hdu Eddy's research II(递推,找规律,打表)
2017-01-12 19:09
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Eddy's research II
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
5
11
递推找规律代码:
打表代码:
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
5
11
递推找规律代码:
#include<stdio.h> int dp[4][1000010]; int dfs(int m,int n) { if(dp[m] ) return dp[m] ; if(m==0) return n+1; if(n==0&&m>0) return dfs(m-1,1); if(n>0&&m>0) return dfs(m-1,dfs(m,n-1)); } int main() { int m,n; while(~scanf("%d%d",&m,&n)) { int ans=dfs(m,n); printf("%d\n",ans); } return 0; }
打表代码:
#include<stdio.h> #define N 1000010 int dp[4] ; void A() { int i; for(i=1; i<N; i++) dp[1][i]=dp[1][i-1]+1; for(i=1; i<N; i++) dp[2][i]=dp[2][i-1]+2; for(i=1; i<=24; i++) dp[3][i]=dp[3][i-1]+(1<<(i+2)); } int main() { dp[1][0]=2; dp[2][0]=3; dp[3][0]=5; A(); int m,n; while(~scanf("%d%d",&m,&n)) { printf("%d\n",dp[m] ); } return 0; }
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