Take an integer `n (n >= 0)` and a digit `d (0 <= d <= 9)` as an integer. Square all numbers `k (0 <= k <= n)` between 0 and n. Count the numbers of 
digits `d` used in the writing of all the `k**2`. Call `nb_dig` (or nbDig or ...) the function taking `n` and `d` as parameters and returning this count.
#Examples:
```
n = 10, d = 1, the k*k are 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
We are using the digit 1 in 1, 16, 81, 100. The total count is then 4.
nb_dig(25, 1):
the numbers of interest are
1, 4, 9, 10, 11, 12, 13, 14, 19, 21 which squared are 1, 16, 81, 100, 121, 144, 169, 196, 361, 441
so there are 11 digits `1` for the squares of numbers between 0 and 25.
```
Note that `121` has twice the digit `1`.
def nb_dig(n, d):
  s = str(d)
  return sum(sum(c == s for c in str(k**2)) for k in range(n+1))
def nb_dig(n, d):
  s = str(d)
  return sum(sum(c == s for c in str(k**2)) for k in range(n+1))