HDU 1165 Eddy's Research II (formula)

Eddy's Research II

Time Limit: 4000/2000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 3122 accepted submission (s): 1137

Problem descriptionas 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 ).

 

Inputeach 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.

 

Outputfor each value of M, N, print out the value of A (m, n ).

 

Sample input1 32 4

 

Sample output511

 

Authoreddy

 

Recommendjgshining sees 0 <m <= 3 for this question. Therefore, we can discuss the push formula based on different situations and try several groups of data to obtain the push formula.

 1 #include<cstdio> 2 #include<cstring> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 __int64 A(int m,int n) 7 { 8     if(n==0)   return A(m-1,1); 9     else if(m==0)  return n+1;10     else if(m==1)  return n+2;11     else if(m==2)   return 2*n+3;12     else if(m==3)   return 2*A(3,n-1)+3;13 }14 int main()15 {16     int m,n;17     while(scanf("%d %d",&m,&n)!=EOF)18     {19         printf("%I64d\n",A(m,n));20     }21     return 0;22 }

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I have not introduced the formula for M = 3, but later I told him that the formula was more optimized than the final game gj.

 1 #include<cstdio> 2 #include<cstring> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 __int64 A(int m,int n) 7 { 8      if(m==0)  return n+1; 9      if(m==1)  return n+2;10      if(m==2)  return 2*n+3;11      if(m==3)  return (1<<(n+3))-3;12 }13 int main()14 {15     int m,n;16     while(scanf("%d %d",&m,&n)!=EOF)17     {18         printf("%I64d\n",A(m,n));19     }20     return 0;21 }

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