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: Accepted)

n = gets.to_s.chomp.chars
k = gets.to_s.to_i

sz = n.size
dpj = Array.new(sz+1){ [0] * (k+1) }
dpl = Array.new(sz+1){ [0] * (k+1) }
dpj[0][0] = 1

n.each_with_index do |c, i|
  
  (k+1).times do |j|
    dpj[i+1][j] += dpj[i][j] if c == '0'
    dpl[i+1][j] += dpj[i][j] if c >  '0'
    dpl[i+1][j] += dpl[i][j]
  end
  
  k.times do |j|
    dpj[i+1][j+1] += dpj[i][j] if c >  '0'
    dpl[i+1][j+1] += dpj[i][j] * [0, c.to_i-1].max
    dpl[i+1][j+1] += dpl[i][j] * 9
  end
end  

puts dpj[sz][k] + dpl[sz][k]