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)

n = gets.chomp
k = gets.to_i
digit = n.length
dp = Array.new(digit+1).map{Array.new(2).map{Array.new(k+1,0)}}
dp[0][0][0] = 1
dp[0][1][0] = 0
zeronum = 0
1.upto(digit) do |i|
  (k+1).times do |j|
    if j == 0
      dp[i][0][j] = n[i-1] == "0" ? dp[i-1][0][j] : 0
      dp[i][1][j] = dp[i-1][1][j] + dp[i-1][0][j]
    else
      dp[i][0][j] = n[i-1] == "0" ? dp[i-1][0][j] : dp[i-1][0][j-1]
      dp[i][1][j] = dp[i-1][1][j-1] * 9 + dp[i-1][1][j] + dp[i-1][0][j-1] * (n[i-1].to_i - 1) + dp[i-1][0][j]
    end
  end
  zeronum += 1 if n[i-1] == "0"
end
p dp[digit][0][k] + dp[digit][1][k] - zeronum