Problem:
Score : 500 points

Problem Statement

Find the number of integers between 1 and N (inclusive) that contains exactly K non-zero digits when written in base ten.

Constraints

1 \leq N < 10^{100}

1 \leq K \leq 3

Input

Input is given from Standard Input in the following format:

N
K

Output

Print the count.

Sample Input 1

100
1

Sample Output 1

19

The following 19 integers satisfy the condition:

1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100

Sample Input 2

25
2

Sample Output 2

14

The following 14 integers satisfy the condition:

11,12,13,14,15,16,17,18,19,21,22,23,24,25

Sample Input 3

314159
2

Sample Output 3

937

Sample Input 4

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
3

Sample Output 4

117879300

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

def factorial(x)
  f = 1
  for i in 1..x do
    f *= i
  end
  return f
end

def combi(p, q)
  return factorial(p) / factorial(q) / factorial(p-q)
end

ni = gets.chomp.reverse.split("").map(&:to_i)
m = ni.length
k = gets.to_i
count = 0
case k
when 1
  if m >= 2
    count += 9 * combi(m-1, 1)
  end
  if m >= 1
    count += 9 * (m-1) + ni[m-1]
  end
when 2
  if m >= 3
    count += 81 * combi(m-1, 2)
  end
  if m >= 2
    count += 9 * (m-1) * (ni[m-1] - 1)
    count += 9 * (m-2) + ni[m-2]
  end
when 3
  if m >= 4
    count += 729 * combi(m-1, 3)
    count += 81 * combi(m-2, 2)
  end
  if m >= 3
    count += 81 * combi(m-1, 2) * (ni[m-1] - 1)
    count += 9 * (m-2) * (ni[m-2] - 1)
    count += 9 * (m-3) + ni[m-3]
  end
end
puts count