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: Wrong Answer)

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

from math import factorial

str_n = str(N)
len_n = len(str(str_n))

def conb(n, r):
    return factorial(n)//factorial(r)//factorial(n-r)
    
def get_num(top, k_val, depth):
    if depth>len_n:
        return 0
    elif k_val>(len_n - depth + 1):
        return 0
    elif k_val==0:
        return 1
    elif top:
        if int(str_n[depth - 1])==0:
            return get_num(top, k_val, depth + 1)
        else:
            b1 = get_num(top, k_val - 1, depth + 1)
            b2 = (int(str_n[depth-1])-1) * get_num(False, k_val-1, depth+1)
            b3 = get_num(False, k_val, depth + 1)
            return b1+b2+b3
    else:
        return 9**(k_val) * conb(len_n - depth + 1, k_val)
    
    
a1 = get_num(True, K-1, 2)
a2 = (int(str_n[0])-1)*get_num(False, K-1, 2)
a3 = get_num(False, K, 2)

#print(a1, a2, a3)
print(a1+a2+a3)