C language-mathematics-statistical problem (Hdu 2563)

C language-mathematics-statistical problem (Hdu 2563)
Problem Description in an infinitely large two-dimensional plane, we make the following assumptions:
1. Only one cell can be moved at a time;
2. You cannot go backward (If your destination is "up", you can go left, right, or up, but not down );
3. The passing grid collapsed immediately and could not go for the second time;

Calculate the number of different solutions in n steps (two steps are considered different if one step is different ).

Input first gives a positive integer C, indicating that there are group C Test Data
In row C, each row contains an integer n (n <= 20), which indicates that n steps are required.

For the Output, Please program the total number of different solutions that take n steps;
The output of each group occupies one row.

Sample Input

212

Here we always define the forward direction.
We use A [I]Indicates the type of the forward step. We use B [I]Indicates the type of step I to the left (right ). A [1] = 3, B [1] = 2;
Then the idea of sub-Governance: Moving Forward I step (I> 1) is equivalent to moving forward the type of the I-1 step plus to the left, to the right to take the type of the I-1 step. The mathematical expression is: A [I] = a [I-1] + 2 * B [I-1];The type of step I (I> 1) to the left (to the right) is equivalent to the type of the forward I-1 step plus the type of the right (to the left) to the I-1 step: B [I] = a [I-1] + B [I-1];
Then, it is obtained by recursion.

The Code is as follows:

#include 
 
  int main(){    int a[21]={0,3},b[21]={0,2},i;    for(i=2;i<21;i++)    {        b[i]=a[i-1]+b[i-1];        a[i]=a[i-1]+2*b[i-1];    }    int m,N;    scanf("%d",&N);    while(N--)    {        scanf("%d",&m);        printf("%d\n",a[m]);    }    return 0;}