P1044 stack, P1044

P1044 stack, P1044
Background

Stack is a classic data structure in computers. Simply put, stack is a linear table that restricts the insertion and deletion at one end.

Stack has two most important operations: pop (an element pops up from the top of the stack) and push (putting an element into the stack ).

The importance of stacks is self-evident. Any course on data structures will introduce stacks. While reviewing the basic concepts of the stack, Ning thought of a question not mentioned in the book, and he could not give the answer himself, so he needed your help.

Description

Ning considers the problem that the depth of stack A is greater than n in an operand sequence, from 1, 2, to n (1 to 3 in the figure below.

You can perform either of the following operations,

1. Move a number from the first end of the operand sequence to the first end of the stack (corresponding to the push operation of the data structure stack)

2. Move a number from the beginning of the stack to the end of the output sequence (corresponding to the pop operation of the data structure stack)

Using these two operations, a series of output sequences can be obtained from an operand sequence, as shown in the process of generating sequence 2 3 1 for 1 2 3.

(The original status is shown in)

Your program will calculate and output the given n from the operand sequence 1, 2 ,..., N the total number of output sequences that may be obtained after the operation.

Input/Output Format Input Format:

The input file contains only one integer n (1 ≤ n ≤ 18)

Output Format:

The output file has only one row, that is, the total number of possible output sequences.

Input and Output sample Input example #1:

3

Output sample #1:

5

This is a bare catlan number.
However, you can also use dp to do this,
Use dp [I] [j] to indicate the number of I solutions in the stack and j outside the stack.
Transfer equation:
Dp [j] [I] = max (dp [j] [I], dp [J-1] [I] + dp [j + 1] [I-1])

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 int read(int & n) 7 { 8     char p='+';int x=0; 9     while(p<'0'||p>'9')10         p=getchar();11     while(p>='0'&&p<='9')12     x=x*10+p-48,p=getchar();13     n=x;14 }15 int dp[20][20];16 int main()17 {18     int ans=0;19     int n;read(n);20     for(int i=0;i<=n;i++)21     {22         for(int j=0;j<=n;j++)23         {24             if(i==0)25                 dp[j][0]=1;26             else if(j==0)27                 dp[0][i]=dp[1][i-1];28             else29                 dp[j][i]=max(dp[j][i],dp[j-1][i]+dp[j+1][i-1]);30         }31     }32     cout<<dp[0][n];33     return 0;34 }