n",n+2);
28         else if(m==2) printf("%d\n",n*2+3);
29         else printf("%I64d\n",a[n]);
30     }
31     return 0;
32 }