title: Count Sum Of Digits In Numbers From 1 To N
signature: function countSumOfDigitsInNumbersFrom1ToN($n) {
source_py: def count_sum_of_digits_in_numbers_from_1_to_n(n):
      if (n < 10):
          return ((n * (n + 1)) / 2)
      d = int(math.log10(n))
      a = ([0] * (d + 1))
      a[0] = 0
      a[1] = 45
      for i in range(2, (d + 1)):
          a[i] = ((a[(i - 1)] * 10) + (45 * int(math.ceil(math.pow(10, (i - 1))))))
      p = int(math.ceil(math.pow(10, d)))
      msd = (n // p)
      return int(((((msd * a[d]) + (((msd * (msd - 1)) // 2) * p)) + (msd * (1 + (n % p)))) + count_sum_of_digits_in_numbers_from_1_to_n((n % p))))
source_cpp: int count_sum_of_digits_in_numbers_from_1_to_n(int n) {
    if ( n < 10 ) return n * ( n + 1 ) / 2;
    int d = log10 ( n );
    int * a = new int [ d + 1 ];
    a [ 0 ] = 0, a [ 1 ] = 45;
    for ( int i = 2;
    i <= d;
    i ++ ) a [ i ] = a [ i - 1 ] * 10 + 45 * ceil ( pow ( 10, i - 1 ) );
    int p = ceil ( pow ( 10, d ) );
    int msd = n / p;
    return msd * a [ d ] + ( msd * ( msd - 1 ) / 2 ) * p + msd * ( 1 + n % p ) + count_sum_of_digits_in_numbers_from_1_to_n ( n % p );
  }