Hackerrank would not accept my code due to timeout, how to optimize this piece?

Question

The problem statement is:

Watson gives an integer NN to Sherlock and asks him: What is the number of divisors of NN that are divisible by 2?.

Input Format

First line contains TT, the number of testcases. This is followed by TT lines each containing an integer NN.

Output Format

For each testcase, print the required answer in one line.

Constraints

1≤T≤1001≤T≤100 
1≤N≤109

Sample Input :

2\n
9\n
8

Sample Output:

0\n
3

I have typed '\n' which means the next integer will be in a new line. Hacker Rank would not accept my code due to timeout. Please help me optimise this C code.

My C code is :

int main() {

      int length,number,i,count =0;
      scanf("%d",&length);

      while(length--){
           scanf("%d",&number);
           for(i=2;i<=number/2;i++){
                 if(number % i == 0 && i % 2 == 0){
                   count = count + 1;
                  }
           }
           if(number % 2 == 0){
             count = count + 1;
            }
           printf("%d\n",count);
           count =0;
        }
    return 0;
}

Answer 1

Below is a algo to solve this problem:

If the number is even:
Run the loop from i=2 till square root of the number ie: sqrt(number) and increase the count in two cases below if i divides the number(number%i==0) :

• if i is even(i%2==0)

• for q where q=number/i if i not equal to q(i!=q) and q is even(q%2==0) because if i divides the number it means q will also divide(number/q=i) it.

At last increase the count by 1 as number also divides itself.

If the number is odd:
count will be 0 as no i(even) would divide an odd number.

int main() 
{
  int length,number,i,count;
  scanf("%d",&length);

  while(length--)
  {
      scanf("%d",&number);
      count =0;i=2;
      if(number%2==0)
      {
          while(i*i<=number)
          {
              if(number%i==0)
              {
                  if(i%2==0)
                      count++;
                  int q=number / i;
                  if (i !=q && q%2==0)
                      count++;
              }
              i++;
          }
          count++;
      }
      printf("%d\n",count);
  }
  return 0;
}