Cyclic shifts

Problem:-

You are given a number $N$ represented as a binary representation of $X=16$ bits. You are also given a number $m$ and a character $c(LorR)$.
Determine a number $M$ that is generated after cyclically shifting the binary representation of $N$ by $m$ positions either left if $c=L$ or right is $c=R$.
Input format 

1. The first line contains an integer $T$ representing the number of queries.
2. The next $T$ lines contain $Nmc$ as mentioned in the problem statement.

Output format
Print $T$ integers in a separate line representing the answer to each query.
Constraints
$$\begin{gathered} 1\le T\le 1e4 \\ 1\le N\le 65535 \\ 1\le m\le 15 \end{gathered}$$
 

SAMPLE INPUT

 

2
7881 5 L
7881 3 R

SAMPLE OUTPUT

 

55587
9177

Explanation

For first case :  N in binary is 0001 1110 1100 1001 and shifting it left by 5 position, it becomes 1101 1001 0010 0011 which in decimal system is 55587

For second case : N in binary is 0001 1110 1100 1001 and shifted 3 position to right it becomes 0010 0011 1101 1001 which in decimal system is 9177

Time Limit:1.0 sec(s) for each input file.

Memory Limit:256 MB

Source Limit:1024 KB

solution:-

#include<stdio.h>

#include<math.h>

void main()

{

     int t,i,m,x,a[16],b[16],j=0,d[16],p;

     char c,c1;

     long int n,s=0;

     scanf("%d",&t);

     for(i=1;i<=t;i++)

     {

         for(j=0;j<16;j++)

          a[j]=0;

         scanf("%ld%d%c%c",&n,&m,&c,&c1);

         j=0;

         while(n!=0)

         {

            x=n%2;

            n=n/2;

            a[j]=x;

            j++;

         }

         for(j=0;j<16;j++)

         b[j]=a[15-j];

         if(c1=='R')

         {

            for(j=15;j>=0;j--)

            {

             if((j-m)>=0)

             d[j]=b[(j-m)];

             else

            {

                 d[j]=b[(j-m)+16];

            }

            }

         }   

         if(c1=='L')

         {

         for(j=0;j<16;j++)

         d[j]=b[(j+m)%16];

         }

         p=15;

         for(j=0;j<16;j++)

         {

            s=s+d[j]*pow(2,p);

            p--;

         }

         printf("%ld\n",s);

         s=0;

     }

}