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 Python (3.4.3) (judge verdict: Accepted)

N = input()
K = int(input())

digit = len(N)
dp = [[[0]*(K+1) for _ in range(2)] for _ in range(digit)]

di = int(N[0])
dp[0][0][0] = 0
dp[0][0][1] = 1
dp[0][1][0] = 1
dp[0][1][1] = di-1

for i in range(1, digit):
    dp[i][1][0] = 1
    for k in range(1, K+1):
        di = int(N[i])
        if di == 0:
            dp[i][0][k] = dp[i-1][0][k]
            dp[i][1][k] = dp[i-1][1][k] + dp[i-1][1][k-1]*9
        else:
            dp[i][0][k] = dp[i - 1][0][k - 1]
            dp[i][1][k] = dp[i-1][0][k-1]*(di-1)+dp[i-1][0][k]\
                      +dp[i-1][1][k]+dp[i-1][1][k-1]*9


print(dp[-1][0][K]+dp[-1][1][K])