HackerEarth Bricks and Building problem solution

1. Buildings were made of identical bricks and all the bricks were equal-sized.
2. All the buildings and bricks were 2D rectangles of some height and an equal width.
3. A building is formed by putting bricks one on the another. The height of the building is the sum of the heights of bricks. None of the bricks used are ever broken or rotated.

#include<bits/stdc++.h>
#define LL long long int
#define M 1000000007
#define MM 1000000009
#define reset(a) memset(a,0,sizeof(a))
#define rep(i,j,k) for(i=j;i<=k;++i)
#define per(i,j,k) for(i=j;i>=k;--i)
#define print(a,start,end) for(i=start;i<=end;++i) cout<<a[i];
#define endl "n"
#define inf 100000000000000
LL pow(LL a,LL b,LL m){LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}
LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}
LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}
using namespace std;
int a[100011];
int ct[1000001];
void update(int val)
  {
    int lim = sqrt(val);
    for(int i = 1; i <= lim ; i++)
      {
        if(val % i == 0)
          {
            ct[i]++;
            if(val != i*i)
              {
                ct[val/i]++;
              }
          }
      }
  }
int main()
  {
    ios_base::sync_with_stdio(0);
    int n;
    cin >> n;
    for(int i = 1; i <= n ; i++)
      {
        cin >> a[i];
        update(a[i]);
      }
      
    int q;
    cin >> q;
    for(int i = 1; i <= q; i++)
      {
        int k;
        cin >> k;
        cout << ct[k] << endl;
      }
  }