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 = list(map(int, list(input())))
k = int(input())

digits = len(n)

dp_le = [[0 for j in range(k+1)] for i in range(digits)]
dp_lt = [[0 for j in range(k+1)] for i in range(digits)]

dp_le[0][0] = 0
dp_le[0][1] = 1
dp_lt[0][0] = 1
dp_lt[0][1] = n[0]-1

for i in range(1, digits):
    for j in range(k+1):
        if n[i] == 0:
            dp_le[i][j] = dp_le[i-1][j]
            dp_lt[i][j] = dp_lt[i-1][j]
            if j > 0:
                dp_lt[i][j] += dp_lt[i-1][j-1]*9
        else:
            dp_lt[i][j] = dp_lt[i-1][j] + dp_le[i-1][j]
            if j > 0:
                dp_le[i][j] += dp_le[i-1][j-1]
                dp_lt[i][j] += dp_lt[i-1][j-1]*9 + dp_le[i-1][j-1] * (n[i] - 1)

print(dp_le[digits-1][k] + dp_lt[digits-1][k])