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)

from math import factorial as fac

def comb(n,r):
    if ( r<0 or r>n ):
        return 0
    return fac(n)/(fac(r)*fac(n-r))


n=int(input())
k=int(input())

nn=str(n)
keta=len(nn)

ans=0

if keta<k:
    print(0)
    exit()

# zero=0

# for i in range(keta):
#     if int(nn[i])==0:zero+=1
#     ans+=comb(keta-i-1,k-i+zero)*9**(k-i)
#     if int(nn[i])-1>=0:
#         ans+=(int(nn[i])-1)*comb(keta-i-1,k-i-1+zero)*9**(k-i-1+zero)
#
# if keta-zero==k:ans+=1

def search(n,k):
    if k<=0:return
    global ans
    nn = str(n)
    keta = len(nn)
    ans += comb(keta - 1, k ) * 9 ** k
    if int(nn[0])-1>0:
        ans+=(int(nn[0])-1)*comb(keta-1,k-1)*9**(k-1)
    ne=n-int(nn[0])*10**(keta-1)
    search(ne,k-1)

search(n,k)
print(int(ans)+1)