hdu 1165 Eddy's research II(递推+击表)

hdu 1165 Eddy's research II(递推+打表)

Eddy's research II

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3476    Accepted Submission(s): 1285


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 Eddy's research II(递推+击表)

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


写个程序找出规律就OK!

附个代码:

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
using namespace std;
#define N 1000005
#define ll __int64
const int inf=0x7fffffff;
int f[4][N];       //A(m,n)
/*void inti()
{
    int i;
    memset(f,-1,sizeof(f));
    for(i=0;i<N;i++)
        f[0][i]=i+1;

}
int solve(int m,int n)
{
    if(f[m][n]!=-1)
        return f[m][n];
    if(m==0)
        return n+1;
    if(n==0)
        return f[m][n]=solve(m-1,1);
    return f[m][n]=solve(m-1,solve(m,n-1));
}
int main()
{
    inti();
    int m,n;
    while(scanf("%d%d",&m,&n)!=-1)
    {
        printf("%d\n",solve(m,n));
    }
    return 0;
}*/
int fun(int i)
{
    if(i==0)
        return 5;
    if(i==1)
        return 13;
    return 15;
}
void inti()
{
    int i,t;
    for(i=0;i<N;i++)
        f[0][i]=i+1;
    for(i=0;i<N;i++)
        f[1][i]=i+2;
    for(i=0;i<N;i++)
        f[2][i]=i*2+3;
    f[3][0]=5;
    for(i=1,t=8;i<30;i++)
    {
        f[3][i]=f[3][i-1]+t;
        t*=2;
    }
}
int main()
{
    int m,n;
    inti();
    while(~scanf("%d%d",&m,&n))
        printf("%d\n",f[m][n]);

    return 0;
}